1. Post hoc tests are used in ANOVA to determine which specific group means differ significantly when an omnibus F-test is significant and there are three or more groups.
2. Three common post hoc tests are described - the LSD test, Tukey's HSD test, and Scheffe's test. They differ in how conservative they are and how much they control for Type 1 error from multiple comparisons.
3. Tukey's HSD test is generally recommended when all pairwise comparisons between groups are of interest, as it maintains the familywise error rate while having more statistical power than Scheffe's very conservative test.
In statistics, regression analysis is a statistical process for estimating the relationships among variables. It includes many techniques for modeling and analyzing several variables, when the focus is on the relationship between a dependent variable and one or more independent variables. More specifically, regression analysis helps one understand how the typical value of the dependent variable (or 'Criterion Variable') changes when any one of the independent variables is varied, while the other independent variables are held fixed. Most commonly, regression analysis estimates the conditional expectation of the dependent variable given the independent variables – that is, the average value of the dependent variable when the independent variables are fixed. Less commonly, the focus is on a quantile, or other location parameter of the conditional distribution of the dependent variable given the independent variables. In all cases, the estimation target is a function of the independent variables called the regression function. In regression analysis, it is also of interest to characterize the variation of the dependent variable around the regression function which can be described by a probability distribution.
When you run Analysis of Variance (ANOVA), the results will tell you if there is a difference in means. However, it won’t pinpoint the pairs of means that are different.
5 essential steps for sample size determination in clinical trials slidesharenQuery
In this free webinar hosted by nQuery Researcher & Statistician Eimear Keyes, we map out the 5 essential steps for sample size determination in clinical trials. At each step, Eimear will highlight the important function it plays and how to avoid the errors that will negatively impact your sample size determination and therefore your study.
Watch the Video: https://www.statsols.com/webinar/the-5-essential-steps-for-sample-size-determination
Analysis of variance (ANOVA) everything you need to knowStat Analytica
Most of the students may struggle with the analysis of variance (ANOVA). Here in this presentation you can clear all your doubts in analysis of variance with suitable examples.
In statistics, regression analysis is a statistical process for estimating the relationships among variables. It includes many techniques for modeling and analyzing several variables, when the focus is on the relationship between a dependent variable and one or more independent variables. More specifically, regression analysis helps one understand how the typical value of the dependent variable (or 'Criterion Variable') changes when any one of the independent variables is varied, while the other independent variables are held fixed. Most commonly, regression analysis estimates the conditional expectation of the dependent variable given the independent variables – that is, the average value of the dependent variable when the independent variables are fixed. Less commonly, the focus is on a quantile, or other location parameter of the conditional distribution of the dependent variable given the independent variables. In all cases, the estimation target is a function of the independent variables called the regression function. In regression analysis, it is also of interest to characterize the variation of the dependent variable around the regression function which can be described by a probability distribution.
When you run Analysis of Variance (ANOVA), the results will tell you if there is a difference in means. However, it won’t pinpoint the pairs of means that are different.
5 essential steps for sample size determination in clinical trials slidesharenQuery
In this free webinar hosted by nQuery Researcher & Statistician Eimear Keyes, we map out the 5 essential steps for sample size determination in clinical trials. At each step, Eimear will highlight the important function it plays and how to avoid the errors that will negatively impact your sample size determination and therefore your study.
Watch the Video: https://www.statsols.com/webinar/the-5-essential-steps-for-sample-size-determination
Analysis of variance (ANOVA) everything you need to knowStat Analytica
Most of the students may struggle with the analysis of variance (ANOVA). Here in this presentation you can clear all your doubts in analysis of variance with suitable examples.
A brief description of F Test and ANOVA for Msc Life Science students. I have taken the example slides from youtube where an excellent explanation is available.
Here is the link : https://www.youtube.com/watch?v=-yQb_ZJnFXw
Parametric vs Nonparametric Tests: When to use whichGönenç Dalgıç
There are several statistical tests which can be categorized as parametric and nonparametric. This presentation will help the readers to identify which type of tests can be appropriate regarding particular data features.
this ppt gives you adequate information about Karl Pearsonscoefficient correlation and its calculation. its the widely used to calculate a relationship between two variables. The correlation shows a specific value of the degree of a linear relationship between the X and Y variables. it is also called as The Karl Pearson‘s product-moment correlation coefficient. the value of r is alwys lies between -1 to +1. + 0.1 shows Lower degree of +ve correlation, +0.8 shows Higher degree of +ve correlation.-0.1 shows Lower degree of -ve correlation. -0.8 shows Higher degree of -ve correlation.
A hypothesis is a testable statement about the relationship between two or more variables and errors reveal about the rejection and acceptance of the statement.
Please like, comment and share
A brief description of F Test and ANOVA for Msc Life Science students. I have taken the example slides from youtube where an excellent explanation is available.
Here is the link : https://www.youtube.com/watch?v=-yQb_ZJnFXw
Parametric vs Nonparametric Tests: When to use whichGönenç Dalgıç
There are several statistical tests which can be categorized as parametric and nonparametric. This presentation will help the readers to identify which type of tests can be appropriate regarding particular data features.
this ppt gives you adequate information about Karl Pearsonscoefficient correlation and its calculation. its the widely used to calculate a relationship between two variables. The correlation shows a specific value of the degree of a linear relationship between the X and Y variables. it is also called as The Karl Pearson‘s product-moment correlation coefficient. the value of r is alwys lies between -1 to +1. + 0.1 shows Lower degree of +ve correlation, +0.8 shows Higher degree of +ve correlation.-0.1 shows Lower degree of -ve correlation. -0.8 shows Higher degree of -ve correlation.
A hypothesis is a testable statement about the relationship between two or more variables and errors reveal about the rejection and acceptance of the statement.
Please like, comment and share
In general, a factorial experiment involves several variables.
One variable is the response variable, which is sometimes called the outcome variable or the dependent variable.
The other variables are called factors.
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 .
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.
These are slides I use when teaching my second year undergraduate statistics course. They are designed more for conceptual understanding, and do not have syntax for programs like SPSS or R. So it is a more conceptual and mathematical review, rather than a "how-to" computer guide.
Happiness Data SetAuthor Jackson, S.L. (2017) Statistics plain ShainaBoling829
Happiness Data Set
Author: Jackson, S.L. (2017) Statistics plain and simple. (4th ed.). Boston, MA: Cengage Learning.
I attach the previous essay so you have idea on how to do this assignment. It is similar to the assignment last week.
Assignment Content
1.
Top of Form
As you get closer to the final project in Week 6, you should have a better idea of the role of statistics in research. This week, you will calculate a one-way ANOVA for the independent groups. Reading and interpreting the output correctly is highly important. Most people who read research articles never see the actual output or data; they read the results statements by the researcher, which is why your summary must be accurate.
Consider your hypothesis statements you created in Part 2.
Calculate a one-way ANOVA, including a Tukey's HSD for the data from the Happiness and Engagement Dataset.
Write a 125- to 175-word summary of your interpretation of the results of the ANOVA, and describe how using an ANOVA was more advantageous than using multiple t tests to compare your independent variable on the outcome. Copy and paste your Microsoft® Excel® output below the summary.
Format your summary according to APA format.
Submit your summary, including the Microsoft® Excel® output to the assignment.
Reference/Module:
Module 13: Comparing More Than Two Groups
Using Designs with Three or More Levels of an Independent Variable
Comparing More than Two Kinds of Treatment in One Study
Comparing Two or More Kinds of Treatment with a Control Group
Comparing a Placebo Group to the Control and Experimental Groups
Analyzing the Multiple-Group Design
One-Way Between-Subjects ANOVA: What It Is and What It Does
Review of Key Terms
Module Exercises
Critical Thinking Check AnswersModule 14: One-Way Between-Subjects Analysis of Variance (ANOVA)
Calculations for the One-Way Between-Subjects ANOVA
Interpreting the One-Way Between-Subjects ANOVA
Graphing the Means and Effect Size
Assumptions of the One-Way Between-Subjects ANOVA
Tukey's Post Hoc Test
Review of Key Terms
Module Exercises
Critical Thinking Check AnswersChapter 7 Summary and ReviewChapter 7 Statistical Software Resources
In this chapter, we discuss the common types of statistical analyses used with designs involving more than two groups. The inferential statistics discussed in this chapter differ from those presented in the previous two chapters. In Chapter 5, single samples were being compared to populations (z test and t test), and in Chapter 6, two independent or correlated samples were being compared. In this chapter, the statistics are designed to test differences between more than two equivalent groups of subjects.
Several factors influence which statistic should be used to analyze the data collected. For example, the type of data collected and the number of groups being compared must be considered. Moreover, the statistic used to analyze the data will vary depending on whether the study involves a between-subjects design (designs in ...
26 Ch. 3 Organizing and Graphing DataAssignment 2ME.docxeugeniadean34240
26 Ch. 3: Organizing and Graphing Data
Assignment 2
MEASURES OF CENTRAL TENDENCY
Fill in the blanks:
4.1. The score that repeats the most often in a distribution is called the ______.
4.2. The descriptive statistic used the most in inferential statistics as a measure of central tendency is the _________.
4.3. The measure of central tendency used with nominal scale data is the _______.
4.4. To find the mean of a sample, thethe sum of the scoresas is divided by ______.
Circle the correct answer:
4.5. In a positively skewed distribution, the majority of the scores cluster above/below the ________.
4.6. The mode and the mean have the same values in distributions that are normal/negatively skewed.
4.7. Distributions with few scores are more/less likely to have a mode than distributions with many scores.
Answer the following questions:
4.8. Which measure of central tendency would be the most appropriate for summarizing the following test scores? Explain your choice.
13, 14, 10, 38, 11, 12, 16, 15
4.9. What is the difference between and ? How are they related to each other?
4.10. A distribution of 10 scores has a mean of 6. Following are 9 scores of this distribution. Which score is missing (remember that the mean should be 6)?
4, 8, 10, 5, 9, 3, 6, 7, 3
4.11. When the sum of a group of scores is 280 and the mean of the scores is 7, how many scores are in the distribution?
4.12. Find the mode, median, and mean of the distribution depicted in the following histogram:
Frequency
Scores
MEASURES OF VARIABILITY
Circle the correct answer:
5.1. The distance between the highest and the lowest scores is called the range/variance.
5.2. The SD is equal to the square root of the mean/variance.
5.3. A test with 30 items is likely to have a higher/lower standard deviation that a test with 80 items.
5.4. The mean of the squared deviation scores is called the variance/standard deviation.
5.5. The SD of the number of errors found by an auditor in a sample of accounts of one company is likely to be higher/lower than the SD of the number of errors found in samples taken from a number of different companies.
5.6. The SD is/is not sensitive to extreme scores.
5.7. The variance of the population is represented by S2/2.
5.8. In most cases, the variance is larger/smaller than the SD.
5.9. The measure of variability that takes into consideration every score in the distribution is the range/standard deviation.
Answer/compute the following questions:
5.10. Study the following three distributions. What are the similarities and differences between the three distributions in terms of their means, ranges, and standard deviations? (Note: Assume the three distributions to be samples if you decide to compute their standard deviations.)
Distribution A: 8, 9, 6, 12, 5
Distribution B: 7, 10, 11, 8, 4
Distribution C: 7, 9, 8, 9, 7
5.11. Three statistics classes (Sections A, B and C), each with 26 students, took the same test. The SD o.
Please Subscribe to this Channel for more solutions and lectures
http://www.youtube.com/onlineteaching
Chapter 9: Inferences from Two Samples
9.4: Two Variances or Standard Deviations
Week 5 Lecture 14 The Chi Square Test Quite often, pat.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).
In a left-tailed test comparing two means with variances unknown b.docxbradburgess22840
In a left-tailed test comparing two means with variances unknown but assumed to be equal, the sample sizes were n1 = 8 and n2 = 12. At α = .05, the critical value would be:
-1.645
-2.101
-1.734
-1.960
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(.
Predicting breast cancer: Adrian VallesAdrián Vallés
Performed and compared predictive modelling approaches (classification tree, logistic regression and random forest) to predict benign vs malignant breast cancers using R for the Data mining class (BANA 4080)
StarCompliance is a leading firm specializing in the recovery of stolen cryptocurrency. Our comprehensive services are designed to assist individuals and organizations in navigating the complex process of fraud reporting, investigation, and fund recovery. We combine cutting-edge technology with expert legal support to provide a robust solution for victims of crypto theft.
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At StarCompliance, we understand the urgency and stress involved in dealing with cryptocurrency theft. Our dedicated team works quickly and efficiently to provide you with the support and expertise needed to recover your assets. Trust us to be your partner in navigating the complexities of the crypto world and safeguarding your investments.
Chatty Kathy - UNC Bootcamp Final Project Presentation - Final Version - 5.23...John Andrews
SlideShare Description for "Chatty Kathy - UNC Bootcamp Final Project Presentation"
Title: Chatty Kathy: Enhancing Physical Activity Among Older Adults
Description:
Discover how Chatty Kathy, an innovative project developed at the UNC Bootcamp, aims to tackle the challenge of low physical activity among older adults. Our AI-driven solution uses peer interaction to boost and sustain exercise levels, significantly improving health outcomes. This presentation covers our problem statement, the rationale behind Chatty Kathy, synthetic data and persona creation, model performance metrics, a visual demonstration of the project, and potential future developments. Join us for an insightful Q&A session to explore the potential of this groundbreaking project.
Project Team: Jay Requarth, Jana Avery, John Andrews, Dr. Dick Davis II, Nee Buntoum, Nam Yeongjin & Mat Nicholas
Levelwise PageRank with Loop-Based Dead End Handling Strategy : SHORT REPORT ...Subhajit Sahu
Abstract — Levelwise PageRank is an alternative method of PageRank computation which decomposes the input graph into a directed acyclic block-graph of strongly connected components, and processes them in topological order, one level at a time. This enables calculation for ranks in a distributed fashion without per-iteration communication, unlike the standard method where all vertices are processed in each iteration. It however comes with a precondition of the absence of dead ends in the input graph. Here, the native non-distributed performance of Levelwise PageRank was compared against Monolithic PageRank on a CPU as well as a GPU. To ensure a fair comparison, Monolithic PageRank was also performed on a graph where vertices were split by components. Results indicate that Levelwise PageRank is about as fast as Monolithic PageRank on the CPU, but quite a bit slower on the GPU. Slowdown on the GPU is likely caused by a large submission of small workloads, and expected to be non-issue when the computation is performed on massive graphs.
Levelwise PageRank with Loop-Based Dead End Handling Strategy : SHORT REPORT ...
Posthoc
1. 1
Post Hoc Tests in ANOVA
This handout provides information on the use of post hoc tests in the Analysis of Variance
(ANOVA). Post hoc tests are designed for situations in which the researcher has already obtained
a significant omnibus F-test with a factor that consists of three or more means and additional
exploration of the differences among means is needed to provide specific information on which
means are significantly different from each other.
For example, the data file “posthoc.por” (available on the web site), contains two factors, gender
and experience and one dependent measure, spatial ability score errors. Applying the GLM-
Unianova procedure in SPSS produces the following ANOVA source table:
Tests of Between-Subjects Effects
Dependent Variable: Spatial Ability Score Errors
Source Type III
Sum of
Squares
df Mean
Square
F Sig.
Corrected
Model
102.575 7 14.654 15.029 .000
Intercept 403.225 1 403.225 413.564 .000
GENDER 60.025 1 60.025 61.564 .000
EXPER 28.275 3 9.425 9.667 .000
GENDER *
EXPER
14.275 3 4.758 4.880 .007
Error 31.200 32 .975
Total 537.000 40
Corrected
Total
133.775 39
a R Squared = .767 (Adjusted R Squared = .716)
Inspection of the source table shows that both the main effects and the interaction effect are
significant. The gender effect can be interpreted directly since there are only two levels of the
factor. Interpretation of either the Experience main effect or the Gender by Experience interaction
is ambiguous, however, since there are multiple means in each effect. We will delay testing and
interpretation of the interaction effect for a later handout. The concern now is how to determine
which of the means for the four Experience groups (see table below) are significantly different
from the others.
The first post hoc, the LSD test. T he original solution to this problem, developed by Fisher,
was to explore all possible pair-wise comparisons of means comprising a factor using the
equivalent of multiple t-tests. This procedure was named the Least Significant Difference (LSD)
test. The least significant difference between two means is calculated by:
_________
LSD = t o 2MSE / n*
where t is the critical, tabled value of the t-distribution with the df associated with MSE, the
2. 2
denominator of the F statistic and n* is the number of scores used to calculate the means of
criticalinterest. In our example, t at " = .05, two-tailed, with df = 32 is 2.0369, MSE from the
source table above is 0.975, and n* is 10 scores per mean.
_________ __________
LSD = t o 2MSE / n* = 2.0369 o 2 (.975 / 10 = 0.6360
So the LSD or minimum difference between a pair of means necessary for statistical significance is
0.636.
In order to compare this critical value or difference for all our means it is useful to organize the
means in a table. First, the number of pair-wise comparisons among means can be calculated using
the formula: k(k-1)/2, where k = the number of means or levels of the factor being tested. In our
present example, the experience factor has four levels so k = 4 and there are k(k-1)/2 = 4(3)/2 = 6
unique pairs of means that can be compared.
Experience with Mechanical Problems
Dependent Variable: Spatial Ability Score Errors
Mean Std. Error 95%
Confidence
Interval
Experience Lower Bound Upper Bound
A lot 2.100 .312 1.464 2.736
Fair Amount 2.700 .312 2.064 3.336
Some 3.600 .312 2.964 4.236
Little to none 4.300 .312 3.664 4.936
The table we will construct is a table showing the obtained means on the rows and columns and
subtracted differences between each pair of means in the interior cells producing a table of
absolute mean differences to use in evaluating the post hoc tests. To construct the table follow
these steps: 1) rank the means from largest to smallest, 2) create table rows beginning with the
largest mean and going through the next-to-smallest mean, 3) create table columns starting with
the smallest mean and going through the next-to-largest mean, 4) compute the absolute difference
between each row and column intersection/mean. In the present example this results in the table of
absolute mean differences below:
2.1 2.7 3.6
4.3 2.2 1.6 0.7
3.6 1.5 0.9
2.7 0.6
Now applying our LSD value of .636 to the mean differences in the table, it can be seen that all
differences among the means are significant at " = .05 except the last difference between the
3. 3
means of 2.1 and 2.7. Unfortunately, the p-values associated with these multiple LSD tests are
inaccurate. Since the sampling distribution for t assumes only one t-test from any given sample,
substantial alpha slippage has occurred because 6 tests have been performed on the same sample.
The true alpha level given multiple tests or comparisons can be estimated as 1 - (1 - " ) , where c
c
= the total number of comparisons, contrasts, or tests performed. In the present example
1 - (1 - " ) = 1 - (1 - .05) = .2649. Given multiple testing in this situation, the true value of
c 6
alpha is approximately .26 rather than .05.
A number of different solutions and corrections have been developed to deal with this problem
and produce post hoc tests that correct for multiple tests so that a correct alpha level is
maintained even though multiple tests or comparisons are computed. Several of these approaches
are discussed below.
Tukey’s HSD test. Tukey’s test was developed in reaction to the LSD test and studies have
shown the procedure accurately maintains alpha levels at their intended values as long as
statistical model assumptions are met (i.e., normality, homogeneity, independence). Tukey’s HSD
was designed for a situation with equal sample sizes per group, but can be adapted to unequal
sample sizes as well (the simplest adaptation uses the harmonic mean of n-sizes as n*). The
formula for Tukey’s is:
_________
HSD = q o MSE / n*
where q = the relevant critical value of the studentized range statistic and n* is the number of
scores used in calculating the group means of interest. Calculation of Tukey’s for the present
example produces the following:
_________ ________
HSD = q o MSE / n* = 3.83 o .975 / 10 = 1.1957
The q value of 3.83 is obtained by reference to the studentized range statistic table looking up the
q value for an alpha of .05, df = < = 32, k = p = r = 4. Thus the differences in the table of mean
differences below that are marked by the asterisks exceed the HSD critical difference and are
significant at p < .05. Note that two differences significant with LSD are now not significant.
2.1 2.7 3.6
4.3 2.2* 1.6* 0.7
3.6 1.5* 0.9
2.7 0.6
Scheffe’s test. Scheffe’s procedure is perhaps the most popular of the post hoc procedures, the
most flexible, and the most conservative. Scheffe’s procedure corrects alpha for all pair-wise or
simple comparisons of means, but also for all complex comparisons of means as well. Complex
comparisons involve contrasts of more than two means at a time. As a result, Scheffe’s is also the