CHI-SQUARE TESTS

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 ?2 (read as chi-square).

3
Figure 11.1 Three chi-square distribution curves.
4
Example 11-1

• Find the value of ?² for 7 degrees of freedom and
  an area of .10 in the right tail of the
  chi-square distribution curve.

5
Table 11.1 ?2 for df 7 and .10 Area in the
Right Tail
Area in the Right Tail Under the Chi-Square Distribution Curve Area in the Right Tail Under the Chi-Square Distribution Curve Area in the Right Tail Under the Chi-Square Distribution Curve Area in the Right Tail Under the Chi-Square Distribution Curve 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 12.017 118.498 7.879 10.597 20.278 140.169
Required value of ?²
6
Figure 11.2
df 7
.10
12.017
0
?²
7
Example 11-2

• Find the value of ?² 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 ?2 for df 12 and .95 Area in the
Right Tail
Area in the Right Tail Under the Chi-Square Distribution Curve Area in the Right Tail Under the Chi-Square Distribution Curve Area in the Right Tail Under the Chi-Square Distribution Curve Area in the Right Tail Under the Chi-Square Distribution Curve 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 28.300 140.169
Required value of ?²
10
Figure 11.3
df 12
Shaded area .95
.05
5.226
0
?²
11
A GOODNESS-OF-FIT TEST

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

12
Multinomial Experiment cont.

1. It consists of n identical trials (repetitions).
2. Each trial results in one of k possible outcomes
   (or categories), where k gt 2.
3. The trials are independent.
4. 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
  ?2 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 a .01
• k number of categories 5
• df k 1 5 1 4
• The critical value of ?2 13.277

21
Figure 11.4
Reject H0
Do not reject H0
a .01
13.277
?2
Critical value of ?2
22
Table 11.3
Category (Day) Observed Frequency O p Expected Frequency E np (O E) (O E)2
Monday Tuesday Wednesday Thursday Friday 253 197 204 279 267 .20 .20 .20 .20 .20 1200(.20) 240 1200(.20) 240 1200(.20) 240 1200(.20) 240 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 23.184
23
Solution 11-3

• All the required calculations to find the value
  of the test statistic ?2 are shown in Table 11.3.

24
Solution 11.3

• The value of the test statistic ?2 23.184 is
  larger than the critical value of ?2 13.277
• 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 a .025
• k number of categories 4
• df k 1 4 1 3
• The critical value of ?2 9.348

30
Figure 11.5
Reject H0
Do not reject H0
a .025
9.348
?2
Critical value of ?2
31
Table 11.4
Category Observed Frequency O p Expected Frequency E np (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 ?2 are shown in Table 11.4.

33
Solution 11-4

• The value of the test statistic ?2 5.782 is
  less than the critical value of ?2 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 Independence
• 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 ?2 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 11-5. 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 11-5. 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

• a .01
• df (R 1)(C 1) (2 1)(3 1) 2
• The critical value of ?2 9.210

46
Figure 11.6
Reject H0
Do not reject H0
a .01
9.210
?2
Critical value of ?2
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 ?2 8.252
• It is less than the critical value of ?2
• 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
• a .05.
• df (R 1)(C 1) (2 1)(2 1) 1
• The critical value of ?2 3.841

54
Figure 11.7
Reject H0
Do not reject H0
a .05
3.841
?2
Critical value of ?2
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 ?2 21.445
• It is larger than the critical value of ?2
• 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

• a .025
• df (R 1)(C 1) (3 1)(2 1) 2
• The critical value of ?2 7.378

63
Figure 11.7
Reject H0
Do not reject H0
a .025
7.378
?2
Critical value of ?2
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 ?2 4.339
• It is less than the critical value of ?2
• 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 / s2
• 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 a)100 confidence interval
  for the population variance s2 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
• a 1 - .95 .05
• a / 2 .05 / 2 .025
• 1 a / 2 1 .025 .975
• df n 1 25 1 24
• ?2 for 24 df and .025 area in the right tail
  39.364
• ?2 for 24 df and .975 area in the right tail
  12.401

74
Figure 11.9
df 24
.025
39.364
?2
Value of
75
Figure 11.9
df 24
.025
?2
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 .0177 and .0561 square ounce.

78
Hypothesis Tests About the Population Variance

• The value of the test statistic ?2 is calculated
  as
• where s2 is the sample variance, s2 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 a .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 s2 .015
• The population variance is within the acceptable
  limit
• H1 s2 gt.015
• The population variance exceeds the acceptable
  limit

83
Solution 11-10

• a .01
• df n 1 25 1 24
• The critical value of ?2 42.980

84
Figure 11.10
Reject H0
Do not reject H0
a .01
42.980
?2
Critical value of ?2
85
Solution 11-10
From H0
86
Solution 11-10

• The value of the test statistic ?2 46.400
• It is greater than the critical value of ? 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 2002. 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 s2 150
• The population variance is not different from 150
• H1 s2 ? 150
• The population variance is different from 150

89
Solution 11-11

• a .05
• Area in the each tail .025
• df n 1 20 1 19
• The critical values of ?2 32.852 and 8.907

90
Figure 11.11
Reject H0
Do not reject H0
Reject H0
a /2 .025
a /2 .025
8.907
32.852
Two critical values of ?2
91
Solution 11-11
From H0
92
Solution 11-11

• The value of the test statistic ?2 21.533
• It is between the two critical values of ?2
• It falls in the nonrejection region
• Consequently, we fail to reject H0.