Download presentation

1
CHAPTER 11: CHI-SQUARE TESTS

2
**THE CHI-SQUARE DISTRIBUTION**

Definition The chi-square distribution has only one parameter called the degrees of freedom. The shape of a chi-squared distribution curve is skewed to the right for small df and becomes symmetric for large df. The entire chi-square distribution curve lies to the right of the vertical axis. The chi-square distribution assumes nonnegative values only, and these are denoted by the symbol χ2 (read as “chi-square”).

3
**Figure 11.1 Three chi-square distribution curves.**

4
Example 11-1 Find the value of χ² for 7 degrees of freedom and an area of .10 in the right tail of the chi-square distribution curve.

5
**Table 11.1 χ2 for df = 7 and .10 Area in the Right Tail**

Area in the Right Tail Under the Chi-Square Distribution Curve df .995 … .100 .005 1 2 . 7 100 .000 .010 .989 67.328 2.706 4.605 12.017 7.879 10.597 20.278 Required value of χ²

6
Figure 11.2 df = 7 .10 12.017 χ²

7
Example 11-2 Find the value of χ² for 12 degrees of freedom and area of .05 in the left tail of the chi-square distribution curve.

8
Solution 11-2 Area in the right tail = 1 – Area in the left tail = 1 – .05 = .95

9
**Table 11.2 χ2 for df = 12 and .95 Area in the Right Tail**

Area in the Right Tail Under the Chi-Square Distribution Curve df .995 … .950 .005 1 2 . 12 100 .000 .010 3.074 67.328 .004 .103 5.226 77.929 7.879 10.597 28.300 Required value of χ²

10
Figure 11.3 df = 12 Shaded area = .95 .05 5.226 χ²

11
**A GOODNESS-OF-FIT TEST**

Definition An experiment with the following characteristics is called a multinomial experiment.

12
**Multinomial Experiment cont.**

It consists of n identical trials (repetitions). Each trial results in one of k possible outcomes (or categories), where k > 2. The trials are independent. The probabilities of the various outcomes remain constant for each trial.

13
**A GOODNESS-OF-FIT TEST cont.**

Definition The frequencies obtained from the performance of an experiment are called the observed frequencies and are denoted by O. The expected frequencies, denoted by E, are the frequencies that we expect to obtain if the null hypothesis is true. The expected frequency for a category is obtained as E = np Where n is the sample size and p is the probability that an element belongs to that category if the null hypothesis is true.

14
**A GOODNESS-OF-FIT TEST cont.**

Degrees of Freedom for a Goodness-of-Fit Test In a goodness-of-fit test, the degrees of freedom are df = k – 1 where k denotes the number of possible outcomes (or categories) for the experiment.

15
**Test Statistic for a Goodness-of-Fit Test**

The test statistic for a goodness-of-fit test is χ2 and its value is calculated as where O = observed frequency for a category E = expected frequency for a category = np Remember that a chi-square goodness-of-fit test is always right-tailed.

16
Example 11-3 A bank has an ATM installed inside the bank, and it is available to its customers only from 7 AM to 6 PM Monday through Friday. The manager of the bank wanted to investigate if the percentage of transactions made on this ATM is the same for each of the five days (Monday through Friday) of the week. She randomly selected one week and counted the number of transactions made on this ATM on each of the five days during this week. The information she obtained is given in the following table, where the number of users represents the number of transactions on this ATM on these days. For convenience, we will refer to these transactions as “people” or “users.”

17
Example 11-3 At the 1% level of significance, can we reject the null hypothesis that the proportion of people who use this ATM each of the five days of the week is the same? Assume that this week is typical of all weeks in regard to the use of this ATM. Day Monday Tuesday Wednesday Thursday Friday Number of users 253 197 204 179 267

18
**Solution 11-3 H0 : p1 = p2 = p3 = p4 = p5 = .20**

H1 : At least two of the five proportions are not equal to .20

19
**Solution 11.3 There are five categories**

Five days on which the ATM is used Multinomial experiment We use the chi-square distribution to make this test.

20
**Solution 11-3 Area in the right tail = α = .01**

k = number of categories = 5 df = k – 1 = 5 – 1 = 4 The critical value of χ2 =

21
**Figure 11.4 α = .01 χ2 Do not reject H0 Reject H0 13.277**

Critical value of χ2

22
**Table 11.3 O p E = np Category (Day) Observed Frequency**

Expected Frequency E = np (O – E) (O – E)2 Monday Tuesday Wednesday Thursday Friday 253 197 204 279 267 .20 1200(.20) = 240 13 -43 -36 39 27 169 1849 1296 1521 729 .704 7.704 5.400 6.338 3.038 n = 1200 Sum =

23
Solution 11-3 All the required calculations to find the value of the test statistic χ2 are shown in Table 11.3.

24
Solution 11.3 The value of the test statistic χ2 = is larger than the critical value of χ2 = It falls in the rejection region Hence, we reject the null hypothesis

25
Example 11-4 In a National Public Transportation survey conducted in 1995 on the modes of transportation used to commute to work, 79.6% of the respondents said that they drive alone, 11.1% car pool, 5.1% use public transit, and 4.2% depend on other modes of transportation (USA TODAY, April 14, 1999). Assume that these percentages hold true for the 1995 population of all commuting workers. Recently 1000 randomly selected workers were asked what mode of transportation they use to commute to work. The following table lists the results of this survey.

26
Example 11-4 Mode of transportation Drive alone Carpool Public transit Other Number of workers 812 102 57 29 Test at the 2.5% significance level whether the current pattern of use of transportation modes is different from that for 1995.

27
Solution 11-4 H0: The current percentage distribution of the use of transportation modes is the same as that for 1995. H1: The current percentage distribution of the use of transportation modes is different from that for 1995.

28
**Solution 11-4 There are four categories**

Drive alone, carpool, public transit, and other Multinomial experiment We use the chi-square distribution to make the test.

29
**Solution 11-4 Area in the right tail = α = .025**

k = number of categories = 4 df = k – 1 = 4 – 1 = 3 The critical value of χ2 = 9.348

30
**Figure 11.5 α = .025 χ2 Do not reject H0 Reject H0 9.348**

Critical value of χ2

31
**Table 11.4 O p E = np Category Observed Frequency Expected Frequency**

(O – E) (O – E)2 Drive alone Car pool Public transit Other 812 102 57 29 .796 .111 .051 .042 1000(.796) = 796 1000(.111) = 111 1000(.051) = 51 1000(.042) = 42 16 -9 6 -13 256 81 36 169 .322 .730 .706 4.024 n = 1000 Sum = 5.782

32
Solution 11-4 All the required calculations to find the value of the test statistic χ2 are shown in Table 11.4.

33
Solution 11-4 The value of the test statistic χ2 = is less than the critical value of χ2 = 9.348 It falls in the nonrejection region Hence, we fail to reject the null hypothesis.

34
**CONTINGENCY TABLES Table 11.5 Total 2002 Enrollment at a University**

Full-Time Part-Time Male 6768 2615 Female 7658 3717 Students who are male and enrolled part-time

35
**A TEST OF INDEPENDENCE OR HOMOGENEITY**

A Test of Homogeneity

36
**A Test of Independence Definition**

A test of independence involves a test of the null hypothesis that two attributes of a population are not related. The degrees of freedom for a test of independence are df = (R – 1)(C – 1) Where R and C are the number of rows and the number of columns, respectively, in the given contingency table.

37
**A Test of Independence cont.**

Test Statistic for a Test of Independence The value of the test statistic χ2 for a test of independence is calculated as where O and E are the observed and expected frequencies, respectively, for a cell.

38
Example 11-5 Violence and lack of discipline have become major problems in schools in the United States. A random sample of 300 adults was selected, and they were asked if they favor giving more freedom to schoolteachers to punish students for violence and lack of discipline. The two-way classification of the responses of these adults is represented in the following table.

39
Example 11-5 Calculate the expected frequencies for this table assuming that the two attributes, gender and opinions on the issue, are independent. In Favor (F) Against (A) No Opinions (N) Men (M) Women (W) 93 87 70 32 12 6

40
**Table 11.6 Solution 11-5 In Favor (F) Against (A) No Opinion (N)**

Row Totals Men (M) 93 70 12 175 Women (W) 87 32 6 125 Column Totals 180 102 18 300

41
**Expected Frequencies for a Test of Independence**

The expected frequency E for a cell is calculated as

42
**Table 11.7 Solution 11-5 In Favor (F) Against (A) No Opinion (O)**

Row Totals Men (M) 93 (105.00) 70 (59.50) 12 (10.50) 175 Women (W) 87 (75.00) 32 (42.50) 6 (7.50) 125 Column Totals 180 102 18 300

43
Example 11-6 Reconsider the two-way classification table given in Example In that example, a random sample of 300 adults was selected, and they were asked if they favor giving more freedom to schoolteachers to punish students for violence and lack of discipline. Based on the results of the survey, a two-way classification table was prepared and presented in Example Does the sample provide sufficient information to conclude that the two attributes, gender and opinions of adults, are dependent? Use a 1% significance level.

44
**Solution 11-6 H0: Gender and opinions of adults are independent**

H1: Gender and opinions of adults are dependent

45
**Solution 11-6 α = .01 df = (R – 1)(C – 1) = (2 – 1)(3 – 1) = 2**

The critical value of χ2 = 9.210

46
**Figure 11.6 α = .01 χ2 Do not reject H0 Reject H0 9.210**

Critical value of χ2

47
**Table 11.8 In Favor (F) Against (A) No Opinion (N) Row Totals Men (M)**

93 (105.00) 70 (59.50) 12 (10.50) 175 Women (W) 87 (75.00) 32 (42.50) 6 (7.50) 125 Column Totals 180 102 18 300

48
Solution 11-6

49
**Solution 11-6 The value of the test statistic χ2 = 8.252**

It is less than the critical value of χ2 It falls in the nonrejection region Hence, we fail to reject the null hypothesis

50
Example 11-7 A researcher wanted to study the relationship between gender and owning cell phones. She took a sample of 2000 adults and obtained the information given in the following table.

51
Example 11-7 At the 5% level of significance, can you conclude that gender and owning cell phones are related for all adults? Own Cell Phones Do Not Own Cell Phones Men Women 640 440 450 470

52
**Solution 11-7 H0: Gender and owning a cell phone are not related**

H1: Gender and owning a cell phone are related

53
**Solution 11-7 We are performing a test of independence**

We use the chi-square distribution α = .05. df = (R – 1)(C – 1) = (2 – 1)(2 – 1) = 1 The critical value of χ2 = 3.841

54
**Figure 11.7 α = .05 χ2 Do not reject H0 Reject H0 3.841**

Critical value of χ2

55
**Table 11.9 Own Cell Phones (Y) Do Not Own Cell Phones (N) Row Totals**

Men (M) 640 (588.60) 450 (501.40) 1090 Women (W) 440 (491.40) 470 (418.60) 910 Column Totals 1080 920 2000

56
Solution 11-7

57
**Solution 11-7 The value of the test statistic χ2 = 21.445**

It is larger than the critical value of χ2 It falls in the rejection region Hence, we reject the null hypothesis

58
**A Test of Homogeneity Definition**

A test of homogeneity involves testing the null hypothesis that the proportions of elements with certain characteristics in two or more different populations are the same against the alternative hypothesis that these proportions are not the same.

59
Example 11-8 Consider the data on income distributions for households in California and Wisconsin given in following table: California Wisconsin Row Totals High Income 70 34 104 Medium Income 80 40 120 Low Income 100 76 176 Column Totals 250 150 400

60
Example 11-8 Using the 2.5% significance level, test the null hypothesis that the distribution of households with regard to income levels is similar (homogeneous) for the two states.

61
Solution 11-8 H0: The proportions of households that belong to different income groups are the same in both states H1: The proportions of households that belong to different income groups are not the same in both states

62
**Solution 11-8 α = .025 df = (R – 1)(C – 1) = (3 – 1)(2 – 1) = 2**

The critical value of χ2 = 7.378

63
**Figure 11.7 α = .025 χ2 Do not reject H0 Reject H0 7.378**

Critical value of χ2

64
**Table 11.11 California Wisconsin Row Totals High income 70 (65) 34**

(39) 104 Medium income 80 (75) 40 (45) 120 Low income 100 (110) 76 (66) 176 Column Totals 250 150 400

65
Solution 11-8

66
**Solution 11-8 The value of the test statistic χ2 = 4.339**

It is less than the critical value of χ2 It falls in the nonrejection region Hence, we fail to reject the null hypothesis

67
**INFERENCES ABOUT THE POPULATION VARIANCE**

Estimation of the Population Variance Hypothesis Tests About the Population Variance

68
**INFERENCES ABOUT THE POPULATION VARIANCE cont.**

Sampling Distribution of (n – 1)s2 / σ2 If the population from which the sample is selected is (approximately) normally distributed, then has a chi-square distribution with n – 1 degrees of freedom.

69
**Estimation of the Population Variance**

Assuming that the population from which the sample is selected is (approximately) normally distributed, the (1 – α)100% confidence interval for the population variance σ2 is

70
Example 11-9 One type of cookie manufactured by Haddad Food Company is Cocoa Cookies. The machine that fills packages of these cookies is set up in such a way that the average net weight of these packages is 32 ounces with a variance of .015 square ounce.

71
Example 11-9 From time to time the quality control inspector at the company selects a sample of a few such packages, calculates the variance of the net weights of these packages, and construct a 95% confidence interval for the population variance. If either both or one of the two limits of this confidence interval is not the interval .008 to .030, the machine is stopped and adjusted.

72
Example 11-9 A recently taken random sample of 25 packages from the production line gave a sample variance of .029 square ounce. Based on this sample information, do you think the machine needs an adjustment? Assume that the net weights of cookies in all packages are normally distributed.

73
Solution 11-9 n = 25 s2 = .029 α = = .05 α / 2 = .05 / 2 = .025 1 – α / 2 = 1 – .025 = .975 df = n – 1 = 25 – 1 = 24 χ2 for 24 df and .025 area in the right tail = χ2 for 24 df and .975 area in the right tail =

74
Figure 11.9 df = 24 = .025 39.364 χ2 Value of

75
Figure 11.9 df = 24 = .025 χ2 12.401 Value of

76
Solution 11-9

77
Solution 11-9 Thus, with 95% confidence, we can state that the variance for all packages of Cocoa Cookies lies between and square ounce.

78
**Hypothesis Tests About the Population Variance**

The value of the test statistic χ2 is calculated as where s2 is the sample variance, σ2 is the hypothesized value of the population variance, and n – 1 represents the degrees of freedom. The population from which the sample is selected is assumed to be (approximately) normally distributed.

79
Example 11-10 One type of cookie manufactured by Haddad Food Company is Cocoa Cookies. The machine that fills packages of these cookies is set up in such a way that the average net weight of these packages is 32 ounces with a variance of .015 square ounce. From time to time the quality control inspector at the company selects a sample of a few such packages, calculates the variance of the net weights of these packages, and makes a test of hypothesis about the population variance.

80
Example 11-10 She always uses α = .01. The acceptable value of the population variance is .015 square ounce or less. If the conclusion from the test of hypothesis is that the population variance is not within the acceptable limit, the machine is stopped and adjusted.

81
Example 11-10 A recently taken random sample of 25 packages from the production line gave a sample variance of .029 square ounce. Based on this sample information, do you think the machine needs an adjustment? Assume that the net weights of cookies in all packages are normally distributed.

82
**Solution 11-10 H0 :σ2 ≤ .015 H1: σ2 >.015**

The population variance is within the acceptable limit H1: σ2 >.015 The population variance exceeds the acceptable limit

83
Solution 11-10 α = .01 df = n – 1 = 25 – 1 = 24 The critical value of χ2 =

84
**Figure 11.10 α = .01 χ2 Do not reject H0 Reject H0 42.980**

Critical value of χ2

85
Solution 11-10 From H0

86
**Solution 11-10 The value of the test statistic χ2 = 46.400**

It is greater than the critical value of χ 2 It falls in the rejection region Hence, we reject the null hypothesis H0 We conclude that the population variance is not within the acceptable limit The machine should be stopped and adjusted

87
Example 11-11 The variance of scores on a standardized mathematics test for all high school seniors was 150 in A sample of scores for 20 high school seniors who took this test this year gave a variance of 170. Test at the 5% significance level if the variance of current scores of all high school seniors on this test is different from 150. Assume that the scores of all high school seniors on this test are (approximately) normally distributed.

88
Solution 11-11 H0: σ2 = 150 The population variance is not different from 150 H1: σ2 ≠ 150 The population variance is different from 150

89
**Solution 11-11 α = .05 Area in the each tail = .025**

df = n – 1 = 20 – 1 = 19 The critical values of χ and 8.907

90
**Figure 11.11 Do not reject H0 Reject H0 Reject H0 α /2 = .025**

8.907 32.852 Two critical values of χ2

91
Solution 11-11 From H0

92
**Solution 11-11 The value of the test statistic χ2 = 21.533**

It is between the two critical values of χ2 It falls in the nonrejection region Consequently, we fail to reject H0.

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google