1. Other Chi-Square Tests
C H A P T E R 1 1
Other Chi-Square Tests
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2. Copyright © 2012 The McGraw-Hill Companies, Inc.
Other Chi-Square Tests
CHAPTER 11
1.1
Outline
Descriptive and Inferential Statistics
11-1
Test for Goodness of Fit
11-2
Tests Using Contingency Tables
Copyright © 2015 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Copyright © 2012 The McGraw-Hill Companies, Inc.
Slide 2

3. Learning Objectives
1.1 1
Test a distribution for goodness of fit, using chi-square. 2
Test two variables for independence, using chi-square. 3
Test proportions for homogeneity, using chi-square.
Descriptive and inferential statistics
Copyright © 2015 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

4. 11.1 Test for Goodness of Fit
The chi-square statistic can be used to see whether a frequency distribution fits a specific pattern. This is referred to as the chi-square goodness-of-fit test.
Bluman Chapter 11

5. Observed & Expected Frequencies
Since the frequencies for each flavor were obtained from a sample, these actual frequencies are called the observed frequencies.
The frequencies obtained by calculation (as if there were no preference) are called the expected frequencies.
Bluman Chapter 11

6. Test for Goodness of Fit
Formula for the test for goodness of fit:
where
d.f. = number of categories minus 1
O = observed frequency
E = expected frequency
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7. Assumptions for Goodness of Fit
The data are obtained from a random sample.
The expected frequency for each category must be 5 or more.
Bluman Chapter 11

8. Chapter 11 Other Chi-Square Tests
Section 11-1
Example 11-1
Page #612
Bluman Chapter 11

9. Example 11-1: Fruit Soda Flavors
A market analyst wished to see whether consumers have any preference among five flavors of a new fruit soda. A sample of 100 people provided the following data. Is there enough evidence to reject the claim that there is no preference in the selection of fruit soda flavors, using the data shown previously? Let α = 0.05.
Cherry
Strawberry
Orange
Lime
Grape
Observed 32 28 16 14 10
Expected 20
Step 1: State the hypotheses and identify the claim.
H0: Consumers show no preference (claim).
H1: Consumers show a preference.
Bluman Chapter 11

10. Example 11-1: Fruit Soda Flavors
Cherry
Strawberry
Orange
Lime
Grape
Observed 32 28 16 14 10
Expected 20
Step 2: Find the critical value.
D.f. = 5 – 1 = 4, and α = CV =
Step 3: Compute the test value.
Bluman Chapter 11

11. Example 11-1: Fruit Soda Flavors
Step 4: Make the decision.
The decision is to reject the null hypothesis, since 18.0 >
Step 5: Summarize the results.
There is enough evidence to reject the claim that consumers show no preference for the flavors.
Bluman Chapter 11

12. Chapter 11 Other Chi-Square Tests
Section 11-1
Example 11-2
Page #614
Bluman Chapter 11

13. Example 11-2: Education Level of Adults
The Census Bureau of the U.S. government found that 13% of adults did not finish high school, 30% graduated from high school only, 29% had some college education but did not obtain a bachelor’s degree, and 28% were college graduates. To see if these proportions were consistent with those people who lived in the Lincoln County area, a local researcher selected a random sample of 300 adults and found that 43 did not finish high school, 76 were high school graduates only, 96 had some college education, and 85 were college graduates. At α = 0.10, test the claim that the proportions are the same for the adults in Lincoln County as those stated by the Census Bureau.
Bluman Chapter 11

14. Example 11-2: Education Level of Adults
Step 1: State the hypotheses and identify the claim.
H0: The proportion of people in each category is as follows: 13% did not finish high school, 30% were high school graduates only, 29% had some college education but did not graduate, and 28% had a college degree (claim).
H1: The distribution is not the same as stated in the null hypothesis.
Step 2: Find the critical value.
Since α = 0.10 and the degrees of freedom are
4 – 1 = 3, the critical value is
Bluman Chapter 11

15. Example 11-2: Education Level of Adults
Step 3: Compute the test value.
First, we must calculate the expected values. Multiply the total number of people surveyed (300) by the percentages of people in each category.
Bluman Chapter 11

16. Example 11-2: Education Level of Adults
Step 4: Make the decision. Since 3.531˂ 6.251, the decision is not to reject the null hypothesis. See Figure 11–4.
Step 5: Summarize the results. There is not enough evidence to reject the claim. It can be concluded that the percentages are not significantly different from those given in the null hypothesis. That is, the proportions are not significantly different from those stated by the U.S. Census Bureau.
Bluman Chapter 11

17. Chapter 11 Other Chi-Square Tests
Section 11-1
Example 11-3
Page #615
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18. Example 11-3: Firearm Deaths
A researcher read that firearm-related deaths for people aged 1 to 18 were distributed as follows: 74% were accidental, 16% were homicides, and 10% were suicides. In her district, there were 68 accidental deaths, 27 homicides, and 5 suicides during the past year. At α = 0.10, test the claim that the percentages are equal.
Accidental
Homicides
Suicides
Observed 68 27 5
Expected 74 16 10
Bluman Chapter 11

19. Example 11-3: Firearm Deaths
Accidental
Homicides
Suicides
Observed 68 27 5
Expected 74 16 10
Step 1: State the hypotheses and identify the claim.
H0: Deaths due to firearms for people aged 1 through 18 are distributed as follows: 74% accidental, 16% homicides, and 10% suicides (claim).
H1: The distribution is not the same as stated in the null hypothesis.
Bluman Chapter 11

20. Example 11-3: Firearm Deaths
Accidental
Homicides
Suicides
Observed 68 27 5
Expected 74 16 10
Step 2: Find the critical value.
D.f. = 3 – 1 = 2, and α = CV =
Step 3: Compute the test value.
Bluman Chapter 11

21. Example 11-3: Firearm Deaths
Step 4: Make the decision.
Reject the null hypothesis, since >
Step 5: Summarize the results.
There is enough evidence to reject the claim that the distribution is 74% accidental, 16% homicides, and 10% suicides.
Bluman Chapter 11

22. Test for Normality (Optional)
The chi-square goodness-of-fit test can be used to test a variable to see if it is normally distributed.
The hypotheses are:
H0: The variable is normally distributed.
H1: The variable is not normally distributed.
The procedure is somewhat complicated. The calculations are shown in example 11-4 on page 616 in the text.
Bluman Chapter 11

23. 11.2 Tests Using Contingency Tables
When data can be tabulated in table form in terms of frequencies, several types of hypotheses can be tested by using the chi-square test.
The test of independence of variables is used to determine whether two variables are independent of or related to each other when a single sample is selected.
The test of homogeneity of proportions is used to determine whether the proportions for a variable are equal when several samples are selected from different populations.
Bluman Chapter 11

24. Test for Independence
The chi-square goodness-of-fit test can be used to test the independence of two variables.
The hypotheses are:
H0: There is no relationship between two variables.
H1: There is a relationship between two variables.
If the null hypothesis is rejected, there is some relationship between the variables.
Bluman Chapter 11

25. Test for Independence
In order to test the null hypothesis, one must compute the expected frequencies, assuming the null hypothesis is true.
When data are arranged in table form for the independence test, the table is called a contingency table.
Bluman Chapter 11

26. Contingency Tables
The degrees of freedom for any contingency table are d.f. = (rows – 1) (columns – 1) = (R – 1)(C – 1).
Bluman Chapter 11

27. Test for Independence The formula for the test for independence: where
d.f. = (R – 1)(C – 1)
O = observed frequency
E = expected frequency =
Bluman Chapter 11

28. Chapter 11 Other Chi-Square Tests
Section 11-2
Example 11-5
Page #627
Bluman Chapter 11

29. Example 11-5: Hospitals and Infections
A researcher wishes to see if there is a relationship between the hospital and the number of patient infections. A sample of 3 hospitals was selected, and the number of infections for a specific year has been reported. The data are shown next.
Bluman Chapter 11

30. Example 11-5: Hospitals and Infections
Step 1: State the hypotheses and identify the claim.
H0: The number of infections is independent of the hospital.
H1: The number of infections is dependent on the hospital (claim).
Step 2: Find the critical value.
The critical value at α = 0.05 with (3 – 1)(3 – 1) =
(2)(2) = 4 degrees of freedom is
Bluman Chapter 11

31. Example 11-5: Hospitals and Infections
Step 3: Compute the test value. First find the expected values.
Bluman Chapter 11

32. Example 11-5: Hospitals and Infections
Bluman Chapter 11

33. Example 11-5: Hospitals and Infections
Bluman Chapter 11

34. Example 11-5: Hospitals and Infections
Step 4: Make the decision.
The decision is to reject the null hypothesis since
>
Bluman Chapter 11

35. Example 11-5: Hospitals and Infections
Step 5: Summarize the results.
There is enough evidence to support the claim that the number of infections is related to the hospital where they occurred.
Bluman Chapter 11

36. Chapter 11 Other Chi-Square Tests
Section 11-2
Example 11-6
Page #629
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37. Example 11-6: Sports Preference
A researcher wished to see if there is a difference in the favorite sport of males and the favorite sport of females. She selected a sample of 32 males and 48 females and asked them which of three sports was their favorite. The results are shown.
Bluman Chapter 11

38. Example 11-6: Sports Preference
Step 1: State the hypotheses and identify the claim.
H0: Sports preference is independent of the gender of the person.
H1: Sports preference is related to the gender of the person (claim).
Step 2: Find the critical value.
Find the critical value. The critical value is 4.605, since the degrees of freedom are (2 – 1)(3 – 1)= 2.
Bluman Chapter 11

39. Example 11-6: Sports Preference
Compute the expected values.
Gender
Alcohol Consumption
Total
Low
Moderate
High
Male 10 9 8 27
Female 13 16 12 41 23 25 20 68
(9.13)
(9.93)
(7.94)
(13.87)
(15.07)
(12.06)
Bluman Chapter 11

40. Example 11-6: Sports Preference
Step 3: Compute the test value.
Bluman Chapter 11

41. Example 11-6: Sports Preference
Step 4: Make the decision.
Do not reject the null hypothesis, since <
Step 5: Summarize the results.
Summarize the results. There is not enough evidence to support the claim that sports preference is related to the gender of the person.
Bluman Chapter 11

42. Test for Homogeneity of Proportions
Homogeneity of proportions test is used when samples are selected from several different populations and the researcher is interested in determining whether the proportions of elements that have a common characteristic are the same for each population.
Bluman Chapter 11

43. Test for Homogeneity of Proportions
The hypotheses are:
H0: p1= p2= p3 =p4
H1: At least one proportion is different from the others.
When the null hypothesis is rejected, it can be assumed that the proportions are not all equal.
Bluman Chapter 11

44. Assumptions for Homogeneity of Proportions
The data are obtained from a random sample.
The expected frequency for each category must be 5 or more.
Bluman Chapter 11

45. Chapter 11 Other Chi-Square Tests
Section 11-2
Example 11-7
Page #631
Bluman Chapter 11

46. Example 11-7: Money and Happiness
A psychologist randomly selected 100 people from each of four income groups and asked them if they were “very happy.” For people who made less than $30,000, 24% responded yes. For people who made $30,000 to $74,999, 33% responded yes. For people who made $75,000 to $90,999, 38% responded yes, and for people who made $100,000 or more, 49% responded yes. At α=0.05, test the claim that there is no difference in the proportion of people in each economic group who were very happy.
Bluman Chapter 11

47. Example 11-7: Money and Happiness
Step 1: State the hypotheses.
H0: p1 = p2 = p3 = p4 (claim)
H1: At least one mean differs from the other.
Step 2: Find the critical value.
Find the critical value. The formula for the degrees of freedom is the same as before:
(R – 1)(C – 1) = (2 – 1)(4 – 1) = 1(3) = 3.
The critical value is
Bluman Chapter 11

48. Example 11-7: Money and Happiness
Step 3: Compute the test value.
Since we want to test the claim that the proportions are equal, we use the expected value
as and the formula
Bluman Chapter 11

49. Example 11-7: Money and Happiness
Step 3: Compute the test value.
Bluman Chapter 11

50. Example 11-7: Money and Happiness
Step 4: Make the decision.
Reject the null hypothesis since >
Bluman Chapter 11

51. Example 11-7: Money and Happiness
Step 5: Summarize the results.
There is enough evidence to reject the claim that there is no difference in the proportions. Hence the incomes seem to make a difference in the proportions.
Bluman Chapter 11