1. The document discusses Chi-Square tests and how they can be used to analyze relationships between nominal variables through frequency data. Chi-Square tests the independence of two variables by comparing observed and expected frequencies.
2. The steps of a Chi-Square test are outlined: determine the test, establish a significance level, state the hypotheses, calculate the test statistic by summing the squared differences between observed and expected frequencies, determine the degrees of freedom, and compare the test statistic to a critical value.
3. An example compares the voting preferences of Democrats and Republicans on gun control. The Chi-Square test statistic is calculated and found to exceed the critical value, so the null hypothesis of independence is rejected.
This article is used to give a basic information regarding the change points that occur in excel and in other files. The detection methods are proposed and they are analyzed with a real time example. The features and application of the change point is also discussed in the later. Copy the link given below and paste it in new browser window to get more information on Change Point:- http://www.transtutors.com/homework-help/statistics/change-point.aspx
Categorical Data and Statistical AnalysisMichael770443
In this presentation, we will introduce two tests and hypothesis testing based on it, and different non-parametric methods such as the Kolmogorov-Smirnov test, the Wilcoxon’s signed-rank test, the Mann-Whitney U test, and the Kruskal-Wallis test.
A chi-squared test is a statistical hypothesis test that is valid to perform when the test statistic is chi-squared distributed under the null hypothesis, specifically Pearson's chi-squared test and variants
differences between the observed values
This article is used to give a basic information regarding the change points that occur in excel and in other files. The detection methods are proposed and they are analyzed with a real time example. The features and application of the change point is also discussed in the later. Copy the link given below and paste it in new browser window to get more information on Change Point:- http://www.transtutors.com/homework-help/statistics/change-point.aspx
Categorical Data and Statistical AnalysisMichael770443
In this presentation, we will introduce two tests and hypothesis testing based on it, and different non-parametric methods such as the Kolmogorov-Smirnov test, the Wilcoxon’s signed-rank test, the Mann-Whitney U test, and the Kruskal-Wallis test.
A chi-squared test is a statistical hypothesis test that is valid to perform when the test statistic is chi-squared distributed under the null hypothesis, specifically Pearson's chi-squared test and variants
differences between the observed values
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Chapter 11: Goodness-of-Fit and Contingency Tables
11.1: Goodness of Fit Notation
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
Statistics practice for finalBe sure to review the following.docxdessiechisomjj4
Statistics practice for final
Be sure to review the following and have this information handy when taking Final GHA:
· Calculating z alpha/2 and t alpha/2 on Tables II and IV
· Find sample size for estimating population mean. Formula 8.3 p. 321 OCR.
· Stating H0 and H1 claims about variation and about the mean. Chapter 9 OCR.
· Type I and Type II errors p. 345 OCR.
· Confidence Interval for difference between two population means. Chapter 10 OCR p. 428
· Pooled sample standard deviation. Chapter 10 OCR p. 397
· What do Chi-Square tests measure? How are their degrees of freedom calculated? Chapter 12 OCR
· Find F test statistic using One-Way ANOVA.xls Be sure to enable editing and change values to match your problem. One-Way ANOVA.xls
· Find equation of regression line used to predict. To do on Excel, go to a blank worksheet, enter x values in one column and their matching y values in another column. Click Insert – Select Scatterplot. Right click any one of the points (diamonds) on the graph. Left click “Add a Trendline.” Check “Display Equation on Chart” box. Regression equation will appear on chart. Try it here with No. 20 below.
Practice Problems
Chapter 8 Final Review
1) In which of the following situations is it reasonable to use the z-interval
procedure to obtain a confidence interval for the population mean?
Assume that the population standard deviation is known.
A. n = 10, the data contain no outliers, the variable under consideration is
not normally distributed.
B. n = 10, the variable under consideration is normally distributed.
C. n = 18, the data contain no outliers, the variable under consideration is
far from being normally distributed.
D. n = 18, the data contain outliers, the variable under consideration is
normally distributed.
Find the necessary sample size.
2) The weekly earnings of students in one age group are normally
distributed with a standard deviation of 10 dollars. A researcher wishes to
estimate the mean weekly earnings of students in this age group. Find the
sample size needed to assure with 95 percent confidence that the sample
mean will not differ from the population mean by more than 2 dollars.
Find the specified t-value.
3) For a t-curve with df = 6, find the two t-values that divide the area under
the curve into a middle 0.99 area and two outside areas of 0.005.
Provide an appropriate response.
4) Under what conditions would you choose to use the t-interval procedure
instead of the z-interval procedure in order to obtain a confidence
interval for a population mean? What conditions must be satisfied in
order to use the t-interval procedure?
CHAPTER 8 Answers
1) B
2) 97
3) -3.707, 3.707
4) When the population standard deviation is unknown, the t-interval procedure is used instead of the
z-interval procedure. The t-interval procedure works provided that the population is normally
distributed or the.
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2. 2
Different Scales, Different Measures of
Association
Scale of Both
Variables
Measures of
Association
Nominal Scale Pearson Chi-Square:
χ2
Ordinal Scale Spearman’s rho
Interval or Ratio
Scale
Pearson r
3. 3
Chi-Square (χ2
) and Frequency Data
Up to this point, the inference to the population has been
concerned with “scores” on one or more variables, such as
CAT scores, mathematics achievement, and hours spent on
the computer.
We used these scores to make the inferences about
population means. To be sure not all research questions
involve score data.
Today the data that we analyze consists of frequencies;
that is, the number of individuals falling into categories. In
other words, the variables are measured on a nominal
scale.
The test statistic for frequency data is Pearson Chi-Square.
The magnitude of Pearson Chi-Square reflects the amount
of discrepancy between observed frequencies and expected
frequencies.
4. 4
Steps in Test of Hypothesis
1. Determine the appropriate test
2. Establish the level of significance:α
3. Formulate the statistical hypothesis
4. Calculate the test statistic
5. Determine the degree of freedom
6. Compare computed test statistic against a
tabled/critical value
5. 5
1. Determine Appropriate Test
Chi Square is used when both variables are
measured on a nominal scale.
It can be applied to interval or ratio data that have
been categorized into a small number of groups.
It assumes that the observations are randomly
sampled from the population.
All observations are independent (an individual
can appear only once in a table and there are no
overlapping categories).
It does not make any assumptions about the shape
of the distribution nor about the homogeneity of
variances.
6. 6
2. Establish Level of Significance
α is a predetermined value
The convention
• α = .05
• α = .01
• α = .001
7. 7
3. Determine The Hypothesis:
Whether There is an Association
or Not
Ho : The two variables are independent
Ha : The two variables are associated
8. 8
4. Calculating Test Statistics
Contrasts observed frequencies in each cell of a
contingency table with expected frequencies.
The expected frequencies represent the number of
cases that would be found in each cell if the null
hypothesis were true ( i.e. the nominal variables
are unrelated).
Expected frequency of two unrelated events is
product of the row and column frequency divided
by number of cases.
Fe= FrFc / N
12. 12
6. Compare computed test statistic
against a tabled/critical value
The computed value of the Pearson chi-
square statistic is compared with the critical
value to determine if the computed value is
improbable
The critical tabled values are based on
sampling distributions of the Pearson chi-
square statistic
If calculated χ2
is greater than χ2
table value,
reject Ho
13. 13
Example
Suppose a researcher is interested in voting
preferences on gun control issues.
A questionnaire was developed and sent to
a random sample of 90 voters.
The researcher also collects information
about the political party membership of the
sample of 90 respondents.
14. 14
Bivariate Frequency Table or
Contingency Table
Favor Neutral Oppose f row
Democrat 10 10 30 50
Republican 15 15 10 40
f column 25 25 40 n = 90
15. 15
Bivariate Frequency Table or
Contingency Table
Favor Neutral Oppose f row
Democrat 10 10 30 50
Republican 15 15 10 40
f column 25 25 40 n = 90
Observed
frequencies
16. 16
Bivariate Frequency Table or
Contingency Table
Favor Neutral Oppose f row
Democrat 10 10 30 50
Republican 15 15 10 40
f column 25 25 40 n = 90
Rowfrequency
17. 17
Bivariate Frequency Table or
Contingency Table
Favor Neutral Oppose f row
Democrat 10 10 30 50
Republican 15 15 10 40
f column 25 25 40 n = 90
Column frequency
18. 18
1. Determine Appropriate Test
1. Party Membership ( 2 levels) and Nominal
2. Voting Preference ( 3 levels) and Nominal
20. 20
3. Determine The Hypothesis
• Ho : There is no difference between D & R
in their opinion on gun control issue.
• Ha : There is an association between
responses to the gun control survey and the
party membership in the population.
21. 21
4. Calculating Test Statistics
Favor Neutral Oppose f row
Democrat fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Republican fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 n = 90
22. 22
4. Calculating Test Statistics
Favor Neutral Oppose f row
Democrat fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Republican fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 n = 90
= 50*25/90
23. 23
4. Calculating Test Statistics
Favor Neutral Oppose f row
Democrat fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Republican fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 n = 90
= 40* 25/90
27. 27
6. Compare computed test statistic
against a tabled/critical value
α = 0.05
df = 2
Critical tabled value = 5.991
Test statistic, 11.03, exceeds critical value
Null hypothesis is rejected
Democrats & Republicans differ
significantly in their opinions on gun
control issues
28. 28
Example Crosstab: Gender x
Student
Student Not Student Total
Males
46 71
117
Females
37 83
120
Total 83 154 237