Trochim, W. M. K. (2006). Internal validity.
http://www.socialresearchmethods.net/kb/intval.php
Please follow link:^^^^^
Social Work Research: Chi Square
Molly, an administrator with a regional organization that advocates for alternatives to long-term prison sentences for nonviolent offenders, asked a team of researchers to conduct an outcome evaluation of a new vocational rehabilitation program for recently paroled prison inmates. The primary goal of the program is to promote full-time employment among its participants.
To evaluate the program, the evaluators decided to use a quasi-experimental research design. The program enrolled 30 individuals to participate in the new program. Additionally, there was a waiting list of 30 other participants who planned to enroll after the first group completed the program. After the first group of 30 participants completed the vocational program (the “intervention” group), the researchers compared those participants’ levels of employment with the 30 on the waiting list (the “comparison” group).
In order to collect data on employment levels, the probation officers for each of the 60 people in the sample (those in both the intervention and comparison groups) completed a short survey on the status of each client in the sample. The survey contained demographic questions that included an item that inquired about the employment level of the client. This was measured through variables identified as none, part-time, or full-time. A hard copy of the survey was mailed to each probation officer and a stamped, self-addressed envelope was provided for return of the survey to the researchers.
After the surveys were returned, the researchers entered the data into an SPSS program for statistical analysis. Because both the independent variable (participation in the vocational rehabilitation program) and dependent variable (employment outcome) used nominal/categorical measurement, the bivariate statistic selected to compare the outcome of the two groups was the Pearson chi-square.
After all of the information was entered into the SPSS program, the following output charts were generated:
TABLE 1. CASE PROCESSING SUMMARY
Cases
Valid
Missing
Total
N
Percent
N
Percent
N
Percent
Program
Participation
*Employment
59
98.3%
1
1.7%
60
100.0%
TABLE 2. PROGRAM PARTICIPATION *EMPLOYMENT CROSS TABULATION
Employment
Total
None
Part-Time
Full-Time
Program
Participation
Intervention
Group
Count % within Program Participation
5
16.7%
7
23.3%
18
60.0%
30
100.0%
Comparison
Group
Count % within Program Participation
16
55.2%
7
24.1%
6
20.7%
29
100.0%
Total
Count % within Program Participation
21
35.6%
14
23.7%
24
40.7%
59
100.0%
TABLE 3. CHI-SQUARE TESTS
Value
df
Asymp. Sig. (2-sided)
Pearson Chi-Square
11.748a
2
.003
Likelihood Ratio
12.321
2
.002
Linear-by-Linear Association
11.548
1
.001
N of Valid Cases
59
a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 6.88.
The first table, titled Case ...
Calculating Analysis of Variance (ANOVA) and Post Hoc Analyses Follo.docxaman341480
Calculating Analysis of Variance (ANOVA) and Post Hoc Analyses Following ANOVA
Analysis of variance (ANOVA)
is a statistical procedure that compares data between two or more groups or conditions to investigate the presence of differences between those groups on some continuous dependent variable (see
Exercise 18
). In this exercise, we will focus on the
one-way ANOVA
, which involves testing one independent variable and one dependent variable (as opposed to other types of ANOVAs, such as factorial ANOVAs that incorporate multiple independent variables).
Why ANOVA and not a
t
-test? Remember that a
t
-test is formulated to compare two sets of data or two groups at one time (see
Exercise 23
for guidance on selecting appropriate statistics). Thus, data generated from a clinical trial that involves four experimental groups, Treatment 1, Treatment 2, Treatments 1 and 2 combined, and a Control, would require 6
t
-tests. Consequently, the chance of making a Type I error (alpha error) increases substantially (or is inflated) because so many computations are being performed. Specifically, the chance of making a Type I error is the number of comparisons multiplied by the alpha level. Thus, ANOVA is the recommended statistical technique for examining differences between more than two groups (
Zar, 2010
).
ANOVA is a procedure that culminates in a statistic called the
F
statistic. It is this value that is compared against an
F
distribution (see
Appendix C
) in order to determine whether the groups significantly differ from one another on the dependent variable. The formulas for ANOVA actually compute two estimates of variance: One estimate represents differences between the groups/conditions, and the other estimate represents differences among (within) the data.
Research Designs Appropriate for the One-Way ANOVA
Research designs that may utilize the one-way ANOVA include the randomized experimental, quasi-experimental, and comparative designs (
Gliner, Morgan, & Leech, 2009
). The independent variable (the “grouping” variable for the ANOVA) may be active or attributional. An active independent variable refers to an intervention, treatment, or program. An attributional independent variable refers to a characteristic of the participant, such as gender, diagnosis, or ethnicity. The ANOVA can compare two groups or more. In the case of a two-group design, the researcher can either select an independent samples
t
-test or a one-way ANOVA to answer the research question. The results will always yield the same conclusion, regardless of which test is computed; however, when examining differences between more than two groups, the one-way ANOVA is the preferred statistical test.
Example 1: A researcher conducts a randomized experimental study wherein she randomizes participants to receive a high-dosage weight loss pill, a low-dosage weight loss pill, or a placebo. She assesses the number of pounds lost from baseline to post-treatment
378
for the thre ...
This document provides a 100-question practice exam for the QNT 275 final exam. It covers topics in statistics including hypothesis testing, data types, measurement scales, sampling, descriptive statistics, and inferential statistics. Sample questions are multiple choice and cover topics like hypothesis tests, measurement scales, sampling methods, descriptive vs inferential statistics, and data analysis techniques like ANOVA. The practice exam allows students to test their understanding of key statistical concepts.
For more classes visit
www.snaptutorial.com
1
To make tests of hypotheses about more than two population means, we use the:
t distribution
normal distribution
chi-square distribution
analysis of variance distribution
For more classes visit
www.snaptutorial.com
1
To make tests of hypotheses about more than two population means, we use the:
t distribution
normal distribution
chi-square distribution
analysis of variance distribution
2
You randomly select two households and observe whether or not they own a telephone answering machine. Which of the following is a simple event?
At most one of them owns a telephone answering machine.
1. An independent samples t-test was conducted to determine if there were differences in anxiety scores between male and female participants before a major competition.
2. The results of the t-test showed no significant difference between the mean anxiety scores of males (M=17, SD=4.58) and females (M=18, SD=3.16), t(8)=0.41, p>0.05.
3. Therefore, the null hypothesis that there is no difference between male and female anxiety scores before a major competition was not rejected.
This document discusses statistical models and inferential statistics. It defines statistical modeling as using mathematical tools and statistical conclusions to understand real-life situations. There are three main types of statistical models: parametric models which have known parameters; nonparametric models which have flexible parameters; and semi-parametric models which are a blend of the two. Inferential statistics are used to draw conclusions about populations based on samples, while descriptive statistics describe sample characteristics. Common inferential statistics techniques include hypothesis testing, regression analysis, z-tests, t-tests, f-tests, and confidence intervals.
Running head COURSE PROJECT –PHASE 3 COURSE PROJECT –PHASE 3.docxsusanschei
Running head: COURSE PROJECT –PHASE 3
COURSE PROJECT –PHASE 3
Course Project –Phase 3
Name: Rodney Wheeler
Institution: Rasmussen College
Course: STA3215 Section 01 Inferential Statistics and Analytics
Date: 03/04/17
Course Project –Phase 3
The primary goal of statistics is to conduct a hypothesis. A hypothesis is a prediction about something; hypothesis testing is done to ascertain if a sampled proportion differs from a specified population. For the test to be valid eight steps are conducted to ensure the results are up to par (Lora M. and Richard J. Cook., 2009);
Step One -Identify and come up with a research question, this helps the researcher narrow down to what they want to test.For instance, is the number of patients admitted with infectious disease less than 65 years of age? Such questions are important as they help one in looking for the necessary data and conduct the test efficiently
Step Two-Ascertain that some expectations are met: The method of research used is Simple random sampling, the resultant outcome is only one, and the population is triple the sample size in question
Step Three-State the two types of hypothesis: Identify the null and alternative hypothesis. Null hypothesis shows equality while alternative does not.
Step Four-Determine a definite significant level that is the odds of refuting a null hypothesis through use of alpha
Step Five-Calculate the test statistic, this are constant values that are calculated from the available data when conducting a hypothesis test
Step Six-Change the test statistic into a P value; A p-value is the possibility that a selected sample would differ with the obtained one. It differs depending on the test used and is determined by use of the normal distribution table
Step Seven-Choose between the null and alternative hypothesis, this is where one has to determine whether the stated research question is correct. If the p-value is greater than the standardized value, the null hypothesis should be rejected
Step Eight-Creating a conclusion of your Research Question, determine whether or not the set values are sufficient evidence in confirming your research.
The p-value is the better approach as computation of one value is required to conduct the test, the critical approach is cumbersome as one has to compute the test statistic and also find the key value of the significance level
Question two
1. Ho:p>=65;Ha p<65
2. The test is left tailed since the sample proportion is less than the hypothesized population proportion
3. The test statistics to be used is the t test since the standard deviation is unknown.
4. =-2.79
5. Degree of freedom is 60-1=59as observed from the t table the p- value is 0.05
6. 0.5-0.05=0.45 critical value is -1.6
Subtracting alpha from the standard value of 0.5 then looking for the resultant difference in the z table.
7. Reject the null hypothesis since the test statistic is less than -1.6 which is the critical value.
8. There is sufficient evidenc ...
1) This document discusses sampling and sampling distributions, including key terms like population, sample, parameter, statistic, and point estimation.
2) It describes simple random sampling for both finite and infinite populations and introduces the concept of sampling distributions - the probability distributions of sample statistics.
3) The sampling distribution of the mean is discussed, including how it approaches a normal distribution as sample size increases due to the central limit theorem.
Calculating Analysis of Variance (ANOVA) and Post Hoc Analyses Follo.docxaman341480
Calculating Analysis of Variance (ANOVA) and Post Hoc Analyses Following ANOVA
Analysis of variance (ANOVA)
is a statistical procedure that compares data between two or more groups or conditions to investigate the presence of differences between those groups on some continuous dependent variable (see
Exercise 18
). In this exercise, we will focus on the
one-way ANOVA
, which involves testing one independent variable and one dependent variable (as opposed to other types of ANOVAs, such as factorial ANOVAs that incorporate multiple independent variables).
Why ANOVA and not a
t
-test? Remember that a
t
-test is formulated to compare two sets of data or two groups at one time (see
Exercise 23
for guidance on selecting appropriate statistics). Thus, data generated from a clinical trial that involves four experimental groups, Treatment 1, Treatment 2, Treatments 1 and 2 combined, and a Control, would require 6
t
-tests. Consequently, the chance of making a Type I error (alpha error) increases substantially (or is inflated) because so many computations are being performed. Specifically, the chance of making a Type I error is the number of comparisons multiplied by the alpha level. Thus, ANOVA is the recommended statistical technique for examining differences between more than two groups (
Zar, 2010
).
ANOVA is a procedure that culminates in a statistic called the
F
statistic. It is this value that is compared against an
F
distribution (see
Appendix C
) in order to determine whether the groups significantly differ from one another on the dependent variable. The formulas for ANOVA actually compute two estimates of variance: One estimate represents differences between the groups/conditions, and the other estimate represents differences among (within) the data.
Research Designs Appropriate for the One-Way ANOVA
Research designs that may utilize the one-way ANOVA include the randomized experimental, quasi-experimental, and comparative designs (
Gliner, Morgan, & Leech, 2009
). The independent variable (the “grouping” variable for the ANOVA) may be active or attributional. An active independent variable refers to an intervention, treatment, or program. An attributional independent variable refers to a characteristic of the participant, such as gender, diagnosis, or ethnicity. The ANOVA can compare two groups or more. In the case of a two-group design, the researcher can either select an independent samples
t
-test or a one-way ANOVA to answer the research question. The results will always yield the same conclusion, regardless of which test is computed; however, when examining differences between more than two groups, the one-way ANOVA is the preferred statistical test.
Example 1: A researcher conducts a randomized experimental study wherein she randomizes participants to receive a high-dosage weight loss pill, a low-dosage weight loss pill, or a placebo. She assesses the number of pounds lost from baseline to post-treatment
378
for the thre ...
This document provides a 100-question practice exam for the QNT 275 final exam. It covers topics in statistics including hypothesis testing, data types, measurement scales, sampling, descriptive statistics, and inferential statistics. Sample questions are multiple choice and cover topics like hypothesis tests, measurement scales, sampling methods, descriptive vs inferential statistics, and data analysis techniques like ANOVA. The practice exam allows students to test their understanding of key statistical concepts.
For more classes visit
www.snaptutorial.com
1
To make tests of hypotheses about more than two population means, we use the:
t distribution
normal distribution
chi-square distribution
analysis of variance distribution
For more classes visit
www.snaptutorial.com
1
To make tests of hypotheses about more than two population means, we use the:
t distribution
normal distribution
chi-square distribution
analysis of variance distribution
2
You randomly select two households and observe whether or not they own a telephone answering machine. Which of the following is a simple event?
At most one of them owns a telephone answering machine.
1. An independent samples t-test was conducted to determine if there were differences in anxiety scores between male and female participants before a major competition.
2. The results of the t-test showed no significant difference between the mean anxiety scores of males (M=17, SD=4.58) and females (M=18, SD=3.16), t(8)=0.41, p>0.05.
3. Therefore, the null hypothesis that there is no difference between male and female anxiety scores before a major competition was not rejected.
This document discusses statistical models and inferential statistics. It defines statistical modeling as using mathematical tools and statistical conclusions to understand real-life situations. There are three main types of statistical models: parametric models which have known parameters; nonparametric models which have flexible parameters; and semi-parametric models which are a blend of the two. Inferential statistics are used to draw conclusions about populations based on samples, while descriptive statistics describe sample characteristics. Common inferential statistics techniques include hypothesis testing, regression analysis, z-tests, t-tests, f-tests, and confidence intervals.
Running head COURSE PROJECT –PHASE 3 COURSE PROJECT –PHASE 3.docxsusanschei
Running head: COURSE PROJECT –PHASE 3
COURSE PROJECT –PHASE 3
Course Project –Phase 3
Name: Rodney Wheeler
Institution: Rasmussen College
Course: STA3215 Section 01 Inferential Statistics and Analytics
Date: 03/04/17
Course Project –Phase 3
The primary goal of statistics is to conduct a hypothesis. A hypothesis is a prediction about something; hypothesis testing is done to ascertain if a sampled proportion differs from a specified population. For the test to be valid eight steps are conducted to ensure the results are up to par (Lora M. and Richard J. Cook., 2009);
Step One -Identify and come up with a research question, this helps the researcher narrow down to what they want to test.For instance, is the number of patients admitted with infectious disease less than 65 years of age? Such questions are important as they help one in looking for the necessary data and conduct the test efficiently
Step Two-Ascertain that some expectations are met: The method of research used is Simple random sampling, the resultant outcome is only one, and the population is triple the sample size in question
Step Three-State the two types of hypothesis: Identify the null and alternative hypothesis. Null hypothesis shows equality while alternative does not.
Step Four-Determine a definite significant level that is the odds of refuting a null hypothesis through use of alpha
Step Five-Calculate the test statistic, this are constant values that are calculated from the available data when conducting a hypothesis test
Step Six-Change the test statistic into a P value; A p-value is the possibility that a selected sample would differ with the obtained one. It differs depending on the test used and is determined by use of the normal distribution table
Step Seven-Choose between the null and alternative hypothesis, this is where one has to determine whether the stated research question is correct. If the p-value is greater than the standardized value, the null hypothesis should be rejected
Step Eight-Creating a conclusion of your Research Question, determine whether or not the set values are sufficient evidence in confirming your research.
The p-value is the better approach as computation of one value is required to conduct the test, the critical approach is cumbersome as one has to compute the test statistic and also find the key value of the significance level
Question two
1. Ho:p>=65;Ha p<65
2. The test is left tailed since the sample proportion is less than the hypothesized population proportion
3. The test statistics to be used is the t test since the standard deviation is unknown.
4. =-2.79
5. Degree of freedom is 60-1=59as observed from the t table the p- value is 0.05
6. 0.5-0.05=0.45 critical value is -1.6
Subtracting alpha from the standard value of 0.5 then looking for the resultant difference in the z table.
7. Reject the null hypothesis since the test statistic is less than -1.6 which is the critical value.
8. There is sufficient evidenc ...
1) This document discusses sampling and sampling distributions, including key terms like population, sample, parameter, statistic, and point estimation.
2) It describes simple random sampling for both finite and infinite populations and introduces the concept of sampling distributions - the probability distributions of sample statistics.
3) The sampling distribution of the mean is discussed, including how it approaches a normal distribution as sample size increases due to the central limit theorem.
- Sampling distribution describes the distribution of sample statistics like means or proportions drawn from a population. It allows making statistical inferences about the population.
- The central limit theorem states that sampling distributions of sample means will be approximately normally distributed regardless of the population distribution, if the sample size is large.
- Standard error measures the amount of variability in values of a sample statistic across different samples. It is used to construct confidence intervals for population parameters.
Biostatistics is the science of collecting, summarizing, analyzing, and interpreting data in the fields of medicine, biology, and public health. It involves both descriptive and inferential statistics. Descriptive statistics summarize data through measures of central tendency like mean, median, and mode, and measures of dispersion like range and standard deviation. Inferential statistics allow generalization from samples to populations through techniques like hypothesis testing, confidence intervals, and estimation. Sample size determination and random sampling help ensure validity and minimize errors in statistical analyses.
This document provides an overview of sampling and sampling distributions. It discusses how random samples are used to make statistical inferences about populations. The key points covered include:
- Sampling is the process of selecting a portion of a population to estimate characteristics of the whole population.
- There are two main types of sampling: probability sampling and non-probability sampling.
- The distribution of sample statistics (like the sample mean) is called the sampling distribution.
- According to the central limit theorem, the sampling distribution of the mean will follow a normal distribution, even if the population is not normally distributed, as long as the sample size is sufficiently large.
- The standard error of the sampling distribution of the
The document discusses performing a correlation analysis on selected numerical and categorical variables from a data set to identify highly correlated variables. A heat map was generated from the correlation analysis. Two numerical variables, total sales by branch and dairy sales total, were identified as highly correlated with other variables and removed from further analysis. Stepwise regression was then performed on the remaining variables to further reduce the number of predictor variables.
Week 5 Lecture 14 The Chi Square TestQuite often, patterns of .docxcockekeshia
Week 5 Lecture 14
The Chi Square Test
Quite often, patterns of responses or measures give us a lot of information. Patterns are generally the result of counting how many things fit into a particular category. Whenever we make a histogram, bar, or pie chart we are looking at the pattern of the data. Frequently, changes in these visual patterns will be our first clues that things have changed, and the first clue that we need to initiate a research study (Lind, Marchel, & Wathen, 2008).
One of the most useful test in examining patterns and relationships in data involving counts (how many fit into this category, how many into that, etc.) is the chi-square. It is extremely easy to calculate and has many more uses than we will cover. Examining patterns involves two uses of the Chi-square - the goodness of fit and the contingency table. Both of these uses have a common trait: they involve counts per group. In fact, the chi-square is the only statistic we will look at that we use when we have counts per multiple groups (Tanner & Youssef-Morgan, 2013). Chi Square Goodness of Fit Test
The goodness of fit test checks to see if the data distribution (counts per group) matches some pattern we are interested in. Example: Are the employees in our example company distributed equal across the grades? Or, a more reasonable expectation for a company might be are the employees distributed in a pyramid fashion – most on the bottom and few at the top?
The Chi Square test compares the actual versus a proposed distribution of counts by generating a measure for each cell or count: (actual – expected)2/actual. Summing these for all of the cells or groups provides us with the Chi Square Statistic. As with our other tests, we determine the p-value of getting a result as large or larger to determine if we reject or not reject our null hypothesis. An example will show the approach using Excel.
Regardless of the Chi Square test, the chi square related functions are found in the fx Statistics window rather than the Data Analysis where we found the t and ANOVA test functions. The most important for us are:
· CHISQ.TEST (actual range, expected range) – returns the p-value for the test
· CHISQ.INV.RT(p-value, df) – returns the actual Chi Square value for the p-value or probability value used.
· CHISQ.DIST.RT(X, df) – returns the p-value for a given value.
When we have a table of actual and expected results, using the =CHISQ.TEST(actual range, expected range) will provide us with the p-value of the calculated chi square value (but does not give us the actual calculated chi square value for the test). We can compare this value against our alpha criteria (generally 0.05) to make our decision about rejecting or not rejecting the null hypothesis.
If, after finding the p-value for our chi square test, we want to determine the calculated value of the chi square statistic, we can use the =CHISQ.INV.RT(probability, df) function, the value for probability is .
Researchers use several tools and procedures for analyzing quantitative data obtained from different types of experimental designs. Different designs call for different methods of analysis. This presentation focuses on:
T-test
Analysis of variance (F-test), and
Chi-square test
Marketing Research Hypothesis Testing.pptxxababid981
This document provides an overview of parametric and non-parametric hypothesis tests. It defines parametric tests as those that assume an underlying normal distribution, and lists common parametric tests like the z-test, t-test, F-test, and ANOVA. Non-parametric tests make no distributional assumptions and common examples discussed include the Mann-Whitney U test, chi-square test, and Kruskal-Wallis test. The document provides details on assumptions and procedures for conducting each of these important statistical hypothesis tests.
This document provides an overview of sampling and statistical inference concepts. It defines key terms like population, sample, parameter, and statistic. It discusses reasons for sampling and types of sampling and non-sampling errors. It also explains important sampling distributions like the sampling distribution of the mean, t-distribution, sampling distribution of a proportion, F distribution, and chi-square distribution. It defines concepts like degrees of freedom, standard error, and the central limit theorem.
This document contains a summary of key statistical concepts and methods for analyzing grouped data. It includes 10 topics: (1) names of group members, (2) acknowledgements, (3) an introduction to statistics, (4) grouped data, (5) mean of grouped data, (6) mode of grouped data, (7) median of grouped data, (8) graphical representation of cumulative distribution, (9) conclusion, and (10) bibliography. The document provides examples and explanations of statistical techniques for summarizing and visualizing grouped data, including calculating the mean, mode, and median from frequency tables and constructing cumulative frequency distributions in ogive graphs.
In the t test for independent groups, ____.we estimate µ1 µ2.docxbradburgess22840
In the t test for independent groups, ____.
we estimate µ1 µ2
we estimate 2
we estimate X1-X2
df = N 1
Exhibit 14-1
A professor of women's studies is interested in determining if stress affects the menstrual cycle. Ten women are randomly sampled for an experiment and randomly divided into two groups. One of the groups is subjected to high stress for two months while the other lives in a relatively stress-free environment. The professor measures the menstrual cycle (in days) of each woman during the second month. The following data are obtained.
High stress
20
23
18
19
22
Relatively stress free
26
31
25
26
30
Refer to Exhibit 14-1. The obtained value of the appropriate statistic is ____.
tobt = 4.73
tobt = 4.71
tobt = 3.05
tobt = 0.47
Refer to Exhibit 14-1. The df for determining tcrit are ____.
4
9
8
3
Refer to Exhibit 14-1. Using = .052 tail, tcrit = ____.
+2.162
+2.506
±2.462
±2.306
Refer to Exhibit 14-1. Using = .052 tail, your conclusion is ____.
accept H0; stress does not affect the menstrual cycle
retain H0; we cannot conclude that stress affects the menstrual cycle
retain H0; stress affects the menstrual cycle
reject H0; stress affects the menstrual cycle
Refer to Exhibit 14-1. Estimate the size of the effect. = ____
0.8102
0.6810
0.4322
0.5776
A major advantage to using a two condition experiment (e.g. control and experimental groups) is ____.
the test has more power
the data are easier to analyze
the experiment does not need to know population parameters
the test has less power
Which of the following tests analyzes the difference between the means of two independent samples?
correlated t test
t test for independent groups
sign test
test of variance
If n1 = n2 and n is relatively large, then the t test is relatively robust against ____.
violations of the assumptions of homogeneity of variance and normality
violations of random samples
traffic violations
violations by the forces of evil
Exhibit 14-3
Five students were tested before and after taking a class to improve their study habits. They were given articles to read which contained a known number of facts in each story. After the story each student listed as many facts as he/she could recall. The following data was recorded.
Before
10
12
14
16
12
After
15
14
17
17
20
Refer to Exhibit 14-3. The obtained value of the appropriate statistic is ____.
3.92
3.06
4.12
2.58
Refer to Exhibit 14-3. What do you conclude using = 0.052 tail?
reject H0; the class appeared to improve study habits
retain H0; the class had no effect on study habits
retain H0; we cannot conclude that the class improved study habits
accept H0; the class appeared to improve study habits
Which of the following is (are) assumption(s) underlying the use of the F test?
the raw score populations are normally distributed
the variances of the raw score populations are the same
the mean of the populations differ
the raw score popul.
This document discusses the distribution of sample means when taking samples from a population. It explains that as the sample size increases, the distribution of sample means approaches a normal distribution, even if the population is not normally distributed. The mean of the distribution of sample means is equal to the population mean. The variability of the distribution is measured by the standard error, which depends on the sample size and population standard deviation. Larger sample sizes result in smaller standard errors and distributions of sample means that are nearly normal.
Assignment 2 Tests of SignificanceThroughout this assignment yo.docxrock73
Assignment 2: Tests of Significance
Throughout this assignment you will review mock studies. You will needs to follow the directions outlined in the section using SPSS and decide whether there is significance between the variables. You will need to list the five steps of hypothesis testing (as covered in the lesson for Week 6) to see how every question should be formatted. You will complete all of the problems. Be sure to cut and past the appropriate test result boxes from SPSS under each problem and explain what you will do with your research hypotheses. All calculations should be coming from your SPSS. You will need to submit the SPSS output file to get credit for this assignment. This file will save as a .spv file and will need to be in a single file. In other words, you are not allowed to submit more than one output file for this assignment.
The five steps of hypothesis testing when using SPSS are as follows:
1. State your research hypothesis (H1) and null hypothesis (H0).
2. Identify your confidence interval (.05 or .01)
3. Conduct your analysis using SPSS.
4. Look for the valid score for comparison. This score is usually under ‘Sig 2-tail’ or ‘Sig. 2’. We will call this “p”.
5. Compare the two and apply the following rule:
a. If “p” is < or = confidence interval, than you reject the null.
Be sure to explain to the reader what this means in regards to your study. (Ex: will you recommend counseling services?)
* Be sure that your answers are clearly distinguishable. Perhaps you bold your font or use a different color.
ASSIGNMENT 2(200) WORD MINIUM
1. They allow us to see if our relationship is "statistically significant". (Remember that this only shows us that there is or is not a relationship but does NOT show us if it is big, small, or in-between.)
2. It let's us know if our findings can be generalized to the population which our sample was selected from and represents.
This week you will decide which test of significance you will use for your project. For this class your choices for tests will include one of the following:
· Chi-square
· t Test
· ANOVA
We will be using a process for hypothesis testing which outlines five steps researchers can follow to complete this process:
1. Write your research hypothesis (H1) and your null hypothesis (H0).
2. Identify and record your confidence interval. These are usually .05 (95%) or .01 (99%).
3. Complete the test using SPSS.
4. Identify the number under Sig. (2-tail). This will be represented by "p".
5. Compare the numbers in steps 2 and 4 and apply the following rule:
1. If p < or = confidence interval, than you reject the null hypothesis
Determine what to do with your null and explain this to your reader. Be sure to go beyond the phrase "reject or fail to reject the null" and explain how that impacts your research and best describes the relationship between variables.
TEST QUESTIONS-NEED FULL ANSWERS
Q1
Make up and discuss research examples corresponding to the various ...
This document discusses key concepts in statistical estimation including:
- Estimation involves using sample data to infer properties of the population by calculating point estimates and interval estimates.
- A point estimate is a single value that estimates an unknown population parameter, while an interval estimate provides a range of plausible values for the parameter.
- A confidence interval gives the probability that the interval calculated from the sample data contains the true population parameter. Common confidence intervals are 95% confidence intervals.
- Formulas for confidence intervals depend on whether the population standard deviation is known or unknown, and the sample size.
This document discusses different statistical tests used to analyze experimental research data, including the t-test, analysis of variance (ANOVA), and chi-square test. It provides examples of how to apply each test and interpret the results. The t-test is used to compare the means of two groups, ANOVA is used for comparing more than two groups, and chi-square is used to analyze relationships between categorical variables. Computer programs like SPSS can perform these statistical analyses to help researchers evaluate experimental data.
Chi‑square Test and its Application in Hypothesis Testing
Rakesh Rana, Richa Singhal
Statistical Section, Central Council for Research in Ayurvedic Sciences, Ministry of AYUSH, GOI, New Delhi, India
This document provides an overview of basic statistical concepts for bio science students. It defines measures of central tendency including mean, median, and mode. It also discusses measures of dispersion like range and standard deviation. Common probability distributions such as binomial, Poisson, and normal distributions are explained. Hypothesis testing concepts like p-values and types of statistical tests for different types of data like t-tests for continuous variables and chi-square tests for categorical data are summarized along with examples.
QUESTION 1Question 1 Describe the purpose of ecumenical servic.docxmakdul
This document contains a summary of a research article that examines the relationship between patient satisfaction scores and inpatient admission volumes at teaching and non-teaching hospitals. The study found a statistically significant positive correlation between patient satisfaction and admissions at teaching hospitals, but a non-significant negative correlation at non-teaching hospitals. When combined, teaching and non-teaching hospitals showed a statistically significant negative correlation. The findings suggest patient satisfaction may impact admissions more at teaching hospitals. The conclusion provides recommendations for healthcare organizations to strategically focus on patient satisfaction to strengthen performance.
Forecasting Academic Performance using Multiple Linear Regressionijtsrd
This document discusses using multiple linear regression to forecast academic performance based on intelligence quotient (IQ) and study hours. The authors collected test score, IQ, and study hour data for 10 students and used the Statistical Package for Social Sciences (SPSS) to analyze the data. They found that IQ and study hours significantly predicted test scores, with IQ and study hours explaining 91% of the variance in test scores. For every one unit increase in IQ, test scores increased by 0.509 units on average, and for every one unit increase in study hours, test scores increased by 0.467 units on average. The authors conclude that regression is a useful statistical method for educational research and that this analysis can help students and teachers improve academic performance
1. The document discusses key concepts in statistics including populations, samples, descriptive statistics, inferential statistics, qualitative and quantitative data, and scales of measurement.
2. It provides examples of statistical tests that can be used including one-sample t-tests, two-sample t-tests, paired t-tests, one-way ANOVA tests, and examples of how they can be applied.
3. The guidelines for designing a statistical study are outlined including identifying variables of interest, developing a data collection plan, collecting and describing data, and interpreting results.
EXERCISE 27I WILL SEND THE DATA TO WHOM EVER WILL DO THE ASSIGNMEN.docxAlleneMcclendon878
EXERCISE 27
I WILL SEND THE DATA TO WHOM EVER WILL DO THE ASSIGNMENT NEEDED NO LATER THAN 11pm
Calculating Descriptive Statistics
There are two major classes of statistics: descriptive statistics and inferential statistics. Descriptive statistics are computed to reveal characteristics of the sample data set and to describe study variables. Inferential statistics are computed to gain information about effects and associations in the population being studied. For some types of studies, descriptive statistics will be the only approach to analysis of the data. For other studies, descriptive statistics are the first step in the data analysis process, to be followed by inferential statistics. For all studies that involve numerical data, descriptive statistics are crucial in understanding the fundamental properties of the variables being studied. Exercise 27 focuses only on descriptive statistics and will illustrate the most common descriptive statistics computed in nursing research and provide examples using actual clinical data from empirical publications.
Measures of Central Tendency
A
measure of central tendency
is a statistic that represents the center or middle of a frequency distribution. The three measures of central tendency commonly used in nursing research are the mode, median (
MD
), and mean (
). The mean is the arithmetic average of all of a variable's values. The median is the exact middle value (or the average of the middle two values if there is an even number of observations). The mode is the most commonly occurring value or values (see
Exercise 8
).
The following data have been collected from veterans with rheumatoid arthritis (
Tran, Hooker, Cipher, & Reimold, 2009
). The values in
Table 27-1
were extracted from a larger sample of veterans who had a history of biologic medication use (e.g., infliximab [Remicade], etanercept [Enbrel]).
Table 27-1
contains data collected from 10 veterans who had stopped taking biologic medications, and the variable represents the number of years that each veteran had taken the medication before stopping.
TABLE 27-1
DURATION OF BIOLOGIC USE AMONG VETERANS WITH RHEUMATOID ARTHRITIS (
n
= 10)
Duration of Biologic Use (years)
0.1
0.3
1.3
1.5
1.5
2.0
2.2
3.0
3.0
4.0
Because the number of study subjects represented below is 10, the correct statistical notation to reflect that number is:
Note that the
n
is lowercase, because we are referring to a sample of veterans. If the data being presented represented the entire population of veterans, the correct notation is the uppercase
N.
Because most nursing research is conducted using samples, not populations, all formulas in the subsequent exercises will incorporate the sample notation,
n.
Mode
The
mode
is the numerical value or score that occurs with the greatest frequency; it does not necessarily indicate the center of the data set. The data in
Table 27-1
contain two
292
modes: 1.5 and 3.0. Each of these numbers occurred twice in the data s.
1 PageAPAsources2Today, many more organizations leverage worl.docxcurranalmeta
1 Page
APA
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Today, many more organizations leverage worldwide opportunities by becoming global organizations. Going global opens doors to markets and services, but it can place new demands on a company. List and discuss 3–5 factors a CIO or the IT group must consider with respect to data, technology, networks, or security when a company goes global.
Response Guidelines
Respond to the posts of at least two other learners. Your responses are expected to be substantive in nature and reference the assigned readings, as well as other theoretical, empirical, or professional literature to support your views and writings. In your response, do at least one of the following:
Ask a probing question.
Offer a suggestion.
Elaborate on a particular point.
Provide an alternative opinion.
Email me
[email protected]
with your rates asap.
.
1) A major corporation agrees to sponsor an internal study on sexual.docxcurranalmeta
1) A major corporation agrees to sponsor an internal study on sexual harassment in the workplace in response to concerns expressed by its female employees. How would you handle the following issues: a)The communication approach (self-administered, telephone, personal interview, and/or mixed) b)The purpose: Fact finding, awareness, relationship building, and/or change c)Participant motivation d)Minimization of response and non-response error.
.
More Related Content
Similar to Trochim, W. M. K. (2006). Internal validity.httpwww.socialres
- Sampling distribution describes the distribution of sample statistics like means or proportions drawn from a population. It allows making statistical inferences about the population.
- The central limit theorem states that sampling distributions of sample means will be approximately normally distributed regardless of the population distribution, if the sample size is large.
- Standard error measures the amount of variability in values of a sample statistic across different samples. It is used to construct confidence intervals for population parameters.
Biostatistics is the science of collecting, summarizing, analyzing, and interpreting data in the fields of medicine, biology, and public health. It involves both descriptive and inferential statistics. Descriptive statistics summarize data through measures of central tendency like mean, median, and mode, and measures of dispersion like range and standard deviation. Inferential statistics allow generalization from samples to populations through techniques like hypothesis testing, confidence intervals, and estimation. Sample size determination and random sampling help ensure validity and minimize errors in statistical analyses.
This document provides an overview of sampling and sampling distributions. It discusses how random samples are used to make statistical inferences about populations. The key points covered include:
- Sampling is the process of selecting a portion of a population to estimate characteristics of the whole population.
- There are two main types of sampling: probability sampling and non-probability sampling.
- The distribution of sample statistics (like the sample mean) is called the sampling distribution.
- According to the central limit theorem, the sampling distribution of the mean will follow a normal distribution, even if the population is not normally distributed, as long as the sample size is sufficiently large.
- The standard error of the sampling distribution of the
The document discusses performing a correlation analysis on selected numerical and categorical variables from a data set to identify highly correlated variables. A heat map was generated from the correlation analysis. Two numerical variables, total sales by branch and dairy sales total, were identified as highly correlated with other variables and removed from further analysis. Stepwise regression was then performed on the remaining variables to further reduce the number of predictor variables.
Week 5 Lecture 14 The Chi Square TestQuite often, patterns of .docxcockekeshia
Week 5 Lecture 14
The Chi Square Test
Quite often, patterns of responses or measures give us a lot of information. Patterns are generally the result of counting how many things fit into a particular category. Whenever we make a histogram, bar, or pie chart we are looking at the pattern of the data. Frequently, changes in these visual patterns will be our first clues that things have changed, and the first clue that we need to initiate a research study (Lind, Marchel, & Wathen, 2008).
One of the most useful test in examining patterns and relationships in data involving counts (how many fit into this category, how many into that, etc.) is the chi-square. It is extremely easy to calculate and has many more uses than we will cover. Examining patterns involves two uses of the Chi-square - the goodness of fit and the contingency table. Both of these uses have a common trait: they involve counts per group. In fact, the chi-square is the only statistic we will look at that we use when we have counts per multiple groups (Tanner & Youssef-Morgan, 2013). Chi Square Goodness of Fit Test
The goodness of fit test checks to see if the data distribution (counts per group) matches some pattern we are interested in. Example: Are the employees in our example company distributed equal across the grades? Or, a more reasonable expectation for a company might be are the employees distributed in a pyramid fashion – most on the bottom and few at the top?
The Chi Square test compares the actual versus a proposed distribution of counts by generating a measure for each cell or count: (actual – expected)2/actual. Summing these for all of the cells or groups provides us with the Chi Square Statistic. As with our other tests, we determine the p-value of getting a result as large or larger to determine if we reject or not reject our null hypothesis. An example will show the approach using Excel.
Regardless of the Chi Square test, the chi square related functions are found in the fx Statistics window rather than the Data Analysis where we found the t and ANOVA test functions. The most important for us are:
· CHISQ.TEST (actual range, expected range) – returns the p-value for the test
· CHISQ.INV.RT(p-value, df) – returns the actual Chi Square value for the p-value or probability value used.
· CHISQ.DIST.RT(X, df) – returns the p-value for a given value.
When we have a table of actual and expected results, using the =CHISQ.TEST(actual range, expected range) will provide us with the p-value of the calculated chi square value (but does not give us the actual calculated chi square value for the test). We can compare this value against our alpha criteria (generally 0.05) to make our decision about rejecting or not rejecting the null hypothesis.
If, after finding the p-value for our chi square test, we want to determine the calculated value of the chi square statistic, we can use the =CHISQ.INV.RT(probability, df) function, the value for probability is .
Researchers use several tools and procedures for analyzing quantitative data obtained from different types of experimental designs. Different designs call for different methods of analysis. This presentation focuses on:
T-test
Analysis of variance (F-test), and
Chi-square test
Marketing Research Hypothesis Testing.pptxxababid981
This document provides an overview of parametric and non-parametric hypothesis tests. It defines parametric tests as those that assume an underlying normal distribution, and lists common parametric tests like the z-test, t-test, F-test, and ANOVA. Non-parametric tests make no distributional assumptions and common examples discussed include the Mann-Whitney U test, chi-square test, and Kruskal-Wallis test. The document provides details on assumptions and procedures for conducting each of these important statistical hypothesis tests.
This document provides an overview of sampling and statistical inference concepts. It defines key terms like population, sample, parameter, and statistic. It discusses reasons for sampling and types of sampling and non-sampling errors. It also explains important sampling distributions like the sampling distribution of the mean, t-distribution, sampling distribution of a proportion, F distribution, and chi-square distribution. It defines concepts like degrees of freedom, standard error, and the central limit theorem.
This document contains a summary of key statistical concepts and methods for analyzing grouped data. It includes 10 topics: (1) names of group members, (2) acknowledgements, (3) an introduction to statistics, (4) grouped data, (5) mean of grouped data, (6) mode of grouped data, (7) median of grouped data, (8) graphical representation of cumulative distribution, (9) conclusion, and (10) bibliography. The document provides examples and explanations of statistical techniques for summarizing and visualizing grouped data, including calculating the mean, mode, and median from frequency tables and constructing cumulative frequency distributions in ogive graphs.
In the t test for independent groups, ____.we estimate µ1 µ2.docxbradburgess22840
In the t test for independent groups, ____.
we estimate µ1 µ2
we estimate 2
we estimate X1-X2
df = N 1
Exhibit 14-1
A professor of women's studies is interested in determining if stress affects the menstrual cycle. Ten women are randomly sampled for an experiment and randomly divided into two groups. One of the groups is subjected to high stress for two months while the other lives in a relatively stress-free environment. The professor measures the menstrual cycle (in days) of each woman during the second month. The following data are obtained.
High stress
20
23
18
19
22
Relatively stress free
26
31
25
26
30
Refer to Exhibit 14-1. The obtained value of the appropriate statistic is ____.
tobt = 4.73
tobt = 4.71
tobt = 3.05
tobt = 0.47
Refer to Exhibit 14-1. The df for determining tcrit are ____.
4
9
8
3
Refer to Exhibit 14-1. Using = .052 tail, tcrit = ____.
+2.162
+2.506
±2.462
±2.306
Refer to Exhibit 14-1. Using = .052 tail, your conclusion is ____.
accept H0; stress does not affect the menstrual cycle
retain H0; we cannot conclude that stress affects the menstrual cycle
retain H0; stress affects the menstrual cycle
reject H0; stress affects the menstrual cycle
Refer to Exhibit 14-1. Estimate the size of the effect. = ____
0.8102
0.6810
0.4322
0.5776
A major advantage to using a two condition experiment (e.g. control and experimental groups) is ____.
the test has more power
the data are easier to analyze
the experiment does not need to know population parameters
the test has less power
Which of the following tests analyzes the difference between the means of two independent samples?
correlated t test
t test for independent groups
sign test
test of variance
If n1 = n2 and n is relatively large, then the t test is relatively robust against ____.
violations of the assumptions of homogeneity of variance and normality
violations of random samples
traffic violations
violations by the forces of evil
Exhibit 14-3
Five students were tested before and after taking a class to improve their study habits. They were given articles to read which contained a known number of facts in each story. After the story each student listed as many facts as he/she could recall. The following data was recorded.
Before
10
12
14
16
12
After
15
14
17
17
20
Refer to Exhibit 14-3. The obtained value of the appropriate statistic is ____.
3.92
3.06
4.12
2.58
Refer to Exhibit 14-3. What do you conclude using = 0.052 tail?
reject H0; the class appeared to improve study habits
retain H0; the class had no effect on study habits
retain H0; we cannot conclude that the class improved study habits
accept H0; the class appeared to improve study habits
Which of the following is (are) assumption(s) underlying the use of the F test?
the raw score populations are normally distributed
the variances of the raw score populations are the same
the mean of the populations differ
the raw score popul.
This document discusses the distribution of sample means when taking samples from a population. It explains that as the sample size increases, the distribution of sample means approaches a normal distribution, even if the population is not normally distributed. The mean of the distribution of sample means is equal to the population mean. The variability of the distribution is measured by the standard error, which depends on the sample size and population standard deviation. Larger sample sizes result in smaller standard errors and distributions of sample means that are nearly normal.
Assignment 2 Tests of SignificanceThroughout this assignment yo.docxrock73
Assignment 2: Tests of Significance
Throughout this assignment you will review mock studies. You will needs to follow the directions outlined in the section using SPSS and decide whether there is significance between the variables. You will need to list the five steps of hypothesis testing (as covered in the lesson for Week 6) to see how every question should be formatted. You will complete all of the problems. Be sure to cut and past the appropriate test result boxes from SPSS under each problem and explain what you will do with your research hypotheses. All calculations should be coming from your SPSS. You will need to submit the SPSS output file to get credit for this assignment. This file will save as a .spv file and will need to be in a single file. In other words, you are not allowed to submit more than one output file for this assignment.
The five steps of hypothesis testing when using SPSS are as follows:
1. State your research hypothesis (H1) and null hypothesis (H0).
2. Identify your confidence interval (.05 or .01)
3. Conduct your analysis using SPSS.
4. Look for the valid score for comparison. This score is usually under ‘Sig 2-tail’ or ‘Sig. 2’. We will call this “p”.
5. Compare the two and apply the following rule:
a. If “p” is < or = confidence interval, than you reject the null.
Be sure to explain to the reader what this means in regards to your study. (Ex: will you recommend counseling services?)
* Be sure that your answers are clearly distinguishable. Perhaps you bold your font or use a different color.
ASSIGNMENT 2(200) WORD MINIUM
1. They allow us to see if our relationship is "statistically significant". (Remember that this only shows us that there is or is not a relationship but does NOT show us if it is big, small, or in-between.)
2. It let's us know if our findings can be generalized to the population which our sample was selected from and represents.
This week you will decide which test of significance you will use for your project. For this class your choices for tests will include one of the following:
· Chi-square
· t Test
· ANOVA
We will be using a process for hypothesis testing which outlines five steps researchers can follow to complete this process:
1. Write your research hypothesis (H1) and your null hypothesis (H0).
2. Identify and record your confidence interval. These are usually .05 (95%) or .01 (99%).
3. Complete the test using SPSS.
4. Identify the number under Sig. (2-tail). This will be represented by "p".
5. Compare the numbers in steps 2 and 4 and apply the following rule:
1. If p < or = confidence interval, than you reject the null hypothesis
Determine what to do with your null and explain this to your reader. Be sure to go beyond the phrase "reject or fail to reject the null" and explain how that impacts your research and best describes the relationship between variables.
TEST QUESTIONS-NEED FULL ANSWERS
Q1
Make up and discuss research examples corresponding to the various ...
This document discusses key concepts in statistical estimation including:
- Estimation involves using sample data to infer properties of the population by calculating point estimates and interval estimates.
- A point estimate is a single value that estimates an unknown population parameter, while an interval estimate provides a range of plausible values for the parameter.
- A confidence interval gives the probability that the interval calculated from the sample data contains the true population parameter. Common confidence intervals are 95% confidence intervals.
- Formulas for confidence intervals depend on whether the population standard deviation is known or unknown, and the sample size.
This document discusses different statistical tests used to analyze experimental research data, including the t-test, analysis of variance (ANOVA), and chi-square test. It provides examples of how to apply each test and interpret the results. The t-test is used to compare the means of two groups, ANOVA is used for comparing more than two groups, and chi-square is used to analyze relationships between categorical variables. Computer programs like SPSS can perform these statistical analyses to help researchers evaluate experimental data.
Chi‑square Test and its Application in Hypothesis Testing
Rakesh Rana, Richa Singhal
Statistical Section, Central Council for Research in Ayurvedic Sciences, Ministry of AYUSH, GOI, New Delhi, India
This document provides an overview of basic statistical concepts for bio science students. It defines measures of central tendency including mean, median, and mode. It also discusses measures of dispersion like range and standard deviation. Common probability distributions such as binomial, Poisson, and normal distributions are explained. Hypothesis testing concepts like p-values and types of statistical tests for different types of data like t-tests for continuous variables and chi-square tests for categorical data are summarized along with examples.
QUESTION 1Question 1 Describe the purpose of ecumenical servic.docxmakdul
This document contains a summary of a research article that examines the relationship between patient satisfaction scores and inpatient admission volumes at teaching and non-teaching hospitals. The study found a statistically significant positive correlation between patient satisfaction and admissions at teaching hospitals, but a non-significant negative correlation at non-teaching hospitals. When combined, teaching and non-teaching hospitals showed a statistically significant negative correlation. The findings suggest patient satisfaction may impact admissions more at teaching hospitals. The conclusion provides recommendations for healthcare organizations to strategically focus on patient satisfaction to strengthen performance.
Forecasting Academic Performance using Multiple Linear Regressionijtsrd
This document discusses using multiple linear regression to forecast academic performance based on intelligence quotient (IQ) and study hours. The authors collected test score, IQ, and study hour data for 10 students and used the Statistical Package for Social Sciences (SPSS) to analyze the data. They found that IQ and study hours significantly predicted test scores, with IQ and study hours explaining 91% of the variance in test scores. For every one unit increase in IQ, test scores increased by 0.509 units on average, and for every one unit increase in study hours, test scores increased by 0.467 units on average. The authors conclude that regression is a useful statistical method for educational research and that this analysis can help students and teachers improve academic performance
1. The document discusses key concepts in statistics including populations, samples, descriptive statistics, inferential statistics, qualitative and quantitative data, and scales of measurement.
2. It provides examples of statistical tests that can be used including one-sample t-tests, two-sample t-tests, paired t-tests, one-way ANOVA tests, and examples of how they can be applied.
3. The guidelines for designing a statistical study are outlined including identifying variables of interest, developing a data collection plan, collecting and describing data, and interpreting results.
EXERCISE 27I WILL SEND THE DATA TO WHOM EVER WILL DO THE ASSIGNMEN.docxAlleneMcclendon878
EXERCISE 27
I WILL SEND THE DATA TO WHOM EVER WILL DO THE ASSIGNMENT NEEDED NO LATER THAN 11pm
Calculating Descriptive Statistics
There are two major classes of statistics: descriptive statistics and inferential statistics. Descriptive statistics are computed to reveal characteristics of the sample data set and to describe study variables. Inferential statistics are computed to gain information about effects and associations in the population being studied. For some types of studies, descriptive statistics will be the only approach to analysis of the data. For other studies, descriptive statistics are the first step in the data analysis process, to be followed by inferential statistics. For all studies that involve numerical data, descriptive statistics are crucial in understanding the fundamental properties of the variables being studied. Exercise 27 focuses only on descriptive statistics and will illustrate the most common descriptive statistics computed in nursing research and provide examples using actual clinical data from empirical publications.
Measures of Central Tendency
A
measure of central tendency
is a statistic that represents the center or middle of a frequency distribution. The three measures of central tendency commonly used in nursing research are the mode, median (
MD
), and mean (
). The mean is the arithmetic average of all of a variable's values. The median is the exact middle value (or the average of the middle two values if there is an even number of observations). The mode is the most commonly occurring value or values (see
Exercise 8
).
The following data have been collected from veterans with rheumatoid arthritis (
Tran, Hooker, Cipher, & Reimold, 2009
). The values in
Table 27-1
were extracted from a larger sample of veterans who had a history of biologic medication use (e.g., infliximab [Remicade], etanercept [Enbrel]).
Table 27-1
contains data collected from 10 veterans who had stopped taking biologic medications, and the variable represents the number of years that each veteran had taken the medication before stopping.
TABLE 27-1
DURATION OF BIOLOGIC USE AMONG VETERANS WITH RHEUMATOID ARTHRITIS (
n
= 10)
Duration of Biologic Use (years)
0.1
0.3
1.3
1.5
1.5
2.0
2.2
3.0
3.0
4.0
Because the number of study subjects represented below is 10, the correct statistical notation to reflect that number is:
Note that the
n
is lowercase, because we are referring to a sample of veterans. If the data being presented represented the entire population of veterans, the correct notation is the uppercase
N.
Because most nursing research is conducted using samples, not populations, all formulas in the subsequent exercises will incorporate the sample notation,
n.
Mode
The
mode
is the numerical value or score that occurs with the greatest frequency; it does not necessarily indicate the center of the data set. The data in
Table 27-1
contain two
292
modes: 1.5 and 3.0. Each of these numbers occurred twice in the data s.
Similar to Trochim, W. M. K. (2006). Internal validity.httpwww.socialres (20)
1 PageAPAsources2Today, many more organizations leverage worl.docxcurranalmeta
1 Page
APA
sources:2
Today, many more organizations leverage worldwide opportunities by becoming global organizations. Going global opens doors to markets and services, but it can place new demands on a company. List and discuss 3–5 factors a CIO or the IT group must consider with respect to data, technology, networks, or security when a company goes global.
Response Guidelines
Respond to the posts of at least two other learners. Your responses are expected to be substantive in nature and reference the assigned readings, as well as other theoretical, empirical, or professional literature to support your views and writings. In your response, do at least one of the following:
Ask a probing question.
Offer a suggestion.
Elaborate on a particular point.
Provide an alternative opinion.
Email me
[email protected]
with your rates asap.
.
1) A major corporation agrees to sponsor an internal study on sexual.docxcurranalmeta
1) A major corporation agrees to sponsor an internal study on sexual harassment in the workplace in response to concerns expressed by its female employees. How would you handle the following issues: a)The communication approach (self-administered, telephone, personal interview, and/or mixed) b)The purpose: Fact finding, awareness, relationship building, and/or change c)Participant motivation d)Minimization of response and non-response error.
.
1 Identify a large racial minority group in U.S. history What has.docxcurranalmeta
1: Identify a large racial minority group in U.S. history? What has been the common ancestral background of this group? When did this group become a significant or notable minority group?
2:
In the 1954 Brown v. Board of Education decision, the Supreme Court desegregated schools in the United States and deemed desegregated schools separate but unequal.
Do you think the legislation was an appropriate reaction to segregation in schools? Do you think the legislation is still relevant? Explain your response.
3:Your texts say that according to sociologists, racial categories are misleading and are harmful ways to divide human groups.
Do you agree or disagree with this statement? Why?
4: How are mobile computing, Web 2.0, and social media changing how IT systems are used?
5:
How are mobile devices changing the way IT systems are used? What are the useful aspects and dangers of these devices? What can organizations do to maximize value and minimize risk when users use these devices?
6: Why is new technology often disruptive for existing IT systems? What can be done to prepare these systems for future technology ahead of time?
.
1 With any new technology, it takes time for ethical, social, and.docxcurranalmeta
1: With any new technology, it takes time for ethical, social, and political issues to be resolved. One example of a new technology used in business today for storing and sharing knowledge and insights is a wiki. What are the ethical, social, and political issues that might arise from using a wiki? What policies would you put in place at your organization to address these issues?
2:
I think it can only impact in positive way. Social media, mobile computing, and Web 2.0 can be use as helpful resources. Social media can be used as a marketing tool, I don't think it's a better marketing tool than social media. If one has an established business than earning followers won't be hard for that business on twitter, Facebook, or Instagram. The benefit for a big business that's established is that they can keep customers up to date with the business, new ideas, and ask for feedback. With one click, depending on the amount of followers, that message can reach millions and millions of people. It can wok for a business that's just stating out as well but they would of course need other resources as well. One can post different messages letting all customers or potential customers know why they're company is the best company, why should one convert to them, and other reasons why ones company is the best way to go. With using the mobile, a small computer is in the palm of ones hand.
3:
When talking about mobile computing, I think that it has changed the way businesses run and conduct their day to day activity. For instance, many jobs now have the option to let you work from home some if not most days of the week. Years ago, this was something that was almost never done. Why are we able to do it today? Because of mobile computing. Even in my position in Human Resources, I had a few appointments yesterday and was able to work from home. My email is on my phone, and I have a fax machine and scanner at home, I went to my appointments in the morning, and then went home and did everything at home that I would have been able to do at my office. It is very nice being able to not use your vacation time for things like that, and still get everything done in the day that you would have gotten done in the office.
4:
I completely agree, Mallory. I feel that the question is racial because some people are born with different races including Black, White, Hispanic, and even a mixed race. Taking advantage of who they are going to hire is very wrong in so many levels, and I believe it is against the law as well because it does count against one person trying to get a job that they can do.
5:
One racial group that many people may not think is large is the Native American group. Native Americans went through a lot since the colonization of the states here in America. Native Americans were here before the Europeans, but unfortunately, due to the large weaponry the settlers had, Native Americans had there land taken away from them. They were pushed into non resourcef.
1 paragraph
No plagiarism
2 Sources
Grading Rubric
Basic response to the DB question
50%
Additional research supporting the initial answer to the DB question
30%
Comments, which serve to further the discussion, apply real-world examples, or lend to enhancing the clarity of the concepts
.
1 PageDiscuss capitation payment methodologies between payers .docxcurranalmeta
1 Page
Discuss capitation payment methodologies between payers and providers, and Medicare or Medicaid with commercial Managed Care organizations (MCO).
Explain at least three important aspects that you as a medical business professional would need to understand when negotiating payment contracts either between a provider and the MCO, or negotiating between the MCO and Medicare.
.
1) Anytime an owner removes any asset for personal use it is recor.docxcurranalmeta
1) Anytime an owner removes any asset for personal use it is recorded as:
A) a withdrawal.
B) payment of a liability.
C) an investment.
D) an expense.
2) The increase or decrease in the owner’s equity is reported on the:
A) statement of owner’s equity.
B) income statement.
C) balance sheet.
D) all of these
3) The purpose of the accounting process is to provide financial information on:
A) large corporations.
B) small businesses.
C) individuals.
D) All of these answers are correct.
4) Which of the following is not a business organization form?
A) Corporation
B) Sole proprietorship
C) Operation
D) Partnership
5) Owner’s withdrawals:
A) increase expenses.
B) increase liabilities.
C) increase assets.
D) decrease owner’s equity.
6) Cater Right, with total assets of $50,000, borrows $15,000 from the bank.
Which of the following is a True statement upon borrowing the money?
A) Total assets are now $35,000.
B) Owner’s equity is $15,000 more.
C) Total assets are now $50,000.
D) Total assets are now $65,000.
7) Mark paid $500 rent for the month. Identify the accounts affected.
A) Cash and Rent Expenses increase.
B) Cash and Capital increase.
C) Cash decreases, and Rent Expense decreases.
D) Cash decreases, and Rent Expense increases.
8) If Suite Dream Toys’ revenues are less than its expenses during the account-
ing period:
A) the business will incur a loss.
B) owner’s withdrawals decrease owner’s equity.
C) owner withdrawals decrease net income.
D) net income causes liabilities to decrease.
9) The financial statement that shows revenue and expenses for a period of time
is the:
A) statement of owner’s equity.
B) balance sheet.
C) statement of liabilities and capital.
D) income statement.
10) The purchase of a truck with a down payment was recorded as a pure cash
purchase. This error would cause:
A) liabilities were understated.
B) assets were overstated.
C) owner’s equity was overstated.
D) None of the above are correct.
11) The business provided services to a credit customer.
A) Assets and owner’s equity increase.
B) Assets and revenue increase.
C) Liabilities and owner’s equity increase.
D) None of the above are correct.
12) Which of the following transactions would cause an asset to increase and the
owner’s equity to increase?
A) The business bought supplies on account.
B) The owner invested cash in the business.
C) The business incurred an expense on credit.
D) The owner withdrew cash from the business.
13) Which of the following would result if the business provided services to a
customer collecting cash?
A) Cash would increase and Revenue would decrease.
B) Since the cash was collected there is no need to record this.
C) Cash would increase and Capital would increase.
D) Cash would increase and Revenue would increase.
14) A formal account that has columns for date, explanat.
1 Page Length In this lab, you will observe the time progressi.docxcurranalmeta
1 Page Length
In this lab, you will observe the time progression of industrialization and human development to help you write up a scientific paper that centers on the following:
If current human development does not change, will groundwater sustainability be affected? Explain your observations.
Human Impacts on the Sustainability of Groundwater
Sustainability is based on a simple principle: Everything that is needed for survival and well-being depends either directly or indirectly on the natural environment. Sustainability creates and maintains the conditions under which humans and nature can exist in productive harmony, while also helping to fulfill the social and economic requirements of present and future generations.
Use the
Lab 1 worksheet
for assignment instructions and data collection.
Part 1
:
Background Information
Planet Earth’s surface is over 70% water, but less than 1% of the water on Earth is considered accessible, usable freshwater for sustaining humans’ and other organisms’ lives. Of the accessible freshwater, approximately 99% is located in aquifers, natural underground water chambers, and other groundwater sources. Unfortunately, humans are depleting the aquifers faster than they can be recharged by the hydrological cycle. Therefore, three quarters of groundwater is considered nonrenewable.
Conditions
The main reason we using groundwater resources mainly for drinking and irrigation. As a result, this not only decreases an important source of freshwater—it also can cause pollution of that groundwater by saltwater intrusion. The recharge rate of groundwater is further hindered by land clearing and deforestation caused by human development. When land is cleared for human development, more flooding occurs, the
transpiration rate
(the amount of water that evaporates into the atmosphere from plants) is reduced, and rainwater is inhibited from adequately
percolating
(penetrating the soil) into the ground to allow for aquifers and groundwater to be recharged.
Figure below shows Saltwater Intrusion
:
(Wright & Boorse, 2010)
Impacts
Forty percent of the world’s food is produced via irrigation. As a result, if the current rate of groundwater usage continues, food production could be drastically reduced worldwide. This reduction in food supply would be detrimental in sustaining the projected worldwide human population of over 10 billion within the next 50 years.
Part 2:
Timeline
Use the Hydrologic Cycle Figure below to understand the impact of industrialization and human development on ground water over 3 centuries.
(Wright & Boorse, 2010)
The table below shows the impacts
:
.
1 page orignal answer needed·A number of organizations exist.docxcurranalmeta
Standards organizations play a crucial role in information systems and security by defining best practices that help ensure consistency, compatibility and security. By establishing common guidelines and protocols, standards help to streamline operations and integration of systems while strengthening defenses against vulnerabilities. Adherence to standards is important for organizations seeking to protect their data and networks through a robust security posture informed by industry experts.
1) discuss 2 major political implications of the cold war in europe..docxcurranalmeta
1) discuss 2 major political implications of the cold war in europe.
2) main causes of the fall of state socialism in the soviet union. who were the main players and what were the main phases of transformation? discuss.
3) to what extent the solidarity movement in poland contributed to political changes in that country within the societ alliance in general?
4) how would you describe major components of putin's political changes in the post-soviet russia? discuss policy changes in both domestic and international areas.
5) discuss major political and economic components of the european union.
6) discuss 2 major challenges facing the european union in the 21st century
Please, answer those questions in an easy to understand bullet point matter. i would have to write a 2 -3 page essay about each question. so please include sufficient points for that. thank you
.
1) Discuss the benefits of using comparative analysis of governments.docxcurranalmeta
1) Discuss the benefits of using comparative analysis of governments. 2) Explain why political systems cannot be compared strictly on their structure.
Your initial post should be at least 250 words in length. Support your claims with examples from required material(s) and/or other scholarly resources, and properly cite any references.
.
1) Discuss essential and non-essential amino acids. Explain why prot.docxcurranalmeta
1) Discuss essential and non-essential amino acids. Explain why protein is described as being of primary importance.(250 words)
2)
Explore the concept of protein quality and how it relates to comparing dietary sources of protein. What are some of the other factors that may play into a consumer’s decision with regards to sources of protein (plant versus animal)?(250 words)
References
.
1 Team Assignment (Will let you know which portion to complete).docxcurranalmeta
1: Team Assignment (Will let you know which portion to complete)
Psychology Game
Develop
a 20-minute Jeopardy game presentation that integrates the breadth and depth of core psychology knowledge.
Search
the Internet for a Microsoft
®
PowerPoint
®
jeopardy game template.
Include
the following topics in your game:
Holistic approach to prior learning
Integrating psychology into your personal life
Psychology across all disciplines
Integrating psychology into career and educational decisions
Your personal learning theory
2:
Senior Survey/Program Reflection (See attachments)
Resource:
BSP Program Reflection Handout
Complete
the BSP Senior Survey through the link located in the BSP Program Reflection Handout. Provide your confirmation of survey completion to your instructor by providing a "Screen Shot". Please reference the "Screenshot Instructions" document for directions for capturing a screenshot.
Write
a 150- to 200-word reflection, following the instructions listed in the BSP Program Reflection Handout.
3: Participation (8 times on 3 different days)
.
1) Body of paper must be 4-5 pages 12 pitch Times Roman Double.docxcurranalmeta
1) Body of paper must be 4-5 pages | 12 pitch | Times Roman | Double Spaced Cover Page (not included in body) Reference Page (not included in body) Use APA style
2) Powerpoint Presentation must be between 5 to 10 minutes. You will lose points for going over the time limit. You must meet the minimum time limits also.
I need everything completed by 1pm tomorrow 2/24/2015. Sorry for late notice!
.
1) CBS and the NYT conducted a national poll of 1048 randomly select.docxcurranalmeta
1) CBS and the NYT conducted a national poll of 1048 randomly selected 13 to 17-year olds and counted how many owned smart phones. The reporters were trying to decide whether more than half of teenagers own smart phones.
Random variable:
Hypothesis:
Tail of the test:
state the Type I and Type II errors in complete sentences
type 1 error:
type 2 error:
.
1 page each question! double spaced Times New Roman 12 font CITE.docxcurranalmeta
1 page each question! double spaced Times New Roman 12 font
CITED. REFERENCE PAGE
PLEASE USE THE ARTICLES I ATTACHED TO ANSWER QUESTIONS!!!!!!!!!!!!
FIRST TWO ARTICLES ARE FOR THE FIRST QUESTION
LAST ARTICLE IS FOR THE SECOND QUESTION
Please type out the question then answer it using the articles I provided! Each answer should be 1 page long double spaced cited!
.
1) Backers of a balanced-budget amendment to the Constitution might .docxcurranalmeta
1) Backers of a balanced-budget amendment to the Constitution might consider the following strategy:
A. seeking support for such an amendment through approval by at least 34 state conventions, as almost occurred in the 1980s.
B. mounting a federal court case and seeking appeals to the Supreme Court.
C. pursuing approval of an amendment in the House and Senate, to then be sent to state legislatures.
D. collecting signatures to mount the proposal as a national referendum.
2) Which of the following presidents is credited with creation of the New Federalism?
A. Bill Clinton
B. George W. Bush
C. Ronald Reagan
D. Jimmy Carter
3) Which of the following issues were NOT a major concern of the Anti-Federalists concerning the Constitution?
A. The new powers granted to the national government
B. Ensuring that individual rights and liberties would be protected
C. Preserving the powers of the states
D. Ensuring that state criminal law would continue to be enforced by state courts
4) The primary motivation of the Framers, according to Charles Beard, was to:
A. protect their property rights through creation of a strong national government.
B. provide opportunities for political advancement for the Framers.
C. strengthen states’ rights in order to allow more effective resolution of property disputes.
D. create an egalitarian system of property distribution.
6) A shift toward cooperative federalism was evidenced by President Obama’s support of the stimulus bill that included temporary __________ funding for __________ functions such as education and public safety.
A. federal; federal
B. state; state
C. state; federal
D. federal; state
7) James Bryce believed that the Constitution was a fine example of:
A. mass democracy at work.
B. development of an oligarchical system of government.
C. an incrementally improved system not far removed from the Magna Carta.
D. a governing system that encouraged the proliferation of political parties.
9) Why was the division of power among the three branches of government an insufficient means to balance their powers?
A. The legislature, as the only initiator of laws, could dominate the other two branches.
B. Congress has more members than the other two branches.
C. Supreme Court judges are impeachable.
D. The president could become a dictator.
10) Which of the following is a true statement about the Federalists and Anti-Federalists?
A. Both factions distrusted the power of state governments.
B. Federalists were pessimistic about human nature, and Anti-Federalists were more optimistic about human nature.
C. The Anti-Federalists preferred a nonelected judiciary and indirectly elected president.
D. Both factions did not fear foreign threats.
11) President Bush and Governor Blanco struggled to agree about how to respond to Hurricane Katrina due to:
A. political differences and perceptions of incompetence by both.
B. President Bush never having been in a hurricane before.
C. Governor Blanco being a woman.
D. Mayor Nagin’s l.
Trust in Team DynamicsTrust is a key aspect in team dynamicurranalmeta
Trust in Team Dynamics
Trust is a key aspect in team dynamics. It is one of the basic emotions that brings teams together, and unfortunately it can also destroy a team if trust is lost.
For the first paragraph of your post, address the following bullet point (and sub-bullet points):
Discuss a team that you have been a part of where something has happened and the trust was lost. If you have not experienced this, talk to someone who has and use their situation instead.
Why did the loss of trust occur?
Were all members involved or just select members?
What did leaders do to try to fix the situation?
Was the team able to recover?
What would you have done differently if you were the leader in that situation?
For the second paragraph of your post, choose one of the following bullet points. Where possible, select a different bullet point than others have been selecting (i.e., we should try to spread the bullet points out so that each is getting equal attention during the discussion).
What are the key processes leaders or organizations consciously use to create trust?
Since most teams do not have conscious designs for creating trust among team members, design a strategy for incorporating trust among them.
How do group and individual emotions affect the trust-building, trust-maintaining, and trust-losing processes?
What is the role of the socialization within an organization for enhancing trust?
The final paragraph of your post (3 to 4 sentences) should summarize the one or two main points that you are attempting to make in your post.
500 words
...
Turnitin Plagiarism checker enabled.What is the definition of curranalmeta
Turnitin Plagiarism checker enabled.
What is the definition of information security? What essential protections must be in place to protect information systems from danger?
Define the InfoSec processes of identification, authentication, authorization, and accountability.
Define project management. Why is project management of particular interest in the field of information security?
What are the five basic outcomes that should be achieved through information security governance?
What is a threat in the context of information security? How many categories of threats exist as presented in this chapter?
...
Try to go to a museum that you have not been to before if your asscurranalmeta
This document provides background information and instructions for writing an essay analyzing a work of art visited at a museum. It recommends visiting one of several museums in Fort Worth or Dallas, Texas, and selecting a piece of art to evaluate. The document outlines the sections and questions to address in a three-page essay, including an introduction with the title, artist, and first impressions of the work, followed by body paragraphs analyzing elements of design and the artist's choices. It concludes with interpreting the theme and meaning of the work and reflecting on how the viewer's understanding has changed.
This presentation was provided by Racquel Jemison, Ph.D., Christina MacLaughlin, Ph.D., and Paulomi Majumder. Ph.D., all of the American Chemical Society, for the second session of NISO's 2024 Training Series "DEIA in the Scholarly Landscape." Session Two: 'Expanding Pathways to Publishing Careers,' was held June 13, 2024.
Level 3 NCEA - NZ: A Nation In the Making 1872 - 1900 SML.pptHenry Hollis
The History of NZ 1870-1900.
Making of a Nation.
From the NZ Wars to Liberals,
Richard Seddon, George Grey,
Social Laboratory, New Zealand,
Confiscations, Kotahitanga, Kingitanga, Parliament, Suffrage, Repudiation, Economic Change, Agriculture, Gold Mining, Timber, Flax, Sheep, Dairying,
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptxEduSkills OECD
Iván Bornacelly, Policy Analyst at the OECD Centre for Skills, OECD, presents at the webinar 'Tackling job market gaps with a skills-first approach' on 12 June 2024
How Barcodes Can Be Leveraged Within Odoo 17Celine George
In this presentation, we will explore how barcodes can be leveraged within Odoo 17 to streamline our manufacturing processes. We will cover the configuration steps, how to utilize barcodes in different manufacturing scenarios, and the overall benefits of implementing this technology.
This document provides an overview of wound healing, its functions, stages, mechanisms, factors affecting it, and complications.
A wound is a break in the integrity of the skin or tissues, which may be associated with disruption of the structure and function.
Healing is the body’s response to injury in an attempt to restore normal structure and functions.
Healing can occur in two ways: Regeneration and Repair
There are 4 phases of wound healing: hemostasis, inflammation, proliferation, and remodeling. This document also describes the mechanism of wound healing. Factors that affect healing include infection, uncontrolled diabetes, poor nutrition, age, anemia, the presence of foreign bodies, etc.
Complications of wound healing like infection, hyperpigmentation of scar, contractures, and keloid formation.
Trochim, W. M. K. (2006). Internal validity.httpwww.socialres
1. Trochim, W. M. K. (2006). Internal validity.
http://www.socialresearchmethods.net/kb/intval.php
Please follow link:^^^^^
Social Work Research: Chi Square
Molly, an administrator with a regional organization that
advocates for alternatives to long-term prison sentences for
nonviolent offenders, asked a team of researchers to conduct an
outcome evaluation of a new vocational rehabilitation program
for recently paroled prison inmates. The primary goal of the
program is to promote full-time employment among its
participants.
To evaluate the program, the evaluators decided to use a quasi-
experimental research design. The program enrolled 30
individuals to participate in the new program. Additionally,
there was a waiting list of 30 other participants who planned to
enroll after the first group completed the program. After the
first group of 30 participants completed the vocational program
(the “intervention” group), the researchers compared those
participants’ levels of employment with the 30 on the waiting
list (the “comparison” group).
In order to collect data on employment levels, the probation
officers for each of the 60 people in the sample (those in both
the intervention and comparison groups) completed a short
survey on the status of each client in the sample. The survey
contained demographic questions that included an item that
inquired about the employment level of the client. This was
measured through variables identified as none, part-time, or
full-time. A hard copy of the survey was mailed to each
probation officer and a stamped, self-addressed envelope was
provided for return of the survey to the researchers.
After the surveys were returned, the researchers entered the data
into an SPSS program for statistical analysis. Because both the
independent variable (participation in the vocational
2. rehabilitation program) and dependent variable (employment
outcome) used nominal/categorical measurement, the bivariate
statistic selected to compare the outcome of the two groups was
the Pearson chi-square.
After all of the information was entered into the SPSS program,
the following output charts were generated:
TABLE 1. CASE PROCESSING SUMMARY
Cases
Valid
Missing
Total
N
Percent
N
Percent
N
Percent
Program
Participation
*Employment
59
98.3%
1
1.7%
60
100.0%
TABLE 2. PROGRAM PARTICIPATION *EMPLOYMENT
CROSS TABULATION
Employment
Total
None
4. 40.7%
59
100.0%
TABLE 3. CHI-SQUARE TESTS
Value
df
Asymp. Sig. (2-sided)
Pearson Chi-Square
11.748a
2
.003
Likelihood Ratio
12.321
2
.002
Linear-by-Linear Association
11.548
1
.001
N of Valid Cases
59
a. 0 cells (.0%) have expected count less than 5. The minimum
expected count is 6.88.
The first table, titled Case Processing Summary, provided the
sample size (N = 59). Information for one of the 60 participants
was not available, while the information was collected for all of
the other 59 participants.
The second table, Program Participation Employment Cross
Tabulation, provided the frequency table, which showed that
among participants in the intervention group, 18 or 60% were
found to be employed full time, while 7 or 23% were found to
be employed part time, and 5 or 17% were unemployed. The
corresponding numbers for the comparison group (parolees who
5. had not yet enrolled in the program but were on the waiting list
for admission) showed that only 6 or 21% were employed full-
time, while 7 or 24% were employed part time, and 16 or 55%
were unemployed.
The third table, which provided the outcome of the Pearson chi-
square test, found that the difference between the intervention
and comparison groups were highly significant, with a p value
of .003, which is significantly beyond the usual alpha-level of
.05 that most researchers use to establish significance.
These results indicate that the vocational rehabilitation
intervention program may be effective at promoting full-time
employment among recently paroled inmates. However, there
are multiple limitations to this study, including that 1) no
random assignment was used, and 2) it is possible that
differences between the groups were due to preexisting
differences among the participants (such as selection bias).
Potential future studies could include a matched comparison
group or, if possible, a control group. In addition, future studies
should assess not only whether or not a recently paroled
individual obtains employment but also the degree to which he
or she is able to maintain employment, earn a living wage, and
satisfy other conditions of probation.
(Plummer 63-65)
Plummer, Sara-Beth, Sara Makris, Sally Brocksen. Social Work
Case Studies: Concentration Year. Laureate Publishing,
10/21/13. VitalBook file.
The citation provided is a guideline. Please check each citation
for accuracy before use.
Statistics for Social
Workers
J. Timothy Stocks
6. tatrstrrsrefers to a branch ot mathematics dealing '"'th the direct
de<erip-
tion of sample or population characteristics and the an.ll)'5i• of
popula·
lion characteri>tics b)' inference from samples. It co•·ers J wide
range of
content, including th~ collection, organization, and
interpretJtion of
data. It is divided into two broad categoric>: de;cnptive
>lathrics and
inferential >lJt ost ics.
Descriptive statistics involves the CQnlputation of statistics or
pnr.1meters to describe a
sample' or a popu lation _~ All t he data arc available and used
in <.omputntlon o f t hese
aggregate characteristics. T his may involve reports of central
tendency or v.~r i al>il i ty of
single variables (univariate statistics). ll also may involve
enumeration of the I'Ciation-
sh ips between or among two or moo·e variables' (bivariate or
multivariJte stot istics}.
Descriptiw statistics arc used 10 provide information about a
large m.b> of data in a form
that ma)' be easily understood. The defining characteristic of
descriptive ;tJtistks b that
the product is a report, not .on inference.
Inferential statisti<> imolvc' the construction of a probable
description of the charac·
teristics of a population b•sed on s.unple data. We compute
statistics from .1 pJrtial;et of
the population data (a samplt) to estimate the population
parameters. Thrse t<timates
are not exact, but ·e can mo~k..: reawnable judgments as w
7. hoV preruc our c~lim:ues are.
Included within inferential statiwcs i;, hypothesis testing, a
procedure for U>ing mathe-
m:uics tO provide evidence for the exi<tence of relationships
between o r among variable;.
T bis testing is a form of inferential •"l~umem.
Descriptive Statistics
Measures of Central Tendency
Measures of central tenden')' are individual numbers that typify
the tot.tl set of ~cores.
The three most frequently used mca>urcs of centraltendenq are
the arithmetic mean, the
mode, and the median.
Arir!Jmeric .1ea11. The arithmetic mean usually is simply
called the mca11. It also is called
the m-erage. It is computed b)' adding up all of a set of scores
and dwidmg by the number
of scores in the set. The algebraic representation of this is
75
76 PA11 f I • OuANTifAllVi AffkOAGHU: fouHo~;noM Of
Ot.r"' CO ltf(TIO'J
~, =l:: X ,
11
where 11 represents the popu I at ion mean, X represems an
individual score, and rr is t he
number of scores being adde(l.
8. The formula for the sample mean is the same except t hat the
mean is represented by
the variable lener with a bar above it:
- l:;X X= --.
II
Following are t he numbers of class periods skipped by 20
seventh-graders d uring
I week: {1, 6,2,6, 15,2(),3,20, 17, 11, 15, 18,8,3, 17, 16, 14,
17,0, 101. Wecomputethe
mean by adding up the class periods missed and dh•iding by 20:
l:;X 219 •
J.l = -- = - = 10.9o.
II 20
Mode. The mode is the most frequently appearing score. It
really is not so much a measure
of centrality as it is a measure of typicalness. It is found by o
rganizing scores int o a fre-
quency distribution and determining which score has t he
greatest fre-
TABLE 6 . 1 Truancy Scores
quency. Table 6. 1 displays the truancy scores arranged in a
frequency
distribution.
Score
20
19
10. 0
0
l
I
0
1
0
2
0
0
2
0
Because 17 is the most frequently appearing number, the mode
(or
modal number) of class periods skipped is 17.
Unlike the mean or median, a distribution o f scores can have
more
than one mode.
,llfedinrr. lf we take all the scores in a set of scores, place t hem
in o rder
from least to greatest, and count in to the middle, then the score
in the
middle is the median. This is easy enough if there is an odd
number of
scores. However, if there is an even number of scores, then
there is no
single score in the middle. In this case, t he two middle scores
are
selected, and their average is the median.
11. There a.re 20 scores in the previous example. The median would
be
the a"erage of the lOth and lith scores. We usc t he frequency
table to
find these scores, which are 14 and J 5. T hus, the median is
14.5.
Measures of Variabi li ty
Whereas measures of central tendency are used to estimate a
typical
score in a dimibution, measures of variability may be thought of
ns a
way in which to measure departu re from typic<~lness. They
pro"ide
information on how "spread out" scores in a d istribution are.
J<auge. The range is the easiest measure of variability to
calculate. It is
simply the distance from the minimum ( lowest) score in a
distribution
If
10
R
:.aJ
13
de
c .. ...nu 6 • STAnsnu t<~~ Soc&AL Wouta~ 77
12. to the maximum ( highest) score. h is obtained by subtracting
the 111ini murn score flom
lhe maximum ~cor~.
Let us compute th.- rang.- for the following dJt.l ~ct:
/1, 6, 10, 14, 18,22/.
'T'he n1inimum i!) 2, and tht." tnJximum is 22:
Range = 22 - 2 20.
Sum ofSquaus. The sum of squares is a measure of the total
amount of variability in" set
of scores. Jts na me tells how to wmpute it. Smu ofsqunres is
short (or sum ofsqumed dc1ti
til ion scores. It is represented by the S)'lnbol SS.
The formulas for sample and population sums ot squares are the
same except for sam-
ple and populat•on mean symbob:
SS = I(X ~tl'
Using the dJtJ set fo r t11e range, the sum of squnres would be
computed as in
'ldble6.2.
V.~rinuce. Another name for variance i~ mean square. This is
short for mean of squared
devintron score<. 1l1is is obtained by dividi ng the sum of
squares by the number of scores
(11). It is a me,tsure of the average amount of variabilit y
associated with each score in a set
of scores. The population variance fOI'mu la is
13. ss
a2= -.
n
whc1e cr2 is the syn>bol for populn tion variance, SS is the
symbol fo r sum of squares, and
11 st,uJds for th e number of scores in the population.
The variance for the example we used to compute
sum of squares would be
TAOLE 6.2 Computing the Sum of Squares
X X m
2 tO
6 6
10 ]
l<t 12
18 >6
12 10
NOTE, !X~ 72; n- 6; ~ • 12; l:(X - p)' ~ 780
(X - m)'
100
36
14. 4
4
36
100
2 280
(J --= 46.67.
6
The sample variJnce is not an unbi.as.ed estin1a1o1
of thf population variance. If we compute the vari
anccs for these samples using the SS/11 formula, then
the- san1ple vadn nccs wil1 average o ut smaller than
the population val'iance. For th is rc:~son, the sample
variance is computed differently froru the population
variance:
ss
sl = - - .
II - I
CHA,Ut 6 • Sr"n~nn HJa SOCIAl wouus 77
to the maximum (highc;t) score. h is obtained by subtracting the
minimum scoo·c from
the maximum score.
let us compute the rnnge for the following data set:
15. 12. 6, 10, 14, 18.221 .
The minimum is 2. and the maximum is 22:
Range 22-2 = 20.
Sum of8qo~t~res. The ,um of squares;, a measure of the total
amoun t o f variability in a set
of score~. It> name tells how to compute it. Sum of 51Jo.arcs is
short for ;um of squared dco•i-
atiou scores. It is reprewnt<>tl by the symlxll SS.
The formulas for <.omple and popul.llion sums of squares are
the ~arne except tor S<J m -
p le a nd population mean sym bols:
ss l.(X -X)'
Usi ng the data set for the range, t he su m of squares would be
computed ns i n
T.,b)e 6.2.
~rta11u. Another name for variance is mean square. This is
short for menn of 51JIUtred
devontw11 scores. This os obtained by dividing the sum of
squares by the number of ><.ores
(n). It is a measure of t he averoge ••m ount of var iability
associated w ith each score in a set
of scores. T he popula tio n variance for m11ln is
ss
¢ =- .
n
where o ' is th e symbol foo· population v•o·ia.nc.e, SS is t he
16. symbol fo o· Slim o f squares. a11d
11 stands for the numbet of scores in the population.
The •-..ria nee for the example we used to compute
sum of squar~s would be
TABu 6.2 Computing the Sum of Squares
X X-m
2 - 10
6 -6
10 -2
14 +2
18 +6
22 +10
HOT£: r.x- 72: n; ti; p = 12: l:lX Ill'= 250.
(X- m)'
100
j(,
4
4
J&
17. tOO
280
cr2 =
6
~ 46.67.
The snmple variance is uot Jn Ulbiased estimalor
o f' t he population variance. Jf we com pute t he vari-
ances for these samples using th" SShr formu la, then
the sample variances will average out smaller than
thc population ••ariance. For this reJson, the sample
Vllriance is computed differe ntly from the population
variance:
ss r =-.
n - J
78 PAll I • QuAiuu.ot.nvt A"MACH(S.:. FouHDAIIOif"i Of
O.AIA CoLLfcnow
The n - 1 i> a correction fac tor for this tendency to
undcre>tima te. I t is c.1 lled
degree• of freedon1. If <lur example we1< a sample. then the
,,ariance would be
.1 280
> =--
6 - 1
280 6
5 = 5.
18. Sumdard Deviatron. Although the variance is a measure of
average variability associJtc'<l
wllh each score, it i> on a d ifferent sc.lle from the score itself.
Tlw variance measures avel·
age squared deviation from the mean. To get " me<tstne of
averdgc variabili ty on the ;a rue
scale as the original scores, we ta ke the squa 1·c rc)Ot of the
varia nee. The st<tndard deviation
is the square root of the variance. The fo rmula< are
Using the same .ct of numbers as before, the population
standard deviation would be
cr -/46.67 = 6.83 .
and the sample st.mdard deviation would be
s J56 = 7.'18.
For a normally d istribured set of scores, n ppwximately 68% of
all ;cores will be within
ll •tanrlard deviation of 1 he mean.
Measures of Relationship
T.1ble 6.3 shows the relat iortship between number of >treSsors
experien<ed by a parent during
.1 week and that parent's frequency of U>C of corporal
punishment during the same wee.k.
One can use ,·eg,·cssion procedures to dcrivr the line that best
fo ts the data. This line is
rcfel'l'ed to as a regression line (or line of best ii 1 o r
prediction I inc). Su ch a line bas been
.CJiculated for the example plot. It has a Y ime,·cept of - 3.555
t11id a slope of + 1.279. T his
gives us the prediction equation of
19. Y,_. = 3.555 t 1.279X,
where Yis fi-equ ency o f <Orporal p unishment and X is
stresso1 ~. This is graphically pre
dieted in Figure 6 . 1.
Slope is the ch•ngc in Y for a unit increase in X. So, the slope
of 11.279 meam that''"
increase in stres.ors (X) of 1 will be accomp.ulicd by an
increase in predicted frequency of
~orporal punishment (I') of + 1.279 incidents per week. If the
slope were a negati'e
number, then an increase in X would be accompanied by a pred
ictcd decrease in Y.
The equation does not give the actual value of Y (called the
obt.tined or obserwd
score); rather, it giv~s a prediction of the value of Y for a
certain value of X. Fo r
-
Cu,"na 6 • SrAliSnc<o 10~ So- '"' WOhi•C. 79
r iQUIO 6.1 8
Frequency ol Stre<sors
and Use of Co•poral 7
0
Punishment
~
20. 6 0
c . Y P'td; - 3.555 + 1.279X ..
" 5 0 r:r
e ...
c 4 ..
E
.r:
3 til
·;:
" Q.
2 0
0
0
0 1 2 3 4 5 6 7 8 9
Stressors
example, if X were 3 , rhen we would predi<.t t hal Y would be
- 3.555 + 1.279(3) ~ - 3.555
+ 3.837 ~ 0.282.
Tuu 6 . 3 frequency of
Sttessors and Use of
Corporal Punishment
Sue-ssors Pun1.shm~nt
3 0
4
21. 4 }
s 3
6 4
7 ~
8 6
7
q 8
1() 9
T he regression li ne is the line that predicts Y >UCh t hat t he
error
of p redictio n is minim ized. Error is d efined as the d ifference
between the predicted score and the obtaine<l score. The
equation
for compu ting error is
E= Y Y..,.. ..
~1en X= 4, there arc two obL1ined ''alues of Y: I and 2. The
p redicted value of Y is
Y,,...t = - 3.555 I 1.279( 4) = - 3.555 + S. l l6 ~ 1.56 1.
rhe error of prediction i~ E =I - 1.561 = -0.561 fu r Y = I, and
E - 2 - 1.561 = +0.<139fnr Y=2 .
If we square each error difference score and sum the squares.
then we get a quantity called the enor sum of sq.ure;., which i;.
r~presented b)•
22. SSI: L( Y - Y,..,.,)'.
T he regressi011 line io !he o ne line that give> the sm.11lcst va
lue
fo r SSt.
80 P~oar 1 • QUAtHnAnvE A ,ROACHES: FouNOAHO~r~~$ of
DAtA Conte I!Otf
The SSE is a measure of the lOla I variability of obtained score
values around their pre-
dicted values. There are two other ;un" of squares !hat are
important to undcr>tanding
correlation and regri'SSion.
The total sum of squ.m:s (SS1) i$ a measure of the total
variabilit)' of the obtained
score values around the mean oft he obtained scores. The SST is
represented by
SST = L(Y-Y)'.
The remaining sum of squa 1·cs is coiled the regression sum of
S<Ju:u·cs (SSR) o r the
explained sum of squares. If we squnre each of the differences
between prcdie1 cd scores
and t he mean and then add t hem u p, w·c get the SSR, which is
represented by
SSR L( v, .... - Y)'.
The SSR is a measure of the tot.d variabil ity of the predicted
score values around the
mean of the obtained scores.
23. An important and interesting feature of the>e three sums of
squares is that the sum of
the SSR and SSE is equal to the SS1:
SST SSR- SSE.
This leads us to three o ther imponnnt stat istics: t he proportion
of variance explJined
(I'VE) , the correlation coefficient, ond the standard error of
estim ate.
Proportion of Iarin nee Expluir~ctl. T ht I'VE is a measure of
how good Lhc rcs,·cssion line
p red icts obtained scores. The values of PV£ 1·ange fro m 0 (
no p red ictive value) to I ( pre-
diction with perfect accurJLy). The cqunt ion fo r PV£ is
SSR
J>vE - - ·
SST
There also is a computational equation for the PVE. which is
where
PVE - ( SSXY )'
SSX • SSY'
SSXY is the "co variance" ~um of ;qua res: l.(X - X)( Y - Y ),
SSX is t he sum of squares for vn rinble X: IlX - XJ', and
SSYis the sum of squares for varinblc Y: 2:( Y - Y)'.
The procedure fo r computing these sums of squares is outlined
in Table 6.4.
24. The proportion of v.triance in the freque ncy of corporal
punishment thnl may be
explained by stressors experienced ;,
( 4 6L5)1 3782.25
l'VE = - = = 0 .953.
(48.1)(825) 3968.25
TABLE 6.4 Computation of r2 (PVE)
y Y - y (Y- Y)' X X x (X - X)' (X X)( Y Y)
3 -33 10 .89 0 -4 5 20 .2 5 +1405
4 -2 3 5.29 -lS 12 .25 +80S
4 -23 529 2 -15 6 .25 < 5.75
5 - Ll 1.69 3 1.5 2.25 • 1.95
6 -ol 0 .09 < -o5 0.25 0 IS
7 +0./ 0.49 5 ·10.5 0.25 035
8 + II 2.89 6 ; 1.5 2 .25 • 2.55
7 TO.! 0.49 7 12.5 6 .25 11.75
9 +27 7.29 R t3.5 12.25 -19.45
10 +3 I 13 69 9 "'5 20.25 16.65
NOTE: Y - 6.3; SSY - 48. l; X = 4.5; S5X = 82.5; S5XY • •6 l S
25. The PVEsometimes is en lied th~ coefticient of determination
and is represented by the
symbol r'.
Correlation Co~ffirirm. A correlation coellicient also is a
111easure of th e strength of rela-
tionship between two variables. The correlation cocfficicnt is
represented by the letter r
and can take on values between - 1 and + I inclu~ivc. The
correlation coefficient always has
the same sign a.< the slope. If one squares a correlation
coefficient, then <me will obtain the
PV£ It is computed using the following formula:
SSXY
r = -vr.;;S50sx""•""S;;;S;;o;Y
For our examph: data, the correlation coefficient would be
+61.5 ~ 61.5 +61.5
R --- = = = -0.976 .
./(18.1)(82.5) ¥'3968.25 62.994
Standard Error of Em mate. The standard error of estimate is the
<tandard deviation of the
prediction errors. It i< computed like any other standard
deviation: the: square root of the
SSE divided by the dcRn:es of freedom.
The fi rst s tep is to compute the variance error (s:.J:
..1
'E
26. SSE
n-2
Notice that the value for degrees of freedom is 11 2 rather than
11 - l. The reason why
we subtract 2 in this instance is that variance error (and
standard Cfi'Or of c:stimatc) is a
statistic describing characteristics of two variables. T hey deal
with the error involved in
the prediction of Y (one variable) from X {the other v.triable) .
'l he standard error of estimate is the square root of the variance
error:
Sf.= ...j(ij.
The standard error of estimate tells us hOv spread out scores
are with respect to their
predicted values. If the error· scores ( E = Y- Y,.o~> are
normally distributed around the
prediction line, then about 68% of actual scores will foil
between ±I :;,; of their predicted
values.
We can calculate the standard error of estimate using the
foUowing computing formula:
( n-1) ( I -- r 2)(-------) , u-2
where
s,. is the standard deviation of Y,
r is the correlation coefficient fo r X and Y, and
n is tl1e sample si7.c.
27. for the example dat..1, this would be
S£ = 2.3lli ((J -- .953) :~ = D = 2.311 ((0.47)~)
= 2.311J0.053 = (0.230)(0.727) = 0 .167.
Inferential Statistics: Hypothesis Testing
The Null and Alternative Hypotheses
Classical ;tatistical hypothesis testing is based on the evaluation
of two rival hypothescs:
the null hypothesis and the alrermltive hypothesis.
We try to dete<:t relationsh ips by identifying changes that are
unl ikely to have occurred
simp!)• bccau~e of random fluctuat ions <If dependent
measures. Statistical analysis is the
usual procedure for identil)•ing ;uch relationsh•p>.
The null hypothesis is the hypotltcsis that there is no
relationship between two vari-
ables. This implies that if the null hypothesis is true, then any
apparent relationship in
Mmples i> the resuh of random flu ctuations in the dependent
meas ure or sampling error.
Statistical hypothesis tests arc carried out on samples. for
example, in nn experi-
ment!// two-gro11p posttcst-only design, there would be a
sample whose members
received an intervention and a sample whose members did not.
Both of these would be
probability samples from a larger population. The interven tion
>ample would reprcse>11
28. Figure 6.2
The Null Hypothesis
and Type I Error
C14Anu 6 • StAJtmu f<M' Socw. Wouus 83
the popula tion of all individuals as if they had received the
i.ntervt•ntion. Th e control
sample would be repre<entative of the <ame popuiJtion of
individuals as if the)· had
not recei>·ed the inten-emion.
lf the intervention had no effect, then th e populations would be
iden tical. However, it
would be unlikely that two samples from two ident ical popula
tions would he ident ical. So,
although the sample mea ns would be diffe rent, they would not
rcpre>CtH any effect of t he
independent variable. The apparent difference would be due to
sampling error.
Statistical hypothC$is tests invoh·e e'-aluating evidence from
.amples to make inler-
ences about populations. II is for this reason that the null
hypothe>i> is a statement about
population parameters. For example, o ne null hypothe>iS for I
he previous design cou ld be
stated as
or as
H, : ll = ~to = 0.
H, stands for the null hypothC$iS. It is J letter H with J " ro
29. subscript. It is a statement
t.ha t the m~ans of the experime ntal ( Mean I) and cont rol (
Mean 2) popultnio'ls arc eq ual.
To <:>tablish that a relat ionship exists between th e in
tervention (independent Vilfi:tble)
and the outcome (measure o f the dependent variable), we must
collect eviden<C that
allows us to reject the null h)'J>Othesis.
Strictly speaking, we do not mak~ J decision as to whether the
nul] hypoth eoi:. is
correct. Ve evaluate the evidence to determine the ext<·nL to
which it •cncls to confirn"' or
disconfi rm the null hypothesis. If the evide nce wct·e suc.h that
it is unlikely that an
observed relationship would have ocwrrcd as the re.ult of
sampling e r ror, then we would
reject the null hypothesis. If the eviden«: were more ambiguous,
then we would f.1il to
reject the null hypothesis. The terms re;err and fail to rrjm carry
the implicit under<tand-
ing tlMt our decision might be in ert'or. Th e truth i, th at we n
ever really know whethe r
our decbio11 is correct.
vVhen we reject the n ull hypothesh and it is true, we ltJve
committed a Type I error. By
setting certain statistic•! criteria beforehand, we can ~"tablish
the prombiliry that we "•ill
commit a 'JYpe l error. 'c decide what proportion of the time
we arc willing to commit a
Type l error. This proportion ( proba bility) is called a l1>ha
(o:). If we n1e willing to reject
the null hypothesis when it is true onl)• I in 20 times, thc11 we
set our a level at .05. If' on ly I
30. in 100 time>, then we set it at .0 I.
Tbe probability that we will fail to reje<t the null hy]>Othesis
when it is true (correct
deci;ion) ts 1 - a (Figure 6.2).
Situahon: NULL HYPOTH ESIS TRUE
Deas1on ACSlllt
Reject H, 1'ype I Error
ex • the probability or rejecting the Null Hypo thes is when it is
true
Fail to Reject H, Correct Decision
I a= the probability of not rejecttng the Nun Hypothesis wllcn
11 is true.
84 PAII t I • Qv.umr:.WI~ A PI'IOACHH: Fourwt. lt<m S OF 0
1.1A CotulCI!Oii
Figure G.:Y
The Nu ll Hypothesis
and u Level
The fol!pwing hypothesis would be evaluated by c<>mparing
the difference between
sample means:
If' we carried out multiple samples from populations with
identical. n>eans (the null
hypothesis was true), then we would find that most of the
vallles for the differences
31. between the sample means wou ld not be 0. Figure 6.3
represents a distribm ion of the dif·
fercn ces between sample means drawn from identical
populations.
The mean d ifference for the total distribution of samp le means
is 0, and the standard
deviation is 5. I f the differences are normally distributed, then
approximately 68% of
lhese differences will be between - 5 (z = - 1) and +5 (z= +l).
Fully 95% of the differences
in the distribution will fall between the range of -9.8 ( z =-1
.96} and +9.8 (z = +1 .96). If
we drew a random sa mple from each population, it '~ould not
be unusual to find a di ffer-
ence between sample means of as mnch as 9 .8, even though the
population means were
the same.
On the other hand, we would expect to fin d a difference more
than 9.8 about 1 in 20
times. If we set our criterion fo r rejecting the null hypothesis
such that a mean difference
must be greater than +9.8 or less than - 9.8, tben we would
commit a Type I error only 1
in 20 times (.OS) on average. O ur (J. level ( the probability of
committing a Type l error)
would be set at .05.
The probability that a relationship or a difference of a certain
size would be seen in a
sample if the nuU hypothesis were true is represented by p. To
reject the null hypothesis,
p mu~t be less than or equal to <X. The probability of getting
an effect this large or !~rger if
the null hypothesis were true is less than or equal to the
32. probability of making a Type l
error that we ha,•e decided is acceptable.
1 - u = .95
- 4 - 3 - 1 0 +1 +3 +4
z
- 20 - 15 - 10 - 5 0 +5 +10 +15 +20
X, -x2
a = .05
CH..,tU 6 • Sr.r.nsnu •o• SoctAt Wo~·~ui 85
Rejecting the H0: We believe that it i~ likely that the
relationship in the sample IS gcncr
alizablc to the population.
Not rejutmg the H,; We do not believe that we have >umcient
e1•idence to draw infer-
ences about the populat ion.
For the previous example, let us imagine that we ha-e set a=
.OS. Al;o, imagine thJt we
obtained a difference betwt-en the sample me.ms of 10. The
probability that we would
obtain a difference of +10 or - 10 would be equivalent to the
probability of a z ~core
g reater than +2.0 plus the probabilit y of a z ~core less th.111 -
2.0 o r .0228 + .0228 = .0156.
This is o ur p value; p = .0456. Because p <a, we would reject
the n ull hypothesis.
Some texts create the impression that the alternative (or
33. research or experimental)
hypothes~ b simply tbc opposite of the null hypothesis. In fact,
sometimes d1is nail·c
alternative h)pothesis is used. However, it generally is not
particularly useful to
researchers. Usually. we nrc inrertsted i n defecting an in
lcrvention effccl of a particu l :~r
size. On certnin measu,·c,, we would be interested in .mwll
effects (<:.g., death rate),
whereas on others, o nly l~rger effects would be of interest.
When we are inter<5ted in an effect of a particular •ize. we use
a specific altemnti1e
hypotbesil. that takes the following form:
H, : f.l 1 - ~,.,;:: id I,
where dis a difference of a particular size. If the test is a
nondirectional I<'St, then the dif-
ference in the alternative hypothesis would be expressed as an
absolute value, ldl, to ohnw
that either ,t positive or neg.tt tve differe~tct~ ;, involv~d.
lt is custo mary to exprc>S the mea11 d i ffere nce in an II , in
units of standard deviat ion.
Such scores are called zsco,·es. T he diffe(ence is called an
effect size. Effect sizes frequently
are used in meta-analyse> of outcome studies to compare the
relatic cllicacy of different
t )'Pes of intencntioos acrOS> 'tudies.
Cohen (1988) groups effect sizes into small, medium , and large
cntegorics. The criteda
for each arc al follows:
Small effect >iu (d ~ .2): It is appro:rimatcly the effect size for
34. the average difference in
height (i.e., 0.5 inches and < = 2.1) between 15- and 16 year-old
girls.
Medium effect size (d • .5): It is ap proximately the effect size
fo r t he average differc11ce
in heigh t ( i.e., 1.0 inches and s~ 2.0) bNwccn 14- aud 18·
year-old g ir ls.
Large cff<Xl size (d: .8): rh1s is the same eflect size (tl = .8) as
the avcrJge difference in
height for 13- and 18-year-old girls.
l ntuit iv<:ly. it would se..-m t hat we wo uld want to detect
even ve1y >mall effect si ~t·s in
our research. llo1Vever, t here is a practicdl trade-off involved.
All o ther things being equal.
the consistcllt detection of unaU effect >izc' requires very large
(1l > 200) sample size,,
Because 'cry large sample sizes require resources thdt might not
be readily available,
they might not be practical for all studies. Furthermore. there
are c~rtail1 outcome vari-
ables for which we would not be part icuia l'l y in terested in
small effec t>.
If we rejeCt t he null hypothesis, t hen we implicitly huvc
decided that t he evidence >Up-
ports the alternative hypothesis. If the alttrnative hypothc<is is
true and we reject t he null
hypothesis. then we have m3de a correct decision. However, if
we fail to reject the null
hypothesis and the alternati•e hypothesis is true, then we hJve
committC'd a Type II error.
A Type !I error involves the fa ilure to detect an existing effect
35. (Figure 6.4).
86 P1o11r I • Qt•MmTM •; e A ?PIOAC HtS: Fou NDAti ON)
o, 0.-.tA Contr'fiO'I
Figur• 6 .4
The Null Hypoth<sis
and Typo II Error
Decision
Reject 1io
Fail to Reject
H•
Siluation: ALTERNATIVE HYPOTHESIS TRUE
Result
Correct 0 edslon
1 -13 a t he
Alternative
probabinty of rejecling tho Null Hypothesis when the
Hypothesis is true. The power ot a test.
Type II E n· or
I}~ the p r
Altornatlvo
36. obability of not rejecling the Null Hypothesis w11e 11 the
Hypothesis is true.
Beta(~) is t he probdbility o f committing a Type rr error. This
probability is eStdblished
when we set our criterion for rejecting the null hypothesis. The
probdbility of a correct
decision (I - f3) is an importdnt probability. It is so important
that it has a nJmc~power.
Power refers to the probability t h.u "e will detect an eff«t of
the size we have sckctcd.
We should decide on the power (I - (3) as well as the a level
before we carry out a sta-
tistical test. just as with Type 1 error, we should decide
beforehand how often we are will-
ing to make a Type 11 error (fail to detect a certain effect size).
This is our f3 level. The
procedure for making such determinat ions is discussed in
Cohen ( 1988).
Assumptio ns for Statisti cal Hypothesis Tests
Although assumptions arc diffc •·cm leu different tests, all tests
of the uull hypo1 hcsis shn re
two related assumptions: randomness nud independence.
T he randomness assum ption is t hnt sample members m ust be
randomly selected from
the populatio n being evaluate d. If the sample is being divided
into groups (e.g., trc:>tment
and control), then assignment to gro ups al.<e> must be random.
This is referred to as mn-
rlom selection and random fWigmnem.
The mathematical models that underlie statistical hypothesis
37. testing depend on ran-
dom sampling. If the samples Jre not random. then •<e cannot
compute .111 accurate prob·
ability (p) that the sample could have resulted if the null
hypothesi~ were true.
The independence asswnption t. that one member's score •<ill
not innucncc another
member's score. The only common re!Jtionship among group
scores should be the inter-
vention. One implication of this is t hat members of a group
should not have any contact
with each other so as nut to a !Teet each o ther's scores.
Again, the mathematical models are dependent on the
independence of sample scores.
l f t he scores are not independent, t hen the probability (p) is,
as before. >i mply n number
t h•t has little to do with the p ro babilit)' of a Type I erro r.
Parametric and Nonpara metric Hypothesis Tests
Traditionally. hypothesis tests arc g rouped into parametric and
nonp.trJntCt ric tests. T he
names are misleading given th at one class of test has no more
or less to do with popula-
tion parameters than t he other. T he difference between t he
two tests lies in the mathe
matical assumptions used to compute the likelihood of a Type I
error.
Parametric tests are based on the assumption that t he
populations from whkh the
samples are drown are norm.•lly di~t rihuted. Non parametric
tests do not have this rigid
38. C HAJ>TEJI 6 • STATI 11(~ 1011: SOCIAl WO !U({IS 87
assumption. T hus, a non parametric test can be carr ied out on a
broader range of data
than can a parametric test. Nonparametric lests remain
serviceable even in circumstances
where parametric procedures collapse.
When the populations from which we sample are nor mally
distributed , and when all
the other assumptions of t he parametric test are met, parametric
test~ are slightly more
powerful than non parametr ic tests. However, when the
parametr ic assu mptions are not
met, nonparametric tests are more powerful.
Specific Hypothesis Tests
•Ve now investigate several frequently used hypothesis te.m
and issues surrounding their
appropria te use. Where appropriate, parametric and
nonparametric tes ts are presented
together for ead1 type of design.
Single-Sample Hypothesis Tests
These are tests i n which a single sample is drawn. Comparisons
are made between sample
values and population parameters to see whether the sample
differs in a statistically sig-
nificant way fro m the parent populnt.ion. Occasionally, these
tests are used to determine
~<hether a sample differs from some theoretical population.
39. For example, we might wish to gather evidence as to whether a
particular population
was normally distributed. We would take a randon1 sample from
this population and com·
pare the <l istribution of scores to an artificially constructed,
normally d istr ibuted set of
scores. If there were a statistically significam difference, tben
we would reject the hypothe-
sis tlwt our sample came from~ normally distributed population
(the null hypothesis}.
Typicrully, these tests are not used for experiments. T hey tend
to be used to demonstrate
that certain strata within populations differ from t he population
as a whole.
Here, we investigate two single-sample test~:
L Single-sample rtest (interval or ratio scale)
2. x' (chi-square) goodness of fit test (nominal scale)
TIJe Single-Srmrple t Test. This rest usually is used to sec
whether a strotum of a population
is different on average from the population as a whole (e.g., are
the mean wages received
by social workers in Lansing different from the mean for aU
social workers in M ichigaJJ?) .
The null hypothesis for t his test is t hat the mean wages fo r a
particular strntum
(l ansing social workers) of the population and the population as
a whole ( Michigan
social wor kers) will be the same:
where !lo is the mean wage fo r the population and ~t 1 is the
40. mean wage fo r t he stratum.
The assumptions of the single-sample t test are as follows:
Randomness: Sample members must be randomly drawn from
the pop ulation.
fndeptmdence: Sa mple (X) scores rnust be independent of each
other.
Sct1liug:The dependent m~sure (X scores) must be interval or
ratio.
Norma l distribr<tion:The population of X scores must be nor
mally di&tributed.
88 PAIIT I • QUANnrAnVf At-nOA.t-H£s: Fo u iOAnotn o•
OA t A Cou.£CIION
These asswnprioos are li<ted more or lc:.s in order of
in1portance. VioiJtions of the frrsr
three assumptions are es>entiJIIy "f•tal" ones. E'·en slight
violations of the lir..t two
assumptions can introduce major error into the compmation of p
value~.
Violation of the assumption of,, normal distribution will
introduce >Ome error into
the computation of p vJiues. Unless the population distribution
is markedly different
fro m a normal distribution, rhe erro" will tend to be slight
(e.g., a re ported p v.tlue of.042
Jctu ally will be a p value of .057). This is what is meant whe n
some-one snys t ha t the t test
is a <•robust" test.
41. T he tstatistic fo•· t he sing le sample t te;t is computed by
subtr:ocr ing t he null hypotbe-
• is (popula tion) mean from t h e s"mple mean and dividing by
th e sta ndard error of th e
n1ean.
T he fo rmu la for r...,, (pronOlii1Ced "t obr•ined") is
As the absolute value of '·• get> larger, tht> more unlikely it is
that such a difference
could occur if the null hypothc>sis is true. At a certain point,
tht' probabilit)' (p) of obtam-
ing a t so large becomes sufficiently small (rt'acbt'S the a.
level) that we rcjt'<t the null
hypotbt'Sis.
T he critical value oft (the v.d ue t hat too. must equal or exceed
to reject the null hypoth-
esis) depends o n the degrees of freedom. For a single-sample
rtest,the degree> of freedom
ure df= n - I , whe re" is the s.omp k >itt'.
Let us look at how to compute '"k
v.re know from a statewide SUI'VC)' I hat the average time
taken to complete an outpa-
tient rehabilitation p rogram r-or .o certain injury, X, is 46.6 d
ays. We w ish to see whethe r
clients seen at o u r clinic nrc taking longer o r ;horter than the
state average.
We randomly sa mple 16 fil e< from the pa>t year, We review
these c.1>cS anu dete•mine
the length of program for each of the clients in the sample. The
mean n umber of days to
42. complete rehabilitation a t our clinic is 19.875 days. This is
lower than the populat ion
mean of 46.6 days. The question is whether this result is
statistically significant. I> itlikel)'
that this sample could ha,·e been drawn from a population with
a mean of 46.6?
To determine thi>, we ne..'<lto calculate r.,... The first step in
calculating t,_,. was arriro out
when we computro the sample mean. Tite next step is to
compute the standard error of the
lllt'aO. We begin this by <umpu ung the standard deviation,
which t urns our to be s 11.888.
Th e standard erro r of the lliCJn i> calculated by d ividing the
standard deviation by t he
square root of the sample size or
s;
_s_ = l 1.888 = l 1.888 =
2
_
9
72.
/ii Jl6 4
We take th e fo rmu la for t,,..., Joel p lug in our n umbers 10
obLain
29.875- 46.6
2.972
43. -1 6.725 8
2.972 - 5.62
We look up the tabled t val u e {I., ) at 15 degrees offreroom.
This turns out to be 2. 131
for a nondirectional test at (X .05 (sec • t•ble of the critical
values for the ttt»t, non<li-
rectional, found in most ,tatistie> texts). The absolute , .. Jue of
r.,.. = 5.628. This is greater
than t"" = 2.131, so we reject the n ull hypothesis. The e-.-
idencc suggests thot clicnls in o ur
clinic average fewer days in rehabilitation thon is t he case in
the statewide population.
T he effect size index for a test o f means is d and is computed
as follows fo r a single-
sample t test:
d = ~~o .
s
The effect size for our example would be as follows:
d = 29.875 - 46.6
11.888
which would be classifie d as a large effect.
-16.725
11.888 = 1.4069'
1he x' Cootfne;s-of· Fit Test . Th e.%' goodness- of-fit test is a
single·sam pic test. lL is used in
44. t he evaluation of 11ominal (categorical) variables. The test
involves comparisons between
observed and expected frequencies wi thin strata in a sample.
Expected freq uencies are
derived from either population v-alues or t heoretical values.
Observed frequencie-s are
those derived from the sample.
T he null hypothesis for !he x' test is that the population from
which the s.1mple has
been drawn will have !he same proportion of members in each
category as the empirical
or theoretical null hypothesis population:
where
P., is the proportion o r case~ •.vitbin category kin the null
hypothesis population
(expected), and
P01 is the proportion of cases within category k in the
population from which the test
sample was drawn (observed).
The assumptio n> fo r thet' goodness-of fit test arc as follows:
• Randomness: Sample members m ust be randornly drawn from
the populnt i<)ll.
• Independence: Snmplc scores m ust be independent of each
other. O ne im plication of
this is that categories must be mut ually cxclu;ive (no case may
appear in more than
one category).
• Scaling: The dependent measure (categories) m ust be
45. nominal.
• expected frequenck$: No exl'ected frequency within a category
should be less !han I,
and no more than 20% of the expected frequencies should be
less than 5.
As "ith all tests of !he nuU hypothesis, the x' test begins with
the assumptions of ran ·
domness and independence. Deriving fr o m thc.~c assumptions
is the requirement that the
categor ies in the cross-tabulation must be mutunlly exclusive
and exhaustive.
Mutually exclusive means t hat an individual may not be in
more than one categot)' per
variable. ExiJaustive means that all categories of int ere;t arc
covered.
These assumpliom nrc listed more or less in o rder of
i.n1portance. Violations of the first
three assumptions are essentially "fatal" ones. Even slight
violations of the first two
assumptions can introduce major errors into the computation of
p values.
90 PA~-r l • OVAinllAt•vt Al'tfiOoCI!CS: FouNOoTION<o 01
DAYA C.ouu:.HON
They} goodness-of-fit test is basically a h>rgc-sam plc test.
Whc11 the c·xpectcd frequen
cies are small (expected frequency les.~ thnn I or atlc:1~t 20o,(,
of expected ft·equ,•ncics less
than 5), the probabilities associated with the X' t~St will be in
46. accurate.
The usual pt·occdtu'c in this case is either to increase expc led
frc<1ucncb b)' colbp, ing
adj.>ccnt C<>tcgorics (also called cells) <>r to u.<c '"' ot her
test. Follo<"ing is a concrete
CX:l111 plc.
The workers at the Interdenom ina tional Social Services Center
in St. Win ifre d
Township wanted to see whether they were servi ng people o f
all fniths (and those of no
fit ith) equ:11l)'· The)' had census 11gures indicating that
religious preferences in the town>hip
were as follows: Ch risti~n (64%), Jewish (10%), Muslim (8%),
other religionino preference
(14%). and agnostic/atheist ( 4%).
The workers randomly sampled 50 clients from those seen
during the previous year.
Befor• they drew the sample, they calculated the expected freq
uency for each category. To
obtain rhe expected frequencies for the sample, the)' converted
the percentage for each
preference to a decimal proportion and multiplied ir by 50.
Thus, the expected frequency
for Christians was 64% of 50 or .64 x 50 : 32, the Jewish
category was 10% of 50 or
. 10 x 50 = 5, and so on. Table 6.5 depicts the expected
frequencies.
TABLE 6.5 Expected Frequencies for Religious Preferences
Expected
fr(!q uency
47. Christi (In
J2
Jewish
5
ti1uslim Other/No Preference Agnostic/ Atheist
4 7 2
Two (40%) of our expected frequencies (Muslim and
agnostichlllteist) are less than 5.
Given that the maximum allowable is 20%, we are violating a
test assumption . We can
remedy this by collapsing categories (merging two or more
categories into one) Ot' by
increasing the sample size. However, thet·e is no c.ategoq• that
we could reasonably com·
bir1e with agnostic/atheist. lt would not work to combine this
C<tegory with any of the
other categol'ics because the latter ar• religious individuals,
whereas atheists and agnostics
aJe not religious.
However, we could increase the sample size. To get a sample in
which onl)• one (20%)
o f the expected frequencies was less than 5, we would need a
sample large enough so that
8% ( percentage of the population identifying as Muslim ) of il
would equal 5:
0.08 • 11 = 5
" = - 5- = 62.5 "' 6J.
48. 0.08
So, our sample size would need to be 63, givi11g us th e
expected frcq ucncio.:> show11 in
Table 6.6. On!)' one of live (20%) of the expect«l frequencies is
less I han 5, and nQne of
them is less tha n I, so the s:un ple size assumption is mel. The
results of a random sample
of 63 cases were as found in Table 6.7.
TABLE 6.6 New Expected Frequencies for Religious
Prefere~ce; ' · < · ;. : •: •: •
. . ~ ' * •
Christian Jewish Muslim Other/No P(eference Agn ostic:/
Atheist
--------------------------
~>:pecte.fl
frcq uc:nc;·
~0.32 6.30 5.04 8 82 2 52
TABLE 6.7 Observed and Expected Frequencies for Religious
Preferences
Christian Jewish Muslim Other/ No Preference Agno$tic/
Ath~isl:
Expected 40.3L &.30 5.04 8 .82 2.52
rr~(j ll CrtCy
Obse1·.-cd 49 2 2 9
frequency
49. The null hypothesis fo r this example is th;~ l the p roporlion of
peo ple living in St.
Win ifred T<>wnship who identify 1vith each religious
categor)' will be the sam.: as the pro·
portion of people who have received services at the
Interdenominational Services Center
in St. Winifred 1b w nship who identify wit·h each relig io us
catt:gory.
The null hypoth~sis expresses the expectation that observed and
expected frequencies
will not be differem. Notice the similari ty ben~<.>en the nu ll
hypothesis and the numerator
of the ,,, .•. test statistic:
/v IJ&
X2 = "' (Jo - rd 0 0 1 L- fE .
T he form ula tells us to >U btract the e xpe<ied score from the
observed score (j~ -.0 and
then to square the difference (ffo - f.:]' ) and divide by the
expected score (ff0 - J~l'!f.) for
each observed and expected score pair. •Vhen we are fmished,
we add the answers and
o bta in the X',,, test s~tlist ic (Ta ble 6.&).
The x.,. is evaluated by comparing it to a cr-itical value <x'.,,)
that is obtained from a
table of critical values of the X2 distribution. If X'.,b, is greater
than or equal to x', ... • then
we reject t he null hypot hesis.
For ax' goodness of fit, the degrees of freedom are equal to the
number of ,,ategories
50. (c) min us I or df = c- L In our case, we have five categories
(Christian. Jewish, Muslim,
otherino prefere nce, and agnostic/athe;st), so df = 5- I = 4.
The critical value fo r X' at C< = .05 an d df =4 is X' .," = 9.49.
We have calculllted 7.'.,., as
23. 1295. Because X1<,1>1 is greater than X.~ena , we reject
the null hypothesh:. The evidence .sug-
gests that people of all faiths (and those of no faith) are not
being sec11 proportionately to
their representations in the township.
Earlier, we discussed the use of t he effect size measure d for
the t test. Jt is an appropri-
ale measure of eftect size: fO r a test of means. However, Lhc
X2 test doc,~ not compare
92 PAIT I • Q UAIITI TA.Tivt A PPfiOAW £s: fou~OAliONS
O f DATA Coll.ECTI OM
TABLE 6.8 Computation of x' ...
Observed (f
0
) Expected (f,) fo - fe lfc - f,)' (f.- t,)'
f,
49 4032 +8.68 75.3424 17.4404
2 6-30 -4.30 18.4900 2.9349
2 5 04 - 3.04 9.24 16 1.8337
51. 9 .8.82 - 0. 18 0.0324 0.0037
2.S2 - 1.52 2.310• 0.9!68
!'JOT!.: I
(f, - f,)'
17,4404 + 2.9349 + I 8337 + 0.0037 + 0.9168= :t',, = 23.1295.
f,
means. It compares frequencies (or proportions}. Therefore, a d
ifferent effect size index is
used for the X' test-w. This measure of effect size ranges from 0
to I . Cohen ( !988) clas-
sifi es these effect s izes into three categories:
Small effe<i size: w~ .10
Medium effect size: w ~ .30
Large effect size: w ~ .50
The effect size c.oefficient for a x! goodness-of-fi t test is
computed according to the fol-
lowing formula:
where N = the total sample size.
For the St. Winifred Township example,
IV= J(23.! 295/ 63}- J(0.367l) = 0.6059,
which would be classiGed as a large effect.
Hypothesis Tests for Two Related Samples
These are Jests in which either a single sample is drawn and
52. rneasLtremen ts are taken at
rwo times or two samples are drawn and members of the sample
are individually matched
o n som e altribute. ~vfeasureJDeDts are taken fot each member
of the matched groups.
We· investigate three examples of two related sample tests in
this section:
I. Dependent (matched, paired, correlated) samples t test (in
terval or ratio scale)
2. Wilcoxon matched pairs, signed rank.~ test (ordinal scale)
3. McNemar change test ( nominal scale)
C1MPH~ 6 • Sunsncs FOR Sot-IAt 'IOKKUlS 93
Difference Scores. The dependent r test and the Wilcoxon
matched pairs, signed ranks test
evaluate d ifference scores. These may be differences between
scores f<om measuremenl~
taken m two differen t times on the same individual (pretest and
posttest) or differences
between scores taken on two diffe rent individuals who have
been paired or matched with
each other based on their similarity on some variable or variable
cluster (e.g., gender,
race/etllnicity, socioeconomic status). The formula for a d
ifference score is
x; - X1 =X0 ,
X, is the first of a pair of scores,
53. x; is the second of a pair of scores. and
X
0
is the d ifference between the two.
The null hypothesis for all these tests is that the samples came
from popub tions in
which the expected differences are zero.
Tlte Dependenr. Samples t Test. This also is called the
correlated, paired, or matched t test.
The nu ll hypothesis for this test is that the mean of the
differences between the paired
scores is 0:
where
J.l.xo = the mean diffe rence between the populations from
which the samples were
d rav.1n) and
)!00 "" the mean difference between the populations specified
by the null hypothesis.
Because the null hypotnesis typically Sp<!cifies no difference
(!!00 = 0), the null hypothe-
sis usually is written as
The t statistic for the dependent t test is the mean of the sample
differences divided by
the standard error of the mean difference or
Xo - l'oo
lobt = 5= ·
54. XD
As the absolute va.lue of t. gets larger, the more unlikely it is
that such a difference could
occur if the nnll ll)'pothesis is true. AI a certain point, the p
robability (p) of obtaining at so
large becomes sufficiently small (reaches the alpha level) that
we reject the null hypothesis.
The assumptions of the dependem t test are as follows:
Randomness: Sample members must be randomly d rawn from
the population.
Tndependence: Xvscores must be independen t of each other.
Sca ling: The Mpcndt'nt measure (X
0
scores) must be interval or ratio.
No r·mal distribution: The population of X
0
scores must be normally distributed .
These a>sumptions a re list ed more or less in order of import>l
11cc. Viola tions of the t1 rsl
t hree asswup tions i1re essen t ially "dea th penalty" violation..
Eve n slight violation. "r the
(ht two assumpti011s can intr oduce majo r e rror in to th e
comp ullll ion or p values. Sim i lady,
dilTnence scores computed fro1n ~""'O sel!t of ordi nal data
tnay inwrporate major error.
55. Violation of th~ assu mption of a normal distribution "ill
introduce some error into
the computation of p values. However. Wllcss the population
distribution is markedly dif
fcrent fi-om a normal di>tribu tion, the errors will tend to be
slight (e.g., a reported p value
of .042 actually will be a p value of .057). Th is is what is
ml·an t wh en someone '"YS thnt
the t test is a "'robu~t .. test.
Still, cvm thoug h t he erwr is sli~;ht, the nonp<tr<~metric.
Wikoxon rn;,tch ed ~>t~ irs,
sig ned ranks test (discussed in the next section} prob;,bly will
yield a more accu rate test
when there are viulation~ of this normal dislribution
as.su.mpliun.
Let us look at the proc<"<iure for compuling th<: dependent
grouvs I statistic. We usc an
evaluation uf an intervention for individuals '"ith dcpre..,.inn
problems. The dependent
measure is the Bclk Depression Inventory ( liD I), a reliable and
well 'alidated mea>urc nf
dcpn:s~;un.
Ten clienL~ were rand omly s~kcted r,·om clients seen fo r d ep
ression problcn" a t a (l,un -
m unity cent~r. 'I 'hey were pretested (X,) with t he BDI, r<·cd
ved I he treatment, ;,nd t he n
were posrtested (~)wi th t h e same inst ru111e n1.. The m ean
of the d iffe rence scores (.k0 )
wa.s - L This means that tJ K· aven1ge: chtUl.gC' in BD f
scnrefi fron1 pcelC'Sl tu pn:-:ttest was a
dtcrease of I poinl. The standard deviation of the ditlcrcnce
s.:ort> 'aS l.H .
56. 'I he ne>.'t step is the cnmpntation of the 'landard error ol tllc
mean. Wedhdde the stan-
dard deviation by the square rout of t he s.unpk siu: to get t he
standard c·rror of th e mean:
.< XD = 1.'33/ V 10 - l .;l3j 3 .16 = 0 .•12.
Ve plug the value.< into the formula li>r t.,.:
XI>
r"lobt = -
-'xl'>
- 1
-~ - .1..38
0.42 ..
Fo1· a = .05 and rlf ~ 11 - I = 10 - I -9, r, ... = 2.262 (sec a
t<~nle of critical values for the
1 te,r, nondire.:tional, fo und in m ost stali>Li" texts). Because
lt .... l - 2 .. l8 is greater !loan or
equal tn the critical ';liuc, we reject the null hyp(llhcsis at a=
.05.
The cff~ctsi/e index for tbiotc.,l i' ll and is rom puled a;
foUows:
;
For the depr~ssion intervention cx,unplc,
-1-0 - 1
d = = = - 0.752.
57. 1.33 1.33
w hich wou ld be classifier! ns " medium effect.
CHAI'rER 6 • SI All~ucs Hl!t Socu .. l Woll.~Eas 95
lv'ilc&X011 Matched Pairs, Signed Ranks Test. The Wilcoxon
matched pairs, signed ranks test
is a nonparametric test for the evalua tion of d ifference scores.
The test involves ranking
d ifference scores as 10 how far they are from 0. The difference
score closest to 0 receives
the rank of I, the next score receives the rank of 2, and so on.
The ranks for diffe rence
scores below 0 are given a negative sign, whereas those above 0
are given a positive s ign.
T he null hypothesis is t hat the sample comes from a
population of di fference scores in
"' hich the expected difference score is 0.
The assumptions fo r t he Wilcoxon matched pairs, signed ranks
test are as follows:
• Ratufomness: Sample members must be randomly drawn fro m
the population.
• independence: XD scores 111ust be independen t of each
other.
• Scaling: T he dependent measure (XD scores) must be ordi nal
(interval or ratio dif-
ferences must be converted to ranks).
Let us look at the procedure for computing the Wilcoxon
matched pairs, signed ranks
58. test statistic. We use the same example as for t he t test. The
dependent measure is t he BDI,
a measure of depression. Scores on the BDI are not normally
distributed, tending to be
positively skewed.
Ten clients were randomly selected from clients seen for
deprcs.~ion problems at a com-
mun ity center. They were pretested w·ith the BDI~ received the
treatment, and I hen were
posttested with t he same instrument. We c.ompute the
difference scores (post -pre) fo r
each indi,·idual. We assign a rank to each difference score
based on irs closeness to 0.
Difference scores ofO do not receive a rank. Tied ranks receive
the average nlllk for the tie.
So, if we look at Table 6.9, we see that there is one difference
score of 0 that goes
unranked. There are five difference so::ores of eit her - 1 or +L
These cover t he first five
ranks {I, 2, 3, 4, 5), giving an average rank of 3. T here are
three difference scores of - 2
(and none of +2). T hese cover the next three ranks (6, 7, 8) ,
giving an average rank of 7.
The una! score is - 3, which is given the rank of 9.
TABLE 6.9 Computation of the Wilcoxon T .. ,
Signed Ranks
JD Number Pretcsl Postte.st Difference Rank Positive Negati ve
17 16 - 1 3 3
2 19 t8 -1 3 3
59. 3 18 15 -3 9 9
4 18 17 -1 3 3
s 16 16 0
6 16 17 +1 3 3
7 18 16 - 2 7 7
8 21 19 - 2 7 7
9 18 19 .+1 3 3
10 18 16 - 2 7 7
NOTE: Sum of ranks for less, frequent ~ign ~ 6:
9 6 t-'11111 I • QUAWhlAII11 Al•f'II(IA(tUI: r t i
UNOATI(Hn ()I I)AlA (.OU I CI101i
T he M<l st<·p is to '';ign" the rank. ' I hi< mcJns to place the
rank in eith« 1hc p<hilivc
or 1hc negative <.Oiumnm 1h~ l.thle. depending on whether 1he
differ,·ncc >(Ore wa, PO>i
tivc or ncg.uivc.
We then determine which ,ign (JXl,ithe or neg.ttive) apJl<'ared
1.-s~ fre<JUCOtl)· Jnd add
up rhc r.mks for 1his >~!(n. lkcause th e positive sign ,tppearctf
only twice (comp>rctf to
~even tim~s for lhc ncg:.uivc sill.n)~ w~: add up I he rank~ in
the pO$itivc column .lnd obtain
1>. rhi•" I he IC1 l3l"lic v~lue for the Wil<OM>n mJI<.hed
60. J>J II.,, stgncd r:lnks test.
Th e IC> I. stati>l icis w iled 'f.,1, . This is an 11 ppcrcase T a
nd is not the >flllll' as the >tatistic
us<:d with the (lo'"erc.tse) I distribullon.
There are two other i<sues with re>pect to the Wilwxon 7.1,. •
hat shoul11 be ad,lresscd:
1. The Wilcoxon T..., is cvaluat<·d according to rhe ruombtr of
nontcro differentc
~cores. So, we should subt ract I from the o rigina l 11 fo r each
<liiferenc~ score th ot h
0 10 obtJin a corrected 11 to usc for the critical '~lue table.
2. Unlike most other t~>l &ratistic~. the Wilcoxon T,,, must be
lrss tlta11 or equa l to t he
c ritical value to ,·eject the null hypothc>is.
We consult a table of critica l values for I he W ilcoxon T(scc t
ahlc of .:ritical values for
Wilcoxon Tin any general swristics book) Jnd stt whether obe
result (7.,.. = 6) was sig·
nificant at o. = .05. lle<:ause there wa. one differen ce score
equal to 0, the corrected 11 = 9.
The critical value for the Wilcoxon 7"a t n=9 and a .05 is T.,. =
5. 1:,.. = 6 is not less than
or equ•lto the critic.ol value, so we fail to reject the nuU
h)·polhesi> at o.- .05.
There is n o weD-accepted post h oc measure of effect sizt for
Otd in:d tesL~ of rela ted
scores. One possib le measure would be proportion of
nonoverlapping scores as a measure
of effect. Cohen ( 1988) brieOy discu~s this measure, called U.
61. The p1·ocedure bc:gins with compul ing the miniJuum and
maximum ~cores for each of
the two related g roups. We choose the least maximum and the
greatest minimum. Tbi>
establish es the end points for the overlap range.
We count t he n umber of scores in both groups w ithin this
mngc (including rhe end
JX>ints) and divide by the total number of scores. This gives a
proportion of overlapping
score.o;. Subt ract t his number from I , and wr o btain the p
ropottion of nunoverlapping
$Cores. T his indc.~ ranges from 0 to I. Lower proportions arc
indicative of ~mallcr effects,
and higher on~> are indicative of larg<·r effects.
Cohe11 ( 1988) calcula tes equivalent< between U a nd d, which
would imply the foUow·
ing definition> of strength of effect:
Small ct rect slzr
Uugc ('tfect SIZC
d~ ~
d:.8
u- .IS
u- .33
u ~ 47
f"Or the example da1~, the minimum scooc for th e prctCl wa&
16, and the mnximum
~core w;1~ 2 1. The poSit(!St miuimum and ua.tximllln -;cores
wt:r~ 15 .md llJ. rc-'>petti•cly.
62. 'I h e grc•test minimum is 16 •• md lht lcastm.l.ximum is 19.
Of 20 total '>()1 e.,, 1 ~ f~U with in thi, 1werl.•1> r.onge. The
p ru('<J rt ion of ovcrhop is I ~/20 c.~) .
Tl'te pwportion of nonovcrl•ppings..otc., b u~ 1 -.90 = .10.
hich would be a smJJI cft<:.:t.
CHAnt~ 6 • STAT1srtcs rQR SQetAL Wcnrxus 97
.WcNmmr Change Test. The Mc:-icmar change test is used for
pre- and post intervention
designs "'here the variables in the anai)'Sis arc dichotomously
scored (e.g., improved ~.
not impro,•ed, same,.,_ different, increase 's. decrease).
The layout for the McJ-:emar change test is shown in Figure
6.5. Cell A cont.Un> the
number of indh~dual.s who changed from+ to-. Cell B contains
the number of individ-
uals who recei,ed +on both measu rement>. Cell C contains the
number of individuals
who received - on both measurements. Cell D contains the
number of individullh who
changed from - to +. The null hypot hesis is expressed "'
where
P, is t he proportion of cases shifting from+ to- (decreasing) in
the null hypothesis
population, and
P
0
63. is the proponion of ca,.,; shifting from - to + (increasing) in the
ouU hypothesi'
population.
The assumptions for the McNemar change test are sintilar to
those for the X' test:
Rrmrlomness: Sample members must be randomly drawn from
the population.
Independence: Withi n-group sa111 plc sco•cs must be
independent of each other (although
llerween-group scores [pre· ~nd poM1c~1 ~cores] will
necessarily be dependent).
Smling: The dependent measure (categol'ies) must be nomi nal.
F.xpected frequencies: No expected freq ue11cy within a
category should be less than 5.
A special case of X'..,, b t he test >tatistic for the McNemar
change test:
where
t _ (If,. .fi,f - I ) 2
'"" - f, + fn
J. =the frequency in Cell A, and
fn =the freq uency in Cell D.
Th ·is is a test statistic with df = I , For rlf I , we need to
include s·omcthiug called the
Yates correction for continuity in the equation. This is - I,
which appears in the n ur.-'1~ 1'"
tor of the test statistic.
64. Figure 6.5
McNemar Change
Test layout
Before +
After
A B
c 0
98 PART I • QuAutlfi~T•vt A PI'AOAC HlS! Fou~JDAfiONS
OF Ot.rA CotUCliON
Let us imagine that we are interested in marijuana use among
high school students. We
also are interested in change in marijuana ust over time.
Jmagine that we collected survey
data on a random sample of ninth-graders in 2007.1n 2009, we
surveyed the same sample
that had been in ninth grade in 2007. We fo und that 32 of 65
students said that they used
marijuana during the previous year, as compared 10 23 of 65 in
2009. The results are sum-
marized in Table 6. 10.
TABLE 6.10 Observed and Expected Frequencies for the
McNemar
Change Test
2009
65. None Marijuana
2007
Marijvana 2 (Cell A) 21 (Cell S)
None 31 (Cell C) 11 (Cell 0)
Total 33 32
l'o!<ll
23
42
65
Cell A repn-serm thMe studeitts who had used marijuaM in
2007 hut who had nOf used
it in 2009. Cell B shows the number of students who had used
marijuana in both 2007 and
2009. CeU C shows the number of students who did not use
marijuana either in 2007 or in
2009. Cell D shows the number of students who did not use
marijuana in 2007 but who did
use it in 2009.
So, the sum of Cells A and D is the total number of students
whose patterns of mari-
juano use changed. The nuU hypothesis fo r the McNemar
change test is th at changing from
nonuse to use would be just as likely as changing from use to
nonuse.
In other words, of the I 3 individuals who c.ha11ged their
66. pauern of marijuana usc, "e
would expect half (6.5} to go from not using 10 using and the
other half (6.5) to go from
using to not using if the null hypothesis were true.
Tile calculation of the McNemar change test statistic is shown
in Table 6. 1 L
!'or df ~ 1 and C/. ~ .05, x,, = 3.84 (see a I<Jbe of critical
values of x' fo<md in most sta-
tistics texts). Because x ',., = 4.92, we would reject the null
hypolhesis at u = .OS. We would
conclude that there was in fact aJl increase in marijuana use
between 2007 and 2009.
TABLE 6.11 Computation of the McNemar Change Test
Statistic
( JI~ - f01)-1
2 11 8
NOTE: 7~1 = 4.923.
64
(If. - f. l- 1 I'
f..,. + fl)
4 ,9230767
CHAot1U 6 e STATISTICS fO-. SOCI~l W O'-I(rll 99
The effect size coefficient for a M':-lemar change test is wand
67. is computed according
to the following formula:
For the high school survey,
w = J(4.923/65) "' Jo.o757 = 0.2752,
which wo uld be classified as a medium effect.
Hypothes is Tests fQr Two Ind e p e nde nt S amples
These are tests in '•hich a sam ple is randomly drawn and
individ uals fro m the sample Jrc
rJ.ndomly assigned to one of two experimental conditions.
We investigate three examples of two independent samples
tests:
I. Independent samples (group) /test (interval or ratio scale)
2. vV"dcoxonfMann-Whitney (WfM-W) test (ordinal scale)
3. ;(2 test of independence (2 X k) ( uominal scale)
l11depeudent Samples 1 Test. T his sometimes is CJIIcd the g
roup t test. It is a test of mcJ.ns
whose null hypothesis is fo r mally stated •• follows:
Following are the assum ptions of t he independent t rest:
Randomness: Sample members m usr be randomly drawn from
the populotion and ran·
dom ly assigned to o ne of the '-"0 groups.
ltrdepe11dence: Scores must be independent of e.1ch or her.
68. Scalitrg: The dependenr measure musr be inrervlll or ratio.
Normal distribution: T he populations from which tbe
individuals in the samples were
d r,own must be normally distribured.
Homogeneity of variances (a,'- a ,'): ' f he samples must be
drawn from populatious
whose variances are eq ual.
Equality of sample sizes ( "• = n,): ' I he samples m ust be of
the same sir.e.
As before, these assumptions are listed more or less in o rder of
imp o rtance. T he fir. r
three assumptions are rbe " fa tal" assum pt ion;.
Violation o f the nonnaliry assumption will make for Jess
accurate p val ues. However,
unlc;.s Lhe population dist r iburion is markedly diiTerent from
a normal d isrr iburion, the
errors will tend to be slight. Slill, e"en though the error is
slight. the oonparamcrric W /M-
W test probably will be more accurate when the norma lit)•
assum prion is violated.
The independent groups t tesr alw is fair!)' robu>t .-ith respect
to •iolation of the
homogeneiry of variances assumption and the equal sample size
assumprion. A problem
may .orise when both of these assumptions are violated Jtthe
same time.
100 PAnl I • OUANntAuvt Art~AoAc.ul~ Fou~~rooAT ION>
69. o• 0"'" Ct~ur<TION
If the ,maller variance •~ mthc "11allca >.~mple.then the
probability of,, I ypc II ca ror ( 1101
deteaing an exi;,ting dilfcrcn<c) ia"rC.1«'>.i£ th(' larger
'ariancc is i 11 til<' <mJIIcr .amp!<-, then
1 he probability of a 1Ypc I error (rei<-.:ting the null
hypothc:.i> when it i> true) anne.a'<".
If there is no ..tSsodarion lk·twt-en s.;1mplt"' Mit.' ~lnd
vari:wcc. then ''iol.l1ion of c:.u.h of
thc>e .~S»umptions is not partiCufMiy problem.uic. There may
be fairly ,,ub>t.mtial di~
crrpJncies bctwet•n s. .. mplc si1C!' withnut much effect on
Lhc dtc.ur~cy o i Ottr /' cMim.lttl'!.
Similarly, if e- very other n~~nmption i!) mel, 1hcu a slight
difference in v11riam:c:. will not
h ave a fa rge effect on probability estimates.
T he t stat i~tic for the independent 1 lc<t is the d ifference be
tween the snmpfc 111cans
d ividc<l by the standard e•-roa· ,,r the diffprrnces between
means or
x , - x2
lut-·1 --
Sx 1- ... ~
Be«luse rwo sample mean• arc computed, 2 degrees of freedom
are lost:
df 110 + n, - 2.
where
"• = number of scores for the first group, and
70. 11
2
= number of scores for the seco11d group.
Following is an example ot the ll>e o( the independent t test
statistic. We whh to sec
wl1ethf:r there is a difference i11 ((•vel of soci.al act iv ity in
children depending 011 whether
they are in after-school care <>r h0111c (.(ltc . Because more
childre11 attendc<l the .1fter
school program, a proportional~ stratilied >ample of 16 children
in afteHchoof care
(Group I ) and 14 childien in home care (Group 2) was drawn.
The dcpcnclcnt meJsure
v,•as a score on a socir1 l activity ).CJ )e in whk h lower scores
represent less soc ial aclivity and
higher scores represent more social activity.
We c'aluate tl1is with an independent 1 tc.L The first step in
calculating '·•• i, to com·
pule the sample mean for each group. The next step is to
compute the stJndard error of
the mean. Howe•·er,the pl'()(cdure for doing thi< i~ a little
different from that u«<< before.
A> lou might recall. the standard error of the mean is the
standard dcvi,ation d" aded by
the square root oi the sample 'ire:
$
.,;;; /sl !.. II
This also is equivalent to the squ:HC •·oot· o f the variance
times the inverse of the,., , .
71. p te size (l/11).
Unf{'trtunately) we c:u•not u~t..· lhis IOI'tnuln for t+ae standa
rd error o f lhc mean. It is I he
"ttdnda l'd crroJ' for a sinr,l<.- ... amplt. Bccauo,r we have two
sample:, in ,m iudcpcndt•nt
WOU(JS lCsi, the formula has to he Jitert·tf J bit.
Th~ first difference i in the (orrnuiJ for •he: va ria nce. TIH!
variM1u: i' the Uill o l
..qual'l."> divided b)' the deg~C·c~ of lrct'dom. ll•s tht same
he...- eX(Cpt that we have two
'oms of squan:s (one for Group I and one for Group 2). and o u1
degree< of freedom Jr('
11 1 rt. 2. Thi• gives "' the folfowint: cquJtion:
ss, ss1
" ' I II• 2'
CH.t.PHR 6 • Su.nsncs f OR SOC IAL W ORKERS 101
s; is the pooled estimate of the variance based on two groups,
55
1
is the sum of squares fo r Group I ,
SS, is the sum of squares for Group 2,
n
1
is the number of scores in Group J, and
72. n, is the number of scores in Group 2.
Because there are two groups, we do not multiply s: times (1/n);
rather, we multiply it
by i lin,+ I In,). We take the square root of this and obtain the
pooled standard error of
the mean:
S.'1-Xl = , (I 1) s- - + -P IlL nz .
The means and sums of squares for our example are presented in
Table 6.1 2. Now, let
us tq• computing t..,,.
TABLE 6.12 Group Statistics
Group Mean Sum of Squafcs "
27.8B <1330.40 16
Home care 21.36 17{)7. 16 4
First, we compute the pooled standard error of the mean (also
called the standard
error of the mean difference). We begin by calculating the
pooled variance:
ss, + ssl 43:;0.40 + 1101.16 6037.56
28 = 215.63 . = n, + n2 - 2 16+14-2
From the estimate fo r the pooled vari<Htce, we may calcubte
the standard errol' of the
mean diffe rence:
s2 - +- = ( 1 I) I' tll ll2, 2 15.63 (~ + ~) = ,128.88 = 5.37 16 14
Wt calculate 1
73. 001
:
27.88 - 21.36 6.52
lobt = = -- = 1.213 .
5.37 5.37
For ex = .05 and df = 111 + 112 - 2 = I 6 + L4 - 2 = 28, Ia;, =
2.048. Because 1100,1 = 1.213 is
less than the critical value, we fa il to reject the null hypothesis
at a. = .OS.
102 PAI!.l I • QuANtiTATIVE AI'P~OACHES: Fou ... O-.liOM
Of 0ATA co~UtliO'f
There are two post hoc effe<:t size measures for an independent
t test. The 11m of these
(d) already has lxen di.cmsed:
Note dlatthe numerator is the difference between the two sample
m eanl and that th e
denominator is the pooled c>ti mate oft he standard deviation.
The pooh.'!! •t andard de,•i-
ation is t he square root of the pooled variance that we
calculated earlier:
Sp = fs~ = V215.63 = 14.68.
The effect size for the example would be
d = 27.88 21 36 = 6.52 = 0.44
14 .68 14.68 ,
74. which would be classified .ts a 1mallto medium effect size.
The other measure is Tl • (eta-.quare). n' is the proportion of
variance explained ( Pifl:) .
This is equivalent to the 'quared point-biserial correlation
coefficient and is computed by
2
/<lbt
2 if.
/Obi + d
We '''ere com paring socinl nc tivity in c hild ren in after-school
care vcrMJ> t hose in home
ca re. Children in after-sdtool cure sCC)rcd h igher on social
activity than d id c hild ren in
home care. T he differe nce was not statistically s ignificant for
<> ur chosen ex = .05.
r.,.,. was 1.2 13 with df • 28. Pu tting these numbers in t h e
formu la, we obtain the
following:
l_ ( 1.213)
1
" - ( 1.213)
2 + 28
1.471
29.47 1 = 0'0499'
So, a litde less than 5% of the variability in social activity
among the chlldren was
potentially explained by whether they were in after-school care
75. or home cJre.
Wilcoxon/Mann -Whiwey Test. Statistic> texts used t o reter to
this te>t as t he Mann-
~Vhitney test. Recent ly, th e name of Wilcoxon has been added
to it. The reason t hat
Wilcoxon's n ame has been added is t hat he developed the test
first and published it first
( Wilcoxon, 1945). Unfortunately, m OI'e fo lks noticed the art
ide publishtd by Mann a nd
I•Vhitn ey ( 1947) 2 years later.
Tbe W/M-W test is a nonp a1·ametric test th at involves initia
lly t reating both samples as
one group and ranking scores from lcn;t to most. After this is
done, the freq ue ncies of low
and high ranks between groups arc compared.
The assumptions of the W/M W test are as follows :
Randomness: Sample members must be randomly drawn fr<>m
the popuiJtion of inter-
est and randomly a>Signed to one of the two groups.
C U AI'rtll 6 • S IAHSHCS FOR $o cu._t W ORKU$ 103
Independence: Scores m ust be independent of each othe r.
Scaling: The dependent measure must be ordinal (inter val or
ratio scores must be con-
verted to ranks).
'When the assumptions of the t test are met, the r test will be
slightly more powel'ful
76. than the W!M-W test. However, if the distr ibution of
population scores is even slightly
d iffe rent from normal, t hen theW /M • W test may be t he
more powerful test.
let us look at the procedure for com puti ng t he W/M-W test
statistic. We use the same
exam ple as we d id fo r t he independent r test. We evaluated
level of social activity in
children in arter-school ca re and in home care. T he dependent
measure was a score o n a
social activity scale in which lower scores represent Jess social
activity and higher scores
represent more social activity.
The first step in carrying out the W/M· W test is to assign ranks
to the scores without
respect to which g roup individuals '"ere in. The rank of I goes
to the highest score, t he
rank of2 to the next highest score, and so on . Tied ranks
receive the average rank. We then
sum t he ran ks within each g roup. The summed ranks are
called W1 for G rou p 1 and W,
for Group 2 and are fo und in Table 6.13.
TABLE 6.13 Summed Ranks for the Wilcoxon/ Mann-Whitney
Test
Summed ranks
After-School Care
n
1
= 16
77. w,= 218
Home Care
n
1
= 14
w;-= 247
The test statistic for the W/M-W test is u..,,. We begin by
calculating U statistics for
each according to t he fol lo wing equations:
U
111 + ( 111 + l)
1 = 11J n;z. + lFV1
2
n2 + (n2 + 1)
U2=11rnz+ 2 w,
nt(nt + 1} u, = ,,, tiJ + 2 - w,
= ( 16)( 14) + ( l6)(~6 - I} 2 18 = 126
(]
112(n 2 + I}
2 = , J l'l:z. + -=-'-=,...--'-
2
w, = ( 16}(14) + ( 14}( 14 - l)
78. 2
182
= 224 +-- 247 = 224 + 91 - 247 = 68.
2
We choose the smaller U as u;,.,. Ln this instance, u.,. = u, =
68.
247
u •• , m ust be less tlran or equal to the critical value to reject t
he null h ypothesis.
The critical value for the W/M· W U at n, = 16 and at n, = 14,
and o: = .OS is U"'' = 64.
104 PoIU I • 0uAN11tAT!V( A1'1'110M.Ht~ : FOU'IDATIO.,.S
or OoTA CouH.UO'
U.,..: 142 is not less than or equal to the critical value, so we
fail to rejtct the null hypothe-
sis at CL: .05.
As before, t here is no well-established effect size measure fo r
the W/M-W test. The U
m easure of nonoverlap probably would be the best bet.
For o ur example data, the minimum and maximum fo r t he
after -school care g roup
w ere 2 and 55. whereas they were 7 and 40 for the home care
grout>· The greatest mini -
mum is 7, and the le"'t ma.ximum is 40. All 14 .cores in the
home ca re g roup are within
79. the overlap range, and 12 of l4 scores in the after-school care
group are in t he overlap
range. This gi•es us a proportion of overlap of 26/30: .867. The
proport•on of nonover-
lap is U I .867"' .133. This would be ,, small effect.
X' Test of lmlcpt!m/ence (2 x k). The assumption> fo r d1e x'
test of indCj>Crtdence are as
follows:
/lat~dom/les.: Sample members must be rnndo mly dra"'n from
the 1>opulation.
/Jillependl'!lre: Sample scores m ust be independent of each
other. O ne implication of
this is tha t categories must be mutually exclusi'e (no case m ay
appear in more than
one c.1tegory ).
Scaling: The dependent measure (categories) must be nominal.
Expmcd frequmcie$: No expected frequency within a category
should be less than 1,
and no more d1an 20% of t he exp«tcd freq uencies sho uld be
less t han 5.
As wit h all tests of t he null hypothesis. the x2 test begins with
t he assumptions of ran-
d omness and independence. Deriving from t hese assumptions
is the requirement that the
categories in the cross·L1 bulation be mulllnl/y exclusive and
ex/u~ustive.
Mwunlly rtclusive meaJlS that nn individual may not be ill more
thn n one category per
variable. Bxluwsti•-e means that all possible categories are
80. covered.
let us imagine that we are interested in marijuana use among
high school students and
sp<-cifically whether there are any diffcrcn= in sutb use
between 9th and 12th-graders
in our school di>trict. We conduct • proportionate str atified
samplt in which we ran-
domly s:~mplc oixt)'-five 9th-graders and fifty-five 12th-g
raders from all Mudents in the
district. T he students are surveyed on t heir usc of ((rugs over
the past ye.ar under condi-
tio ns guaranteeing co nfiden tiality of response. Table 6.14
depicts reported marijuana use
f o r t he s tudents in the sam ple o ver the past yenr.
TABLE 6.14 Marijuana Use
None
MatiJuanil
l eta I
Grade
9th 12th
42 33
23 22
65 55
Toto!
81. 75
1 ~0
A higher proport ion of 12th-g raders
than 9th-graders in t his sample used mar-
ijua na at least once during t he past year.
The question we are interested in is
whether it is likely that >uch a sample
could have come from a population in
which the proportion.1 of 9th- and 12th-
graders using mc:1rijuana were identicaL
The usual test used to evaluate such
data is the x: test of i ndepcndcnce. The X1
test evaluates the likelihood that a per·
ccived relationsg1ip between propor tions
in categories (called being dependent)
C HAI'TEII: 6 • STATISTIC-S fOR. Soc•AL Wo~Kflt S 105
co uld have come from a po pulatio n in which no such
relationship existed (call ed
independence) .
The null hypothesis for this example would be that the same
proportion of 9th-graders
as 12th-graders used marijuana during the past year. The null
hypot hesis values for this
test are called the expected frequencies. These expected
frequencies ior marijuana are cal-
culated so as to be proportionately equal for bot h 9th- and 12th
-graders.
82. Because 45 of 120 of the total sample (9th· and 12th-graders)
used marijuana during
the past year, the proportion for t he total sample is 45f!20 =
.375. The expected frequency
of marijuana use for the sixty-live 9th-graders would be
.375(65) = 24 .375. T he expected
marijuana use fo rthe fifty-five 12th-graders would be .375(55)
= 20.625. Table 6.15 shows
the expected frequencies in parentheses.
The%' test evaluates the likelihe>od of the observed frequency
departing from the
expected freq uency. T he null hypothesis is
H,: P"'- P,,= O,
where P
0
, is the pro port ion of cases within category k in the null
hypothesis population
(e.xpected; in this case, this is the expected proportion of
students in each of the two gt·ade
levels [9th and 12th] who fell into o ne or t he other use
category [marijuana use or no
marijuana usc)}; and P,~ is the proportion of cases wi thin
categor y k drawn from the
actual population (observed; in this case, this is the obser ved
[or obtaine.d] proportion of
students in eacb of t he two grade levels [9th and 12th] who fell
into one or the other use
category [marijuana use or no marijuana use]).
The X'.,, test statistic is
Degrees o f freedom for a x' test of independence are computed
83. by multiplying the
number of rows minus I times the n umber of columns min us I
or
df= (Row - I )(Colum ns- 1)
TABlE 6.15 Observed and Expected Frequencies for Marijuana
Use
None
Marijuana
Total
9th
42 (40.625)
23 (24.375}
65
N01'E: Expwcd frequencies are in parentheses.
Grade
12th
33 (34.375)
22 (20.675)
55
Total
84. 75
45
120
For Ollr example, this would be
d/=(2 -1}(2 1)=(1)(1)=1
Re.::all from our dbcussion of the ;'.lcNemar change te:.t that
we include the Yates cor
rection for continuit)· in the formula ,,hen df l . The equation
for the corrected test sta
tistic is as follows:
X
1 = I: (Vo- fr,l - 0.5)
1
ul>• /c
The form of the equ~tion tells us to suhtr.ltt the expected ;core
from the observed
>eore and take the ab:.olute value of the difference (make the
difference positive). Then.
subtract O.S fro m the absolute difference (I/., f. I -0.5) and
square t he result. Next. divide
by t he expected score. T his is re~1eated for ca<h observed and
expe<ted score pair. W hc u
we are finished , we sum the answers and obtnin the corre<ted
x· .. ,. test st.ttistic.
85. The reader might have noticed that t he con ection for the
McNemar c hange test wa,l
I.Q, whereas th e correct ion for the X' test of independence
(and the goodness-ol:fitiCit)
was 0.5. I will not go iuto an)' detail beyond sa)'ing that this is
be.::ause the McNemar
change test uses o nly half of the a••ailable cross-tabulation
cells ( two of four) to computl'
its x.'..,., ••hereas all cells Jre used to compute ;c,.. in the
independence and goodne~< of·
fit tl'sts.
Tnble 6.16 shows how 10 work out the ma rijuJna survey data.
For df= I and ex .05, the critical value fot· x',,.,. is 3.84. Ou r c
alculated value (X',,,l was
0. 1 09. Bec<Juse t he obtuiued (cakuloted) value did not
exceed t he critical value, we wou ld
not reject the null hypothesis at a= .05.
As before, the effe.::t <i>c measure is ";which is wmputed a• a
post h oc measure by
w - Ji.x'/N).
~or a 2 >< 2 tab le, w;, eq ual to the absolute v.tlue of <p (phi),
which i, J true cor relation
cocfticient.. If we sq uare w, t hen we obta in tp' , w h ich is the
propor tion of variance
ex plained (P1£).
T AILE 6.16 Compuution of x' ...
CJb,crved (f0 ) Expected (1, ) (If.- f, J - 0.5)
42 ~() 615 8/~
86. lJ 14 375 81~
23 ]4.375 .875
n 20 62~ 875
NOTE: 7.' = 0.01 9 + 0.02l + 0.031 + 0.037 ~ 0. 109.
bbt
(If. - f, J- 0.5)' (Jf.- f,l - 0 .5)'
f,
0.7651>2'> 0.019
0.76~6lS 0022
0765675 0.031
0.765625 0.037
CHAI'tfft 6 • Sr.t.nsncs FOil So C-I.t.l WOII(US 107
For our example,
w = /(O. J09/t 20) = Jo.ooo90S3 - oo3o i
and
w' = PVE - .0009.
This is an extremely smaU effect size.
f'or 2 x k tabulation, we cannot convert tv to PVE.
Hypothesis Tests fork > 2 Independent Samples
87. Irnaginc that we wert: in terested in ageist attitudes among
sodal 'Orkers. Specificall)'> we
are interested in whether there are any d ifferences in the
magnitudes of ageist attitudes
among (a) hospital social workers. ( b) nursing home social
workers, and (c) adult pro tee-
tive services social workers.
We cotdd conduct independent group tests among aU possible
pair ings: hospital (a) with
nursing home (b), hospital (a) with protective services (c), and
nursing home (b) with pro-
tective services (c).
This gives us three tests. When we conduct o ne test at the ex=
.05 levd, we have a
.05 chance of committing a Type I error (rejecting the null
hypothesis when it is tr ue) and
a .95 chance of making a correct decision (not rejecting the null
hypot~esis when it is
true). If 1ve conduct three tests at u = .05, our chance of commi
tting at least one Type I
error increases to about .15 (the precise probability is .
142625). So, we actually are testing
at around 0'. = . 15.
As the number of comparisons incceases, t·he likelihood of
rejecting the null hypothe-
sis "rhen it is true increases. oVe are ((capitalizing on chattce
.'>
One way of dealing with capitalization on chance would be to
use a stricter alpha
leveL f'o r three co mpa risons, we m ight cond uct our tests at u
"' .05/3 "' .0 167.
88. Unfortunately, if we do th is, then we will reduce the po,ver ( I
- ~) of o ur test to detect a
possible existing effect.
However, there are tests that allow one to detect whether there
are any differences
among groups wiLhout compromising power. This is done by
siJnultaneously eva1U(lting
all groups for any differences. If no d ifferences are detected,
then we fai l to reject the null
hypothesis and stop. No further tests are conducted because w e
already have our ans11w.
The difference> among all gro ups are not sufficien tly large
that we can reject the notion
that all of the samples come from the s ame population.
If significant differences are detected, then further pair
comparisons are conducted to
determine which pairs arc different. T he screening tests do not
tell us whether only one
pair, two pairs, o r all pairs show statistically significant
differences. Screening tests show
only that there are some differences among all possible
comparisons.
lf we conduct our screening test at a ,. .OS, then we will carry
out the pair comparisons
when the null hypothesis is true 1 out of20 times (commit a
Type I error). By conducting
the in itial overall screening in a single test, we protect against
the compounding o f the
alpha level brought on by multiple comparisons.
We look at three examples of screen ing tests fork> 2
independent samples:
89. I. One-way analysis o f variance (ANOVA) (interval or ratio
scale)
2. Kruskal· Wallis (K· W) test (ordinal scale)
3. X1 test of independence (k x k) (nominal scale)
108 '""' I • QUANTITATIVl AmtOA.CIILS : fOU"-DATIOJr.S
Of DA'rA C.olUCltOh'
One· Way A011dysis of'ariance. The AtOVA is a test of
means. The null hypothesis is
where k is the number of population nocans being estimated.
If all of the means are equal, then it fo llows that the voriance
of the means is 0 or
I 10 : &,. = 0.
The test statistic used in A..'OVA is called F and is calculated
as follows:
n_.;
7
where the numerator is the variance of the sample means mu
ltiplied by the sample size,
and the denominator is a pooled estimntc of the score variances
within the samples.
The assumptions underlying o ne-way ANOVA are as follows:
Randomness: Sample members must be randomly drawn from
the population and randomly
90. assigned to one of the k groups.
Indepelltltllct: Scores must be independent of each other.
Scalir~g: The dependent measure must be interval or ratio.
Normnl distribution: The populations from which the
individuals in the sam ples were
drawn must be normally d istributed.
Homoge11ciry of variances (oi = o~ = .. . = o~): The samples
must be drawn from pop·
ulntions whose variances arc equal.
&jualiry of sample sizes (n, = n, = ... = 11,): The samples must
be of the same size.
ANOVA involves taking the variability among scores and
detumining which is vari·
ability due to membership in a particular group (variability
a.~sociated with group means
or between-group variance) and which is variability associated
with unexplained fluctua·
tions (wi thin-group variance).
The totnl variability of scores is divided into one componenl
representing the variability
of treatment group means around an overall mean (sometimes
called a grand mean) and
another component representing the variability of group scores
around their own individ·
ual group means. The variability of group means around the
grand mean is called between·
group variance. The variabiliry of individual scores around their
own group means is called
within-group variance. This division is rep.--nted by the
91. foUowing equation:
{X - X)~ (X -Xl +(X-X).
Total Within Between
The X with two bars represems the grand mean, which is the
mean of all scores with·
out respect to which group they are in. X is a particular score,
and the X with one bar is
the mean of the group to which that score belongs.
C.HAPlUt 6 a STATiiliGS roll: SOCIAl W Oill({fi S 109
This equation illustrates that tbe deviatio n of the particul ar
score fro m t he grand mean
is the sLun of the deviation of the sco re fro m its g roup mean
and the deviation of tbe
g ro up mean fro m t he g rand mean. T his might be a little
dearer if we look at a simple data
set. Let us hlke the exam ple about ageist attit udes among
hospital social workers (Group I),
nursing ho me social workers (Gro up 2), a11d adult protective
services social workers
(Group 3). T be dependent measure quan tifies ageist attitudes
(higher scores represent
n1ore ageist sentiment).
There are k = 3 g ro ups, with each containing n = 4 scores. The
total number of scores
is N= 12. The group means are 3 (Gro up 1 ), 5 (G roup 2), and
9 (Grotlp 3), and the grand
mea n is 5.67.
There are t hree types of sum of squares calculated in AN OVA.
92. T he fo rm ulas fo r the
sums of sq uares are derived fro m t he deviatio n score C<j
uations.
ss, ...
1
is calculated by subtracting the grand mean from each score,
squaring the differ-
ences, and add ing up (summing) the squared differences:
=2
ss,."' = (X - Xl .
ss .... m is calculated by subtracting the group mean fro m each
score within a group,
squaring the differences, a nd adding up (summing) the squared
differences fo r each
g ro up. This gives us t hree s ums of squares: sswoup I'
SSC.,>I>p , . and SS.;ooup>· These are added
up to give us ssv.·ilhin:
- 2 - 2 - 2
ssW'''" = r <x - x,J + r <x - x,) + r <x - x,J .
s~.~ is calculated by subtracting t he g rand mea n from each
group mean, squaring
the diffe rences, and adding up (summing) the squared
differences. Then, we multiply the
to tal by the sample size. This is because this sum of squares
needs to be weighted. Whereas
N = 12 scores ~~ent to make up SS10,.1, and ( k)(n) = (3)(4) =
12 scores went to m ake up
SS., ... ,,,, o nly the k= 3 g roup means went to make upS~""'".
We m ultiply by 11 = •l so that
S~~ will have t he same " 'eig ht as tlte o ther two sums of
93. squares:
S~"'""' = " I (X - X)'.
The sums of squares arc as fo llow·s:
SS,.;,'"' = 20 + 20 + 20 = 60
s~ ..... ,"' (4) 18.667 = 74 .667
ss ... ,, = 134.667.
The to tal sum of squares (SS~,1 ) is t he sum of the within-g ro
up su m of sq <Lares
(SS.,.,.,) and the between-group sum of squares (55,....,,):
o r
134.667 = 60.00 + 74.667.
110 PAtH I a Q u AN11JA1 1V[ APPI0A(H£S: FOUIIOAltO~S
Of 0 AlA COlltCTIO!.'
Each of these sums o f squares is a component o f a d iffere nt
variance. In ANOVA jar-
gon, a variance is called a mean square. Each particular m ean
square ( variance) has its
own degrees of freedom .
Because the total sum o f squares (SS,.,1) involves t he varia
bility o f all scores aro und
o ne grand mean, the degrees of freedom ar e N - l. The within-
groups sum of squares
(SSw"''") involves the variability of all scores wit hin g roups
around k g ro up m eans, where
94. k is the n umber o f g ro ups. So, the within-groups degrees o f
freedo m are N- k. T he
between-groups sum of squares($""""') involves the va riability
of k gr o up m eans
around the grand mea n. So, the between-g roups degrees of
freed om are k - J.
BeCtJase :1 (/tlritlii<'Y:' (meoll sqa,?re) is,? Rllll of square>
diviOed br degrees of freedom,
the fo rmu la fo r a m ean square would be MS ~ SSitlf
Two mean squares are u::;ed to calcnlate the Fubt statistic:
MS~·i!Jun and A-f~,wMn · Their
specific fo rm ulas are as follows:
There are k ~ 3 groups, so df,"""" = k - 1 = 3- 1 = 2. We may
now compute
A•f""'" = i 4.66712 = 3i.333
and
T here are a to tal of N = 12 scores within k = 3. so di,;,,;0 =
12- 3 = 9 and MS .. n ,h;, ~ 60/9
~ 6.667.
These are the two variances u~ed ro m ake up the F ratio (F ••
,): MS.., • ...., and MS,.,,,,.
The fo rm LLla for F •• , is
MSt,.,w..,n
MSwulUn .
l f we plug in t he values from o ur example, t hen we obtain
fo~x = MSb""'"" = 37.333 = S.6s.
MS,,;,hin 6.667
95. This is a bit confusing when presented in bits aJ1d pieces. The
ANOVA sununary table
is a way of p resent ing t he information about the sums of
squares, degrees of freedom,
mean squares, and F statistics in a more easily understood
fashion. Table 6 . 17 uses the
example data.
Once we have computed the Poht' iL is compared to a critical F.
Because two variances
were used to calculate o ur F •• ,. there are two types of degrees
o f freedom asso ciated with
it: n umerator deg rees o f freedom (between g ro u ps) and de;w
.minator d egrees of freedom
(within g roups). T hese are used either to look up values in a
table o f the F distribution or
by computer programs to com pu te p values.
For our example, the n umerator degrees o f freedo m are df = 2
because 2 degr ees of
freedom were used in the calculation o f MS,""'"'' The d
enominator d egrees of freedom
C HJo i'IU 6 • S t ATISTIC.S fO ft S OCtAl 1N CIIUP.S 111
TABLE 6 . 17 ANOVA Summary Table
Source Sum of Squares Degrees of Fceedom Mean Squar~ F
11111
B~tween 74.667 3 - 1 - 2 74.67/2 = 37 333 37..333/6 667 = 5 65
Within
96. Total
60.00
134.667
12 - 3 - 9 60.00/ 9 = 6.667
12- 1 • 11
are df: 9 because 9 degrees of freedom were used in the
calculation of MS . .,,h;, · The criti-
cal value for Fat 2 and 9 degrees of freedom is .t~"' = 4.26.
Because F..,,: 5.6 is greater than
the critical value, we reject the null hypothesis at«= .OS.
Based on these findin gs, it is likely th at at least one pair of
means come from d ifferent
populations. Because we already have screened out other
opportuni ties LO commit'I)'Pe 1
error, further testing would not be capi[aiizing on chance. Thus,
we may carry out the fol-
lowing pair comparisons:
Group l versus Group 2
Group I versus Group 3
Group 2 versus Group 3
The individual pair comparisons may be carried out using any of
a number of multi-
ple comparison tests. One of the more frequently used is the
least significant difference
(LSD) test. The l.SD test is a variant on the t test. However, the
97. standard error of the mean
is calculated from the within-groups mean square (variance)
from the ANOVA:
where
tt, is the nwnber of scores in Group i, and
tt, is the number of scores in Group J.
If the group TIS are equal, then this becomes
For our example,
Sx;-.<_; = )(2}(6 .667)/4 = J3.333 = 0.557.
We now maycarry oul our comparisons evaluating tat df= N - k=
12 - 3 = 9 (Figure 6.6).
In all three instances, we reject the rwll hypothesis at a = .OS.
I
Figure 6 .6
Multiple Comparisons
Hospilal (Group I) vs t - 3 - 5 - 3466 df= 9,«= 05
Nursing Home (Group 2) "' - 0.577 - . / t!tl = 2.262
Reject H.
Hosprtal (Group 1) vs. r.,. •• g;~ = 10399 Clf = 9, a- .05
Adult Protective Services t .. , = 2.262
(Group 3) Reject H.
98. Nursrng Home (Group 2) '-=~5~ = 6.932 Clf = 9,a ~ 05 vs.
Adun Pro!ectrve la.= 2.262
Services (Group 3) Rejecl H.
T here are a number of measure> for effect size for ru'0'A. For
the >.Ike of srmplicity,
we d eal wit h rwo: Cohen'• (1988) J and 1{
The J effect· size mca>ure is eq ual to Lhe stand ard deviatio n
of th e sam ple means divided
by the pooled "ithin group standard devialion. It ranges from a
min imum of 0 to an
rndetinitcly large upper limit. It m~) be estimated from F..,. by
using the following for mula:
f = JnFobr·
11' wa, discussed earlier and defined as a proportion of variance
explarned. It is calcu-
laled by the fo llowing formula:
l S.'itwlwttn
1) =-- - . ss,,,,.,
It also may be calcul.lled from art F.,.:
Cohen ( 1988) categorizes these effect si1-"s into small,
medium, and large categories.
The critcri~ lor each are as folio" s:
Sm all cfYcct size: f :. .lO
Medium efYect size: f; .25
Large effect size: f .40
Using the exarn plr dJLa, 11' is
99. 11' = .0 1
11' ; .06
11'; . 14
z SSt.,,.... 74.667
'l = = 0.554.
ss,"'·'' t 34.667
CHArtfa 6 • Sr.c..nsTIC;.s fQI SociAL WoRKEss 113
which is a very large effect.
Kmskal-Wal!is Test. The K-W test is the k > 2 groups
equivalent o f the W/M -W test.
The test involves iniliall y treating all samples as one gro up
and ranking scores from
least to most. After this is done, the frequenc ies of low and
high ranks among groups <1re
compared.
The assumptions of the K-W test are as follows:
Rat~donmess: Sample members must be randomly drawn from
the population of inter-
est and randomly assigned to one of the k groups.
Independence: Scores must be independent of each other.
Scali?Jg: The dependent measure must be ordi nal (interval or
ratio scores must be con-
100. verted to ranks).
When the assumptions of ANOVA arc mer, the analysis of
variance will be sligh tly
more po<,•erful than the K -W test. However, if the distribution
of population scores is not
normal and/or the population variances are not equal. then the
K-W test might be the
more powerful test.
The K-W test is a screening test. If th ere is no significant
difference foun d, then we stop
testing. If a significant difference is fo und, then we proceed to
test ind ividual pairs with
the W/M -W test.
Our example involves the evaluation of three interven tion
techniques being used with
clients who wish to stop making negative self-statements: (a)
self-disputation,
(b) thought stopping, and (c) identifying the source of the
negative statement (insight). A
total o r 27 clients with this concern were randomly selected
and assigned to one of the
three intervention conditions. On the 28th day of the
intervention, each client counted
the n umber of negative self-statementS that he or she had
made.
The proced ure for tlle K-W test is s imilar to that for the W/M-
W test. We begin by
assigning ranks to the scores without regard to which group
individuals were in. We then
sum the ranks within each group. The sununed ranks are called
W, for Group I, W2 for
Group 2, and W, fo r Group 3 (Table 6 .18).