Chapter 12 - Chi-Square Tests

CHAPTER 12—Chi-Square Tests

12.1 (1) We perform an experiment in which we carry out n identical trials and in which there are k possible outcomes on each trial.

(2) The probabilities of the k outcomes are denoted, where . These probabilities stay the same from trial to trial.

(3) The trials in the experiment are independent.

(4) The results of the experiment are observed counts of the number of trials that result in each of the k possible outcomes. The counts are denoted . That is, is the number of trials resulting in the first possible outcome, is the number of trials resulting in the second possible outcome, and so forth.

12.2 All the expected cell frequencies ( values) are at least 5. Alternatively, k exceeds 4, the average of the values is at least 5, and the smallest value is at least 1.

12.3 Explanations will vary.

12.4 The larger χ2, the more the observed frequencies differ from the expected frequencies.

12.5 Using classes of a histogram or using intervals from the empirical rule.

12.6 a. each expected value is 5

since 50.951 > 9.488, reject , the current market shares differ from those of 1990.

12.7 a. each expected value is 5

since 300.605 > 7.779, reject . The viewing shares of the current rating period differ from those of the last.

12.8 a. each expected value is 5

since 880.22 > 12.592, reject , not consistent.

12.9 a.

Reject .

b. differences between brand preferences

12.10 a. Each (see below)

b. / /
0 / 479 / .87(500) = 435
1 / 10 / .08(500 = 40
2 / 8 / .03(500) = 15
3 / 2 / .01(500) = 5
> 3 / 1 / .01(500) = 5

= 4.4506 + 22.5 + 3.2667 + 1.8 + 3.2
= 35.2173
Since (with 4 degrees of freedom), reject , the systems differ

12.11 a.
(1) [–¥, 10.185]
(2) [10.185, 14.147]
(3) [14.147, 18.108]
(4) [18.108, 22.069]
(5) [22.069, 26.030]
(6) [26.030, ¥]

b. (1) For instance, = .5 – .4772 = .0228
and » 1.5
(2) .1359, » 9
(3) .3413, » 22
(4) .3413, » 22
(5) .1359, » 9
(6) .0228, » 1.5

c. We can use the chi-square test since the average of the values is 10.833 (which is ³ 5) and since the smallest value is 1.482 (which is ³ 1).

d. : the probabilities that a randomly selected payment time will be in intervals 1, 2, 3, 4, 5, and 6 are and versus
: the above null hypothesis is not true.

e. 1, 9, 30, 15, 8, 2
= .167 + 0 + 2.909 + 2.227 + .111 + .167
= 5.581

f. Fail to reject since (with 6 – 1 – 2 = 3 degrees of freedom). Conclude normality.

12.12 / Interval / / /
/ [–¥, 24.135] / 3 / .0228(60) = 1.368 / 1.947
/ [24.135, 27.243] / 7 / .1359(60) = 8.154 / .163
/ [27.243, 30.35] / 15 / .3413(60) = 20.478 / 1.465
/ [30.35, 33.457] / 29 / .3413(60) = 20.478 / 3.546
/ [33.457, 36.565] / 6 / .1359(60) = 8.154 / .569
/ [36.565, ¥] / 0 / .0228(60) = 1.368 / 1.368

Reject since (with 6 – 1 – 2 = 3 degrees of freedom). Conclude ratings are not normal.

12.13 / Interval / / /
/ [–¥, .51] / 1 / .0228(100) = 2.28 / .719
/ [.51, 2.985] / 16 / .1359(100) = 13.59 / .427
/ [2.985, 5.46] / 36 / .3413(100) = 34.13 / .102
/ [5.46, 7.935] / 30 / .3413(100) = 34.13 / .500
/ [7.935, 10.41] / 14 / .1359(100) = 13.59 / .012
/ [10.41, ¥] / 3 / .0228(100) = 2.28 / .227

Since (with 6 – 1 – 2 = 3 degrees of freedom), we do not reject normality. Conclude waiting times normal.

12.14

x / Observed
/ Probability / Expected
/
/ 6 / .0611 / 30(.0611) = 1.833 / 9.473
/ 5 / .2812 / (30)(.2812) = 8.436 / 1.399
/ 7 / .3606 / (30)(.3606) = 10.818 / 1.347
/ 8 / .2105 / (30)(.2105) = 6.315 / .450
8 or more / 4 / .0866 / (30)(.0866) = 2.598 / .757

Since (with 5 – 1 = 4 degrees of freedom), we reject Poisson. Conclude errors are not Poisson with m = 4.5.

12.15 Studying the relationship between two variables

12.16 Computed by assuming independence; first estimate marginal probabilities. These are multiplied together to obtain the joint probabilities arising from the independence hypothesis. These joint probabilities are multiplied by the sample size and compared to the observed cell frequencies.

12.17 a. / 8 / 32 / 40
12 / 48 / 60
20 / 80 / 100

: whether a person drinks wine and whether a person watches tennis are independent versus : dependent. Since (with (2 – 1)(2 – 1) = 1 degree of freedom), we reject Conclude dependent.

d. Explanations will vary. Probably.

12.18 a.
Since (with (3 – 1)(3 – 1) = 4 degrees of freedom), we reject independence. Conclude depreciation method and country are dependent.

The test is valid because there are 9 cells (which exceeds 4), the average of the values is 8.67 (which is at least 5) and the smallest is at least 1.

b. Plots like those in Figure 16.3 should be constructed.

12.19 a. row total

24.24%

22.73%

53.03%

col total 51.515% 48.485%

c. , p-value=.032, Cannot reject at .01 level

d. There is no difference between smokers and nonsmokers at the .01 level, but can reject at α=.05 so a difference is possible.

12.20 a. / For example, for the first (upper-left) cell, Others are as follows:
8.4 / 9.6 / 6
5.6 / 6.4 / 4

Since (with (2 – 1)(3 – 1) = 2 degrees of freedom), reject independence.

b. The test is valid because the number of cells exceeds 4, the average of the values is 6.667 (which is ³ 5) and the smallest value is 4 (which is ³ 1).

c. Yes

d. Plots like those in Figure 16.3 should be constructed. Dependent.

12.21 a.

Since 16.385 > 7.815, reject : independence

12.22
Since (with 4 – 1 = 3 degrees of freedom), we reject ; brands not equally preferred.
95% C.I.:

12.23
= 15 + 16.02 + 22.82 + 6.67 + 5.4
= 65.91
Since (with 5 – 1 = 4 degrees of freedom), we reject ; entrances not equally used.
95% C.I.:

12.24 a. For instance, for the first (upper-left) cell, Others are as follows:
Auditor Expected

18.58 / 124.42
4.42 / 29.58

Since (with (2 – 1)(2 – 1) = 1 degree of freedom), we reject independence.

b. For instance, for the first (upper-left) cell, Others are as follows:
Client Expected

17.15 / 114.85
5.85 / 39.15

= 1.547 + .231 + 4.534 + .677 = 6.989
Since (with (2 – 1)(2 – 1) = 1 degree of freedom), we reject independence.

c. Plots like those in Figure 16.3 should be constructed.

d. Yes

12.25 a.

Since 20.940 > 3.841, reject : independence.

c. Graph not included in this manual. Dependent, explanations will vary.

12.26 For instance, for the first (upper-left) cell, Others are as follows:

27.69 / 151.31
28.31 / 154.69

= .017 + .003 + .017 + .003 = .04
Since (with (2 – 1)(2 – 1) = 1 degree of freedom), we do not reject independence. Conclude relationships not the same.

12.27 Chi-square = 71.476, p-value = .000

Reject : independence

12.28 If one uses the stem-and-leaf display of the satisfaction data from Chapter 2, you get the following results:

Interval / / /
/ [–¥, 37.665] / 1 / .0228(65) = 1.48 / .156
/ [37.665, 40.308] / 12 / .1359(65) = 8.83 / 1.138
/ [40.308, 42.95] / 12 / .3413(65) = 22.18 / 4.672
/ [42.95, 45.592] / 29 / .3413(65) = 22.18 / 2.097
/ [45.592, 48.235] / 11 / .1359(65) = 8.83 / .533
/ [48.235, ¥] / 0 / .0228(65) = 1.48 / 1.48

X2 = .156 + 1.138 + 4.672 + 2.097 + .533 + 1.48 = 10.076

Since 10.076 < degrees of freedom, reject ; not a normal distribution.

Internet Exercise:

12.29 : There is no difference by Ethnic Group

Ha: There is a difference by Ethnic Group

Chi-square Contingency Table Test for Independence
Yes / No / Total
White / Observed / 1285 / 2070 / 3355
% of row / 38.3% / 61.7% / 100.0%
% of column / 84.8% / 78.4% / 80.7%
% of total / 30.9% / 49.8% / 80.7%
Black / Observed / 131 / 344 / 475
% of row / 27.6% / 72.4% / 100.0%
% of column / 8.6% / 13.0% / 11.4%
% of total / 3.2% / 8.3% / 11.4%
Hispanic / Observed / 100 / 227 / 327
% of row / 30.6% / 69.4% / 100.0%
% of column / 6.6% / 8.6% / 7.9%
% of total / 2.4% / 5.5% / 7.9%
Total / Observed / 1516 / 2641 / 4157
% of row / 36.5% / 63.5% / 100.0%
% of column / 100.0% / 100.0% / 100.0%
% of total / 36.5% / 63.5% / 100.0%
25.96 / chi-square
2 / df
2.31E-06 / p-value

Reject . There is a difference by Ethnic Group at the .01 level of significance.

12-1