chapter 11 chi-square tests

Chapter 11 Chi-Square Tests

277

Note that answers may vary throughout due to rounding differences. Section 11.1 11.1 The chi-square distribution has only one

parameter, called the degrees of freedom (df). The shape of a chi-square 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.

11.2 For df = 12 and .025 area in the right tail,

χ2 = 23.337. 11.3 For df = 28 and .05 area in the right tail,

χ2 = 41.337. 11.4 For an area of .10 in the left tail, the area in

the right tail is 1 – .10 = .90. Hence, the value of chi-square for df = 14 and .10 area in the left tail is the same as for df = 14 and .90 area in the right tail. Thus, the chi-square value is χ2 = 7.790.

11.5 For an area of .990 in the left tail, the area in

the right tail is 1 – .990 = .01. Hence, the value of chi-square for df = 23 and .990 area in the left tail is the same as for df = 23 and .01 area in the right tail. Thus, the chi-square value is χ2 = 41.638.

11.6 a. For df = 4 and .005 area in the right tail,

χ2 = 14.860.

b. For an area of .05 in the left tail, the area in the right tail is 1 – .05 = .95. Hence, the value of chi-square for df = 4 and .05 area in the left tail is the same as for df = 4 and .95 area in the right

tail. Thus, the chi-square value is χ2 = .711.

11.7 a. For an area of .025 in the left tail, the

area in the right tail is 1 – .025 = .975. Hence, the value of chi-square for df = 13 and .025 area in the left tail is the same as for df = 13 and .975 area in the right tail. Thus, the chi-square value is χ2 = 5.009.

b. For df = 13 and .995 area in the right tail, χ2 = 3.565.

Section 11.2

11.8 Following are the four characteristics of a

multinomial experiment: 1. It consists of n identical trials. 2. Each trial results in one of k possible

outcomes (or categories) where k > 2. 3. The trials are independent. 4. The probabilities of various outcomes

remain constant for each trial.

11.9 Α goodness–of–fit test compares the observed frequencies from a multinomial experiment with expected frequencies derived from a certain pattern or theoretical distribution. The test evaluates how well the observed frequencies fit the expected frequencies.

11.10 The observed frequencies are the

frequencies obtained from the performance of a multinomial experiment. The expected frequencies are the frequencies that we expect to obtain if the null hypothesis is true.

278 Chapter 11 Chi-Square Tests

11.11 The expected frequency of a category is given by Ε = 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. The degrees of freedom for a goodness–of–fit test are k – 1, where k denotes the number of possible outcomes (or categories) for the experiment.

11.12 The minimum expected frequency for each category should be 5. If this condition is not satisfied, we

may increase the sample size or combine two or more categories to make each expected frequency at least 5.

11.13 Step 1: H0: The die is fair, Η1: The die is not fair.

Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 6, df = k – 1 = 6 – 1 = 5 For α = .05, the critical value of χ2 is 11.070. Step 4: Note that the die will be fair if the probability of each of the six outcomes is the same, which is 1/6.

Outcome O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε)2/Ε 1 7 1/6 10 –3 9 .900 2 12 1/6 10 2 4 .400 3 8 1/6 10 –2 4 .400 4 15 1/6 10 5 25 2.500 5 11 1/6 10 1 1 .100 6 7 1/6 10 –3 9 .900 n = 60 Sum = 5.200

χ2 = Σ (O – E)2/E = 5.200 Step 5: Do not reject H0 since 5.200 < 11.070. Conclude that the die is fair.

11.14 Step 1: H0: The current distribution of exercise frequency is the same as that in 2011. Η1: The current distribution of exercise frequency differs from that of in 2011. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 3, df = k – 1 = 3 – 1 = 2 For α = .05, the critical value of χ2 is 5.991. Step 4:

Exercise Category O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε A (3 or more days) 141 .516 268.32 −127.32 16210.382 60.414 B (1 or 2 days) 131 .188 97.76 33.24 1104.898 11.302 C (at least 1 day) 248 .297 154.44 93.56 8753.474 56.679 n = 520 Sum = 128.395

χ2 = Σ (O – E)2/E = 128.395 Step 5: Reject H0 since 128.395 > 5.991. Conclude that the current distribution of exercise frequency differs from that in 2011.

11.15 Step 1: H0: The current distribution of responses is the same as that of October 2011.

Η1: The current distribution of responses differs from that of October 2011. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .025, the critical value of χ2 is 9.348.

(continued on next page)

Section 11.2 A Goodness-of-Fit Test 279

(continued)

Step 4:

Response O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε 0 hours 2 .03 6.45 −4.45 19.803 3.070 1-2 hours 41 .24 51.60 −10.60 112.360 2.178 3-5 hours 55 .22 47.30 7.70 59.290 1.253 6 or more hours 117 .51 109.65 7.35 54.023 0.493 n = 215 Sum = 6.994

χ2 = Σ (O – E)2/E = 6.994 Step 5: Do not reject H0 since 6.994 < 9.348. Conclude that the current distribution of responses does not differ from that of October 2011.

11.16 Step 1: H0: The genes received by the children are the same as what Mendelian genetics predicts.

Η1: The genes received by the children differ from what Mendelian genetics predicts. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 3, df = k – 1 = 3 – 1 = 2 For α = .05, the critical value of χ2 is 5.991. Step 4:

Child’s genes O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε LL 14 .25 16.25 –2.25 5.063 .312 Lr 31 .50 32.50 –1.50 2.250 .069 rr 20 .25 16.25 3.75 14.063 .865 n = 65 Sum = 1.246

χ2 = Σ (O – E)2/E = 1.246 Step 5: Do not reject H0 since 1.246 < 5.991 Conclude that the genes received by the children is not significantly different from what Mendelian genetics predicts.

11.17 Step 1: H0: The number of boxes sold of each of the five colors is the same.

Η1: The number of boxes sold of each of the five colors differs. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 5, df = k – 1 = 5 – 1 = 4 For α = .01, the critical value of χ2 is 13.277. Step 4:

Color O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε Blue 310 .20 278.8 31.2 973.44 3.492 Green 292 .20 278.8 13.2 174.24 .625 Orange 280 .20 278.8 1.2 1.44 .005 Red 216 .20 278.8 –62.8 3943.84 14.146 Yellow 296 .20 278.8 17.2 295.84 1.061 n = 1394 Sum = 19.329

χ2 = Σ (O – E)2/E = 19.329 Step 5: Reject H0 since 19.329 > 13.277. Conclude that the number of boxes sold of each of the five colors differs.


11.18 Step 1: H0: The distribution of payment methods has not changed. Η1: The distribution of payment methods has changed. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .01, the critical value of χ2 is 11.345. Step 4:

Method O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε C 240 .41 205 35 1225 5.976 CK 104 .24 120 – 16 256 2.133 D 111 .26 130 – 19 361 2.777 N 45 .09 45 0 0 0.000 n = 500 Sum = 10.886

χ2 = Σ (O – E)2/E = 10.886 Step 5: Do not reject H0 since 10.886 < 11.345. Conclude that the distribution of payment methods has not changed.

11.19 Step 1: H0: The orders are evenly distributed over all days of the week.

Η1: The orders are not evenly distributed over all days of the week. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 5, df = k – 1 = 5 – 1 = 4 For α = .05, the critical value of χ2 is 9.488. Step 4:

Day O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε Monday 92 .20 80 12 144 1.800 Tuesday 71 .20 80 – 9 81 1.013 Wednesday 65 .20 80 – 15 225 2.813 Thursday 83 .20 80 3 9 .113 Friday 89 .20 80 9 81 1.013 n = 400 Sum = 6.752

χ2 = Σ (O – E)2/E = 6.752 Step 5: Do not reject H0 since 6.752 < 9.488. Conclude that the orders are evenly distributed over all days of the week.

11.20 Step 1: H0: The sample is random.

Η1: The sample is not random. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .025, the critical value of χ2 is 9.348. Step 4:

Classification O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε Seniors 50 .19 38 12 144 3.789 Juniors 46 .23 46 0 0 .000 Sophomores 55 .27 54 1 1 .019 Freshmen 49 .31 62 –13 169 2.726 n = 200 Sum = 6.534

χ2 = Σ (O – E)2/E = 6.534 Step 5: Do not reject H0 since 6.534 < 9.348. Conclude that the sample is random.

Section 11.3 A Test of Independence or Homogeneity 281

11.21 Step 1: H0: The percentage distribution of users' opinions is unchanged since the product was redesigned. Η1: The percentage distribution of users' opinions has changed since the product was redesigned. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .025, the critical value of χ2 is 9.348. Step 4:

Opinion O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε Excellent 495 .53 424 71 5041 11.889 Satisfactory 255 .31 248 7 49 .198 Unsatisfactory 35 .07 56 –21 441 7.875 No opinion 15 .09 72 –57 3249 45.125 n = 800 Sum = 65.087

χ2 = Σ (O – E)2/E = 65.087 Step 5: Reject H0 since 65.087 > 9.348. Conclude that the percentage distribution of users' opinions has changed since the product was redesigned.

11.22 Step 1: H0: The distribution of defects is the same as when the process is working properly and the

process does not need an adjustment. Η1: The distribution of defects is not the same as when the process is working properly and the process needs an adjustment. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 3, df = k – 1 = 3 – 1 = 2 For α = .01, the critical value of χ2 is 9.210. Step 4:

Defects O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε 0 262 .92 276 –14 196 0.710 1 24 .05 15 9 81 5.400 2 or more 14 .03 9 5 25 2.778 n = 300 Sum = 8.888

χ2 = Σ (O – E)2/E = 8.888 Step 5: Do not reject H0 since 8.888 < 9.210. Conclude that the distribution of defects is the same as when the process is working properly and the process does not need an adjustment.

Section 11.3 11.23 In a test of independence, we test the null hypothesis that two characteristics of the elements in a given

population are not related against the alternative hypothesis that the two characteristics are related. See Example 11−6 in the text. In a test of homogeneity, we test if two (or more) populations are similar with respect to the distribution of certain characteristics. See Example 11−8 in the text.

11.24 The expected frequency for a cell in a contingency table is given by Ε = (Row total)(Column total)/n

where n is the sample size. The degrees of freedom are given by df = (R – 1)(C – 1), where R and C are the number of rows and columns, respectively, in the given contingency table.

11.25 The minimum expected frequency for each cell should be 5. If this condition is not satisfied, we may

increase the sample size or combine some categories. 11.26 a. H0: Row and columns are independent, H1: Rows and columns are dependent


b. The expected frequencies are given in parentheses below the observed frequencies in the table below.

Column 1 Column 2 Column 3 Total Row 1 137 64 105 306 (125.85) (77.67) (102.48) Row 2 98 71 65 234 (96.24) (59.39) (78.37) Row 3 115 81 115 311 (127.91) (78.94) (104.15) Total 350 216 285 851

c. df = (R – 1)(C – 1) = (3 – 1)(3 – 1) = 4

For α = .01, the critical value of χ2 is 13.277.

d. 2

2

2 2 2 2 2

2 2 2 2

( )

(137 125.85) (64 77.67) (105 102.48) (98 96.24) (71 59.39)125.85 77.67 102.48 96.24 59.39

(65 78.37) (115 127.91) (81 78.94) (115 104.15)

78.37 127.91 78.94 104.15.988 2.406 .062 .032 2.

O E

Eχ −=

− − − − −= + + + +

− − − −+ + + +

= + + + +

∑

270 2.281 1.303 .054 1.130 10.526+ + + + =

e. Do not reject H0 since 10.526 < 13.277. Conclude that the rows and columns are independent. 11.27 a. H0: The proportion in each row is the same for all four populations.

Η1: The proportion in each row is not the same for all four populations.

b. The expected frequencies are given in parentheses below the observed frequencies in the table.

Column 1 Column 2 Column 3 Column 4 Total Row 1 24 81 60 121 286 (31.62) (63.95) (89.95) (100.49) Row 2 46 64 91 72 273 (30.18) (61.04) (85.86) (95.92) Row 3 20 37 105 93 255 (28.19) (57.01) (80.20) (89.59) Total 90 182 256 286 814


c. df = (R – 1)(C – 1) = (3 – 1)(4 – 1) = 6 For α = .025, the critical value of χ2 is 14.449.

d. 2

2

2 2 2 2 2 2

2 2 2 2

2

( )

(24 31.62) (81 63.95) (60 89.95) (121 100.49) (46 30.18) (64 61.04)

31.62 63.95 89.95 100.49 30.18 61.04(91 85.86) (72 95.92) (20 28.19) (37 57.01)

85.86 95.92 28.19 57.01(105 80.20)

80.

O E

Eχ −=

− − − − − −= + + + + +

− − − −+ + + +

−+

∑

2(93 89.59)20 89.59

1.836 4.546 9.972 4.186 8.293 .144 .308 5.965 2.379 7.023 7.669 .130 52.451

−+

= + + + + + + + + + + + =

e. Since 52.451 > 14.449, reject H0. 11.28 Step 1: H0: Gender and brand preference are independent.

Η1: Gender and brand preference are dependent. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(2 – 1) = 1 For α = .01, the critical value of χ2 is 6.635. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Name Brand Store Brand Total Men 150 165 315 (139.5) (175.5) Women 160 225 385 (170.5) (214.5) Total 310 390 700

2 2 2 2 2

2 ( ) (150 139.5) (165 175.5) (160 170.5) (225 214.5)

139.5 175.5 170.5 214.5.790 .628 .647 .514 2.579

O E

Eχ − − − − −= = + + +

= + + + =∑

Step 5: Do not reject H0 since 2.579 < 6.635. Conclude that gender and brand preference are independent.

11.29 Step 1: H0: Gender and wearing or not wearing of seat belt are not related.

Η1: Gender and wearing or not wearing of seat belt are related. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(2 – 1) = 1 For α = .025, the critical value of χ2 is 5.024. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Wearing seat belt Not wearing seat belt Total Men 40 15 55 (42.9) (12.1) Women 38 7 45 (35.1) (9.9) Total 78 22 100



(continued)

2 2 2 2 2

2 ( ) (40 42.9) (15 12.1) (38 35.1) (7 9.9)

42.9 12.1 35.1 9.9.196 .695 .240 .849 1.980

O E

Eχ − − − − −= = + + +

= + + + =∑

Step 5: Do not reject H0 since 1.980 < 5.024. Conclude that gender and wearing or not wearing of seat belt are not related.

11.30 Step 1: H0: Gender and responses are independent.

Η1: Gender and responses are dependent. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(3 – 1) = 2 For α = .01, the critical value of χ2 is 9.210. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table .

Yes No Uncertain Total Women 125 59 21 205 (115.54) (70.55) (18.92) Men 101 79 16 196 (110.46) (67.45) (18.08) Total 226 138 37 401

2 2 2 2 2

2

2 2

( ) (125 115.54) (59 70.55) (21 18.92) (101 110.46)

115.54 70.55 18.92 110.46(79 67.45) (16 18.08)

67.45 18.08 .775 1.891 .229 .810 1.978 .239 5.922

O E

Eχ − − − − −= = + + +

− −+ +

= + + + + + =

∑

Step 5: Do not reject H0 since 5.922 < 9.210. Conclude that gender and responses are independent.

11.31 a. Step 1: H0: Approval ratings by own party voters and president are not related.

Η1: Approval ratings by own party voters and president are related. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(2 – 1) = 1 For α = .01, the critical value of χ2 is 6.635. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Own Party Obama Bush Total Approve 693 775 1468 (734) (734) Not sure/ 207 125 332 disapprove (166) (166) Total 900 900 1800

2 2 2 2 2

2 ( ) (693 734) (775 734) (207 166) (125 166)

734 734 166 1662.290 2.290 10.127 10.127 24.834

O E

Eχ − − − − −= = + + +

= + + + =∑

Step 5: Reject H0 since 24.834 > 6.635. Conclude that approval ratings by own party voters and president are related.

b. Step 1: H0: Approval ratings by opposition party voters and president are not related. Η1: Approval ratings by opposition party voters and president are related. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(2 – 1) = 1 For α = .01, the critical value of χ2 is 6.635.



(continued)

Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Opposition Party Obama Bush Total Approve 99 171 270 (135.00) (135.00) Not sure/ 801 729 1530 disapprove (765.00) (765.00) Total 900 900 1800

2 2 2 2 2

2 ( ) (99 135) (171 135) (801 765) (729 765)

135 135 765 765 9.600 9.600 1.694 1.694 22.588

O E

Eχ − − − − −= = + + +

= + + + =∑

Step 5: Reject H0 since 22.588 > 6.635. Conclude that approval ratings by opposition party voters and president are related. 11.32 Step 1: H0: The decision to accept/not accept the offer and age group are not related.

Η1: The decision to accept/not accept the offer and age group are related. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(5 – 1) = 4 For α = .05, the critical value of χ2 is 9.488. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Age Group Total 21-29 30-39 40-49 50-59 60 & over Deal 78 82 89 92 63 404 (77.337) (87.726) (85.994) (89.457) (63.486) No Deal 56 70 60 63 47 296 (56.663) (64.274) (63.006) (65.543) (46.514) Total 134 152 149 155 110 700

2 2 2 2 2

2

2 2 2 2

2 2

( ) (78 77.337) (82 87.726) (89 85.994) (92 89.457)

77.337 87.726 85.994 89.457(63 63.486) (56 56.663) (70 64.274) (60 63.006)

63.486 56.663 64.274 63.006(63 65.543) (47 46.514)

65.543 46.51

O E

Eχ − − − − −= = + + +

− − − −+ + + +

− −+ +

∑

4.006 .374 .105 .072 .004 .008 .510 .143 .099 .005 1.326= + + + + + + + + + =

Step 5: Do not reject H0 since 1.326 < 9.488. Conclude that the decision to accept/not accept the offer and age group are not related.

11.33 Step 1: H0: Region and causes of fire are unrelated.

Η1: Region and causes of fire are related. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(4 – 1) = 3 For α = .05, the critical value of χ2 is 7.815. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Region Arson Accident Lightning Unknown Total A 6 9 6 10 31 (5.30) (9.38) (8.57) (7.75) B 7 14 15 9 45 (7.70) (13.62) (12.43) (11.25) Total 13 23 21 19 76



(continued)

2 2 2 2 2 22

2 2 2

( ) (6 5.30) (9 9.38) (6 8.57) (10 7.75) (7 7.70)

5.30 9.38 8.57 7.75 7.70(14 13.62) (15 12.43) (9 11.25)

13.62 12.43 11.25.092 .015 .771 .653 .064 .011 .531 .450 2.587

O E

Eχ − − − − − −= = + + + +

− − −+ + +

= + + + + + + + =

∑

Step 5: Do not reject H0 since 2.587 < 7.815. Conclude that region and causes of fire are unrelated.

11.34 Step 1: H0: The distribution of good and defective parts is the same for both subsidiaries.

Η1: The distribution of good and defective parts is not the same for both subsidiaries. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(2 – 1) = 1 For α = .05, the critical value of χ2 is 3.841. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Subsidiary A Subsidiary B Total Good 284 381 665 (285) (380) Defective 16 19 35 (15) (20) Total 300 400 700

2 2 2 2 2

2 ( ) (284 285) (381 380) (16 15) (19 20)

285 380 15 20.004 .003 .067 .050 .124

O E

Eχ − − − − −= = + + +

= + + + =∑

Step 5: Do not reject H0 since .124 < 3.841. Conclude that the distribution of good and defective parts is the same for both subsidiaries.

11.35 Step 1: H0: The two drugs are similar in curing the patients.

Η1: The two drugs are not similar in curing the patients. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(2 – 1) = 1 For α = .01, the critical value of χ2 is 6.635. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Cured Not Cured Total Drug I 44 16 60 (37.2) (22.8) Drug II 18 22 40 (24.8) (15.2) Total 62 38 100

2 2 2 2 2

2 ( ) (44 37.2) (16 22.8) (18 24.8) (22 15.2)

37.2 22.8 24.8 15.21.243 2.028 1.865 3.042 8.178

O E

Eχ − − − − −= = + + +

= + + + =∑

Step 5: Reject H0 since 8.178 > 6.635. Conclude that the two drugs are not similar in curing the patients.


11.36 Step 1: H0: The distributions of favorite forms of camping are the same for all four regions. Η1: The distributions of favorite forms of camping are different for at least two regions. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (4 – 1)(4 – 1) = 9 For α = .01, the critical value of χ2 is 21.666. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Midwest Northeast South West Total Camper/ 132 129 129 135 525 trailer (131.25) (131.25) (131.25) (131.25) Family 180 175 168 146 669 style (167.25) (167.25) (167.25) (167.25) Rustic 46 50 59 68 223 (55.75) (55.75) (55.75) (55.75) None 42 46 44 51 183 (45.75) (45.75) (45.75) (45.75) Total 400 400 400 400 1600

22

2 2 2 2

2 2 2 2

2 2

( )

(132 131.25) (129 131.25) (129 131.25) (135 131.25)

131.25 131.25 131.25 131.25(180 167.25) (175 167.25) (168 167.25) (146 167.25)

167.25 167.25 167.25 167.25(46 55.75) (50 55.75)

55.75

O E

Eχ −=

− − − −= + + +

− − − −+ + + +

− −+ +

∑

2 2

2 2 2 2

(59 55.75) (68 55.75)

55.75 55.75 55.75(42 45.75) (46 45.75) (44 45.75) (51 45.75)

45.75 45.75 45.75 45.75.004 .039 .039 .107 .972 .359 .003 2.700 1.705 .593 .189 2.692

.307 .001 .067 .602 10.379

− −+ +

− − − −+ + + +

= + + + + + + + + + + ++ + + +

=

Step 5: Do not reject H0 since 10.379 < 21.666. Conclude that the distributions of favorite forms of camping are the same for all four regions.

11.37 Step 1: H0: The distributions of opinions about government policy toward illegal immigrants are the

same. Η1: The distributions of opinions about government policy toward illegal immigrants are different for at least two of the political affiliations. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (3 – 1)(4 – 1) = 6 For α = .05, the critical value of χ2 is 12.592. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Opinion A Opinion B Opinion C Unsure Total Democrat 55 43 288 8 394 (71.76) (52.83) (260.17) (9.24) Independent 19 25 107 6 157 (28.59) (21.05) (103.67) (3.68) Republican 89 52 196 7 344 (62.65) (46.12) (227.16) (8.07) Total 163 120 591 21 895



(continued)

22

2 2 2 2

2 2 2 2

2 2 2

( )

(55 71.76) (43 52.83) (288 260.17) (8 9.24)

71.76 52.83 260.17 9.24(19 28.59) (25 21.05) (107 103.67) (6 3.68)

28.59 21.05 103.67 3.68(89 62.65) (52 46.12) (196 227.16)

62.65 46.12 227.1

O E

Eχ −=

− − − −= + + +

− − − −+ + + +

− − −+ + +

∑

2(7 8.07)6 8.07

3.914 1.829 2.977 .166 3.217 .741 .107 1.463 11.083 .750 4.274 .14230.663

−+

= + + + + + + + + + + +=

Step 5: Reject H0 since 30.663 > 12.592. Conclude that the distributions of opinions about government policy toward illegal immigrants are different for at least two of the political affiliations.

11.38 Step 1: H0: The grade distributions are homogeneous for the three professors.

Η1: The grade distributions are not homogeneous for the three professors. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (4 – 1)(3 – 1) = 6 For α = .025, the critical value of χ2 is 14.449. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Professor Miller Smith Moore Total A 18 36 20 74 (24.67) (28.07) (21.26) B 25 44 15 84 Grade (28.00) (31.86) (24.14) C 85 73 82 240 (80.00) (91.03) (68.97) D & F 17 12 8 37 (12.33) (14.03) (10.63) Total 145 165 125 435

2

2

2 2 2 2 2

2 2 2 2

2 2

( )

(18 24.67) (36 28.07) (20 21.26) (25 28.00) (44 31.86)

24.67 28.07 21.26 28.00 31.86(15 24.14) (85 80.00) (73 91.03) (82 68.97)

24.14 80.00 91.03 68.97(17 12.33) (12 14.03) (

12.33 14.03

O E

Eχ −=

− − − − −= + + + +

− − − −+ + + +

− −+ + +

∑

28 10.63)10.63

1.803 2.240 .075 .321 4.626 3.461 .313 3.571 2.462 1.769 .294 .65121.586

−

= + + + + + + + + + + +=

Step 5: Reject H0 since 21.586 > 14.449. Conclude that the grade distributions are not homogeneous for the three professors.

Section 11.4 Inferences About the Population Variance 289

11.39 Step 1: H0: The distributions of opinions are homogeneous for the two groups of workers. Η1: The distributions of opinions are not homogeneous for the two groups of workers. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(3 – 1) = 2 For α = .025, the critical value of χ2 is 7.378. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Opinion Favor Oppose Uncertain Total Blue-collar 44 39 12 95 Workers (42.59) (42.59) (9.83) White-collar 21 26 3 50 Workers (22.41) (22.41) (5.17) Total 65 65 15 145

2 2 2 2 22

2 2

( ) (44 42.59) (39 42.59) (12 9.83) (21 22.41)

42.59 42.59 9.83 22.41(26 22.41) (3 5.17)

22.41 5.17.047 .303 .479 .089 .575 .911 2.404

O E

Eχ − − − − −= = + + +

− −+ +

= + + + + + =

∑

Step 5: Do not reject H0 since 2.404 < 7.378. Conclude that the distributions of opinions are homogeneous for the two groups of workers.

Section 11.4 11.40. a. df = n – 1 = 12 – 1 = 11

χ2 for 11 df and .025 area in the right tail = 21.920 χ2 for 11 df and .975 area in the right tail = 3.816 The 95% confidence interval for σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(12 1)(46)

21.920

− to

(12 1)(46)

3.816

− = 23.0839 to 132.5996

b. df = n – 1 = 16 – 1 = 15 χ2 for 15 df and .025 area in the right tail = 27.488 χ2 for 15 df and .975 area in the right tail = 6.262 The 95% confidence interval for σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(16 1)(46)

27.488

− to

(16 1)(46)

6.262

− = 25.1019 to 110.1884

c. df = n – 1 = 25 – 1 = 24 χ2 for 24 df and .025 area in the right tail = 39.364 χ2 for 24 df and .975 area in the right tail = 12.401 The 95% confidence interval for σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(25 1)(46)

39.364

− to

(25 1)(46)

12.401

− = 28.0459 to 89.0251

As the sample size increases, the confidence interval for σ2 decreases in width. 11.41 df = n – 1 = 25 – 1 = 24 a. χ2 for 24 df and .005 area in the right tail = 45.559

χ2 for 24 df and .995 area in the right tail = 9.886 The 99% confidence interval for σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(25 1)(35)

45.559

− to

(25 1)(35)

9.886

− = 18.4376 to 84.9686


b. χ2 for 24 df and .025 area in the right tail = 39.364 χ2 for 24 df and .975 area in the right tail = 12.401 The 95% confidence interval for σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(25 1)(35)

39.364

− to

(25 1)(35)

12.401

− = 21.3393 to 67.7365

c. χ2 for 24 df and .05 area in the right tail = 36.415 χ2 for 24 df and .95 area in the right tail = 13.848 The 90% confidence interval for σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(25 1)(35)

36.415

− to

(25 1)(35)

13.848

− = 23.0674 to 60.6586

As the confidence level decreases, the confidence interval for σ2 decreases in width. 11.42 a. H0: σ2 = 14, H1: σ2

≠ 14

b. df = n – 1 = 22 – 1 = 21 χ2 for 21 df and .025 area in the right tail = 35.479. χ2 for 21 df and .975 area in the right tail = 10.283.

c. χ2 = (n – 1)s2/σ2 = (22 – 1)(18)/14 = 27.000

d. Do not reject H0 since 10.283 < 27.000 < 35.479.

11.43 a. H0: σ2 = 1.75, H1: σ2 > 1.75

b. df = n –1 = 21 – 1 = 20 χ2 for 20 df and .025 area in the right tail = 34.170.

c. χ2 = (n – 1)s2/σ2 = (20 – 1)(1.97)/1.75 = 22.514

d. Do not reject H0 since 22.514 < 34.170.

Section 11.4 Inferences About the Population Variance 291

11.44 a. H0: σ2 = 6.0, H1: σ2 ≠ 6.0

b. df = n – 1 = 30 – 1 = 29 χ2 for 29 df and .025 area in the left tail is the same as χ2 for 24 df and .975 area in the right tail which gives χ2 = 16.047. χ2 for 29 df and .025 area in the right tail = 45.722.

c. χ2 = (n –1 )s2/σ2 = (30 – 1)(5.8)/6.0 = 28.033

d. Do not reject H0 since 16.047 < 28.033 < 45.722. 11.45 a. H0: σ2 = 2.2, H1: σ2 ≠ 2.2

b. df = n – 1 = 18 – 1 = 17 χ2 for 17 df and .025 area in the right tail = 30.191 χ2 for 17 df and .975 area in the right tail = 7.564

c. χ2 = (n –1)s2/σ2 = (18 – 1)(4.6)/2.2 = 35.545

d. Reject H0 since 35.545 > 30.191. 11.46 a. df = n – 1 = 51 – 1 = 50

χ2 for 50 df and .025 area in the right tail = 71.420 χ2 for 50 df and .975 area in the right tail = 32.357 The 95% confidence interval for the population variance σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(51 1)(2.13)

71.420

− to

(51 1)(2.13)

32.357

− = 1.4912 to 3.2914

The 95% confidence interval for σ is 1.4912 to 3.2914 = 1.221 to 1.814.

b. Step 1: H0: σ2 = 2.0, Hl: σ2 > 2.0 Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: For α = .025 with df = 50, the critical value of χ2 is 71.420. Step 4: χ2 = (n –1)s2/σ2 = (51 – 1)(2.13)/2.0 = 53.250 Step 5: Do not reject H0 since 53.250 < 71.420. Conclude that the population variance is not greater than 2.0 square micrometers.


11.47 a. df = n – 1 = 24 – 1 = 23 χ2 for 23 df and .01 area in the right tail = 41.638 χ2 for 23 df and .99 area in the right tail = 10.196 The 98% confidence interval for the population variance σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(24 1)(1.47)

41.638

− to

(24 1)(1.47)

10.196

− = .8120 to 3.3160

The 98% confidence interval for σ is .8120 to 3.3160 = .901 to 1.821

b. Step 1: H0: σ2 = 1.0, Hl: σ2 > 1.0 Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: For α = .01 with df = 23, the critical value of χ2 is 41.638. Step 4: χ2 = (n –1)s2/σ2 = (24 – 1)(1.47)/1.0 = 33.810 Step 5: Do not reject H0 since 33.810 < 41.638. Conclude that the population variance is not greater than 1.0 square grams.

11.48 a. df = n – 1 = 22 – 1 = 21


2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(22 1)(.62)

35.479

− to

(22 1)(.62)

10.283

− = .3670 to 1.2662


b. Step 1: H0: σ2 = .30, Hl: σ2 ≠ .30 Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: For α = .01 with df = 21, the critical values of χ2 are 8.034 and 41.401. Step 4: χ2 = (n –1)s2/σ2 = (22 – 1)(.62)/.30 = 43.400 Step 5: Reject H0 since 43.400 > 41.401. Conclude that the population variance is different from .30 mpg squared.

11.49 a. df = n – 1 = 25 – 1 = 24

χ2 for 24 df and .005 area in the right tail = 45.559 χ2 for 24 df and .995 area in the right tail = 9.886 The 99% confidence interval for σ 2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(25 1)(5200)

45.559

− to

(25 1)(5200)

9.886

− = 2739.3051 to 12,623.9126

The 99% confidence interval for σ is 2739.3051 to 12,623.9126 = 52.338 to 112.356

b. Step 1: H0: σ2 = 4200, Hl: σ2 ≠ 4200 Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: For α = .05 with df = 24, the critical values of χ2 are 12.401 and 39.364. Step 4: χ2 = (n –1)s2/σ2 = (25 – 1)(5200)/4200 = 29.714 Step 5: Do not reject H0 since 12.401 < 29.714 < 39.364. Conclude that the population variance is not different from 4200 square hours.

Chapter 11 Supplementary Exercises 293

Supplementary Exercises 11.50 Step 1: H0: The distribution of recent auto sales is the same as the November 2011 distribution.

Η1: The distribution of recent auto sales is different from the November 2011 distribution. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .025, the critical value of χ2 is 9.348. Step 4:

Segment O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε Cars 374 .4659 340.11 33.89 1148.735 3.378 Light-duty trucks 138 .2023 147.68 −9.68 93.683 0.634 SUVs 65 .1152 84.10 −19.10 364.657 4.336 Crossovers 153 .2166 158.12 −5.12 26.194 0.166 n = 730 Sum = 8.514

χ2 = Σ (O – E)2/E = 8.514 Step 5: Do not reject H0 since 8.514 < 9.348. Conclude that the distribution of recent auto sales is not different from the November 2011 distribution.

11.51 Step 1: H0: The percentage of people who consume total-bran cereal is the same for all four brands.

Η1: The percentage of people who consume total-bran cereal is not the same for all four brands. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .05, the critical value of χ2 is 7.815. Step 4:

Brand O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε A 212 .25 250 –38 1444 5.776 B 284 .25 250 34 1156 4.624 C 254 .25 250 4 16 .064 D 250 .25 250 0 0 .000 n = 1000 Sum = 10.464

χ2 = Σ (O – E)2/E = 10.464 Step 5: Reject H0 since 10.464 > 7.815. Conclude that the percentage of people who consume total-bran cereal is not the same for all four brands.

11.52 Step 1: H0: The 2012 distribution of birth weights for all children born in North Carolina who

shared multiple births is the same as the one for 2009. Η1: The 2012 distribution of birth weights for all children born in North Carolina who shared multiple births is different than the one for 2009. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .025, the critical value of χ2 is 9.348 Step 4:

Weight O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε 0- 500 2 .0145 8.51 −6.51 42.400 4.981 501-1500 60 .1102 64.69 −4.69 21.972 0.340 1501-2500 305 .4923 288.98 16.02 256.637 0.888 2501-8165 220 .3830 224.82 −4.82 23.242 0.103 n = 587 Sum = 6.313

χ2 = Σ (O – E)2/E = 6.313 Step 5: Do not reject H0 since 6.3313 < 9.348. Conclude that the 2011 distribution of birth weights for all children born in North Carolina who shared multiple births is not different than the one for 2009.


11.53 Step 1: H0: The movies are equally preferred. Η1: The movies are not equally preferred. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 5, df = k – 1 = 5 – 1 = 4 For α = .01, the critical value of χ2 is 13.277. Step 4:

Movie O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε It’s a Wonderful Life 247 .20 193.60 53.40 2851.560 14.729 A Christmas Story 237 .20 193.60 43.40 1883.560 9.729 Miracle on 34th Street 226 .20 193.60 32.40 1049.760 5.422 White Christmas 124 .20 193.60 −69.60 4844.160 25.021 A Christmas Carol 134 .20 193.60 −59.60 3552.160 18.348 n = 968 Sum = 73.249

χ2 = Σ (O – E)2/E = 73.25 Step 5: Reject H0 since 73.25 > 13.277 Conclude that the five movies are not equally preferred.

11.54 Step 1: H0: The percentages of investors favoring the four choices are equal.

Η1: The percentages of investors favoring the four choices are not equal. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .025, the critical value of χ2 is 9.348. Step 4:

Favored choice O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε Stocks 46 .25 35 11 121 3.457 Bonds 41 .25 35 6 36 1.029 Real estate 32 .25 35 –3 9 .257 Cash 21 .25 35 –14 196 5.600 n = 140 Sum = 10.343

χ2 = Σ (O – E)2/E = 10.343 Step 5: Reject H0 since 10.343 > 9.348. Conclude that the percentages of investors favoring the four choices are not equal.

11.55 Step 1: H0: The proportions of all allergic persons are the same over the four seasons.

Η1: The proportions of all allergic persons are not the same over the four seasons. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .01, the critical value of χ2 is 11.345. Step 4:

Season O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε Fall 18 .25 25 –7 49 1.960 Winter 13 .25 25 –12 144 5.760 Spring 31 .25 25 6 36 1.440 Summer 38 .25 25 13 169 6.760 n = 100 Sum = 15.920

χ2 = Σ (O – E)2/E = 15.920 Step 5: Reject H0 since 15.920 > 11.345. Conclude that the proportions of all allergic persons are not the same over the four seasons.


11.56 Step 1: H0: Sentencing for shoplifting does not depend on which judge tries the case. Η1: Sentencing for shoplifting depends on which judge tries the case. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(2 – 1) = 1 For α = .05, the critical value of χ2 is 3.841.


Judge Jail Other Sentence Total Stark 27 65 92 (29.64) (62.36) Rivera 31 57 88 (28.36) (59.64) Total 58 122 180

2 2 2 2 2

2 ( ) (27 29.64) (65 62.36) (31 28.36) (57 59.64)

29.64 62.36 28.36 59.64.235 .112 .246 .117 .710

O E

Eχ − − − − −= = + + +

= + + + =∑

Step 5: Do not reject H0 since .710 < 3.841. Conclude that sentencing for shoplifting does not depend on which judge tries the case.

11.57 Step 1: H0: Political affiliation and opinion about the American Healthcare Act are independent.

Η1: Political affiliation and opinion about the American Healthcare Act are dependent. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (3 – 1)(3 – 1) = 4 For α = .05, the critical value of χ2 is 9.488. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Democrat Independent Republican Total Expand it 216 143 99 458 (173.11) (161.89) (123.01) Keep it as is 126 149 129 404 (152.70) (142.80) (108.50) Repeal it or repeal 121 141 101 363 and replace it (137.20) (128.31) (97.49) Total 463 433 329 1225

22

2 2 2 2 2

2 2 2 2

( )

(216 173.11) (143 161.89) (99 123.01) (126 152.70) (149 142.80)

173.11 161.89 123.01 152.70 142.80(129 108.50) (121 137.20) (141 128.31) (101 97.49)

108.50 137.20 128.31 97.4910.626 2.20

O E

Eχ −=

− − − − −= + + + +

− − − −+ + + +

= +

∑

4 4.686 4.669 .269 3.873 1.913 1.255 .12629.622

+ + + + + + +=

Step 5: Reject H0 since 29.622 > 9.488. Conclude that political affiliation and opinion about the American Healthcare Act are dependent.

11.58 Step 1: H0: The percentage of Americans age 18−25 who used illicit drugs is the same in each year.

Η1: The percentage of Americans age 18−25 who used illicit drugs is not the same in each year. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(3 – 1) = 2 For α = .01, the critical value of χ2 is 9.210.



(continued)


2008 2009 2010 Total Yes 706 763 774 2243 (747.67) (747.67) (747.67) No 2894 2837 2826 8557 (2852.33) (2852.33) (2852.33) Total 3600 3600 3600 10,800

2 2 2 22

2 2 2

( ) (706 747.67) (763 747.67) (774 747.67)

747.67 747.67 747.67(2984 2852.33) (2837 2852.33) (2826 2852.33)

2852.33 2852.33 2852.33 2.322 .314 .927 .609 .082 .2434.498

O E

Eχ − − − −= = + +

− − −+ + +

= + + + + +=

∑

Step 5: Do not reject H0 since 4.498 < 9.210. Conclude that the percentage of Americans age 18−25 who used illicit drugs is the same in each year.

11.59 Step 1: H0: Gender and marital status are not related for all persons who hold more than one job.

Η1: Gender and marital status are related for all persons who hold more than one job. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(3 – 1) = 2 For α = .01, the critical value of χ2 is 9.210. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Gender Single Married Other Total Male 72 209 39 320 (67.20) (199.04) (53.76) Female 33 102 45 180 (37.80) (111.96) (30.24) Total 105 311 84 500

22

2 2 2 2

2 2

( )

(72 67.20) (209 199.04) (39 53.76) (33 37.80)67.20 199.04 53.76 37.80

(102 111.96) (45 30.24)

111.96 30.24.343 .498 4.052 .610 .886 7.204 13.593

O E

Eχ −=

− − − −= + + +

− −+ +

= + + + + + =

∑

Step 5: Reject H0 since 13.593 > 9.210. Conclude that gender and marital status are related for all persons who hold more than one job.

11.60 Step 1: H0: The distributions of opinions in regard to ATVs are the same for both age groups.

Η1: The distributions of opinions in regard to ATVs are not the same for both age groups. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(3 – 1) = 2 For α = .025, the critical value of χ2 is 7.378. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.



(continued)

Age group More Restrictions Fewer Restrictions No Change Total 18 to 27 40 92 68 200 (46.34) (78.05) (75.61) 28 to 37 55 68 87 210 (48.66) (81.95) (79.39) Total 95 160 155 410

22

2 2 2 2 2 2

( )

(40 46.34) (92 78.05) (68 75.61) (55 48.66) (68 81.95) (87 79.39)46.34 78.05 75.61 48.66 81.95 79.39

.867 2.493 .766 .826 2.375 .729 8.056

O E

Eχ −=

− − − − − −= + + + + +

= + + + + + =

∑

Step 5: Reject H0 since 8.056 > 7.378. Conclude that the distributions of opinions in regard to ATVs are not the same for both age groups.

11.61 Step 1: H0: The percentages of people with different opinions are similar for all four regions.

Η1: The percentages of people with different opinions are not similar for all four regions. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (4 – 1)(3 – 1) = 6 For α = .01, the critical value of χ2 is 16.812. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Region Favor Oppose Uncertain Total Northeast 56 33 11 100 (63.75) (29.75) (6.50) Midwest 73 23 4 100 (63.75) (29.75) (6.50) South 67 28 5 100 (63.75) (29.75) (6.50) West 59 35 6 100 (63.75) (29.75) (6.50) Total 255 119 26 400

22

2 2 2 2 2

2 2 2 2

2 2 2

( )

(56 63.75) (33 29.75) (11 6.50) (73 63.75) (23 29.75)

63.75 29.75 6.50 63.75 29.75(4 6.50) (67 63.75) (28 29.75) (5 6.50)

6.50 63.75 29.75 6.50(59 63.75) (35 29.75) (6 6.50)

63.75 29.75

O E

Eχ −=

− − − − −= + + + +

− − − −+ + + +

− − −+ + +

∑

6.50.942 .355 3.115 1.342 1.532 .962 .166 .103 .346 .354 .926 .03810.181

= + + + + + + + + + + +=

Step 5: Do not reject H0 since 10.181 < 16.812. Conclude that the percentages of people with different opinions are similar for all four regions.


11.62 a. df = n – 1 = 21 – 1 = 20 χ2 for 20 df and .01 area in the right tail = 37.566 χ2 for 20 df and .99 area in the right tail = 8.260 The 98% confidence interval for the population variance σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(21 1)(9.2)

37.566

− to

(21 1)(9.2)

8.260

− = 4.8980 to 22.2760

The 98% confidence interval for σ is 4.8980 to 22.2760 = 2.213 to 4.720


22/

2)1(

αχsn −

to 2

2/1

2)1(

αχ −

− sn

= 000.32

)7.1)(117( −

to 812.5

)7.1)(117( −

= .8500 to 4.6800

The 98% confidence interval for σ is .8500 to 4.6800 = .922 to 2.163 11.63 a. df = n – 1 = 10 – 1 = 9


2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(10 1)(7.2)

19.023

− to

(10 1)(7.2)

2.700

− = 3.4064 to 24.0000



2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(18 1)(14.8)

30.191

− to

(18 1)(14.8)

7.564

− = 8.3336 to 33.2628

The 95% confidence interval for σ is 8.3336 to 33.2628 = 2.887 to 5.767 11.64 Step 1: H0: σ2 = 6.5, Hl: σ2 ≠ 6.5

Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: df = n – 1 = 21 – 1 = 20 For α = .05, the critical values of χ2 are 9.591 and 34.170. Step 4: χ2 = (n –1)s2/σ2 = (21 – 1)(9.2)/6.5 = 28.308 Step 5: Do not reject H0 since 9.591 < 28.308 < 34.170. Conclude that the population variance is not different from 6.5.

11.65 Step 1: H0: σ2 = 1.1, Hl: σ2 > 1.1

Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: df = n – 1 = 17 – 1 = 16 For α = .025, the critical value of χ2 is 28.845. Step 4: χ2 = (n –1)s2/σ2 = (17– 1)(1.7)/1.1 = 24.727 Step 5: Do not reject H0 since 24.727 < 28.845. Conclude that the population variance is not greater than 1.1.


11.66 Step 1: H0: σ2 = 4.2, Hl: σ2 > 4.2 Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: df = n – 1 = 10 – 1 = 9 For α = .01, the critical value of χ2 is 21.666. Step 4: χ2 = (n –1)s2/σ2 = (10 – 1)(7.2)/4.2 = 15.429 Step 5: Do not reject H0 since 15.429 < 21.666. Conclude that the population variance is not greater than 4.2.

11.67 Step 1: H0: σ2 = 10.4, Hl: σ2 ≠ 10.4

Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: df = n – 1 = 18 – 1 = 17 For α = .05, the critical values of χ2 are 7.564 and 30.191. Step 4: χ2 = (n –1)s2/σ2 = (18 – 1)(14.8)/10.4 = 24.192 Step 5: Do not reject H0 since 7.564 < 24.192 < 30.191. Conclude that the population variance is not different from 10.4.

11.68 a. Step 1: H0: σ2 = 4.0, Hl: σ2 > 4.0

Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: df = n – 1 = 25 – 1 = 24 For α = .01, the critical value of χ2 is 42.980. Step 4: χ2 = (n –1)s2/σ2 = (25 – 1)(8.3)/4.0 = 49.800 Step 5: Reject H0 since 49.800 > 42.980. Conclude that the population variance is greater than 4.0 square minutes.

b. χ2 for 24 df and .005 area in the right tail = 45.559 χ2 for 24 df and .995 area in the right tail = 9.886 The 99% confidence interval for the population variance σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(25 1)(8.3)

45.559

− to

(25 1)(8.3)

9.886

− = 4.3724 to 20.1497

11.69 a. Step 1: H0: σ2 = 5000, Hl: σ2 < 5000

Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: df = n – 1 = 20 – 1 = 19 For α = .025, the critical value of χ2 is 8.907. Step 4: χ2 = (n –1)s2/σ2 = (20 – 1)(3175)/5000 = 12.065 Step 5: Do not reject H0 since 12.065 > 8.907. Conclude that the population variance is not less than 5000.

b. χ2 for 19 df and .01 area in the right tail = 36.191 χ2 for 19 df and .99 area in the right tail = 7.633 The 98% confidence interval for σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(20 1)(3175)

36.191

− to

(20 1)(3175)

7.633

− = 1666.8509 to 7903.1835

The 98% confidence interval for the population standard deviation σ is

1666.8509 to 7903.1835 = 40.827 to 88.900 11.70 a. Step 1: H0: σ2 ≤ .025, Hl: σ2 > .025

Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: df = n – 1 = 23 – 1 = 22 For α = .05, the critical value of χ2 is 33.924. Step 4: χ2 = (n –1)s2/σ2 = (23 – 1)(.034)/.025 = 29.920 Step 5: Do not reject H0 since 29.920 < 33.924. Conclude that the population variance is not greater than .025 square millimeter, and the machine does not need an adjustment.


b. χ2 for 22 df and .025 area in the right tail = 36.781 χ2 for 22 df and .975 area in the right tail = 10.982 The 95% confidence interval for the population variance σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(23 1)(.034)

36.781

− to

(23 1)(.034)

10.982

− = .0203 to .0681

11.71 a. df = n – 1 = 25 – 1 = 24

χ2 for 24 df and .005 area in the right tail = 45.559 χ2 for 24 df and .995 area in the right tail = 9.886 The 99% confidence interval for σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(25 1)(.19)

45.559

− to

(25 1)(.19)

9.886

− = .1001 to .4613

The 99% confidence interval for σ is .1001 to .4613 = .316 to .679

b. Step 1: H0: σ2 = .13, Hl: σ2 ≠ .13 Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: For α = .01 with df = 24, the critical values of χ2 are 9.886 and 45.559. Step 4: χ2 = (n –1)s2/σ2 = (25 – 1)(.19)/.13 = 35.077 Step 5: Do not reject H0 since 9.886 < 35.077 < 45.559. Conclude that the population variance is not different from .13.

11.72 a. ∑x = 278.1, ∑x2 = 11,199.27

( )2 22

2

(278.1)11,199.27

7 25.125714281 7 1

xx

nsn

− −= = =

− −

∑∑

b. df = n – 1 = 7 – 1 = 6 χ2 for 6 df and .01 area in the right tail = 16.812 χ2 for 6 df and .99 area in the right tail = .872 The 98% confidence interval for σ2 is

2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(7 1)(25.12571428)

16.812

− to

(7 1)(25.12571428)

.872

− = 8.9671 to 172.8834


c. Step 1: H0: σ2 = 20, Hl: σ2 > 20 Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: For α = .05 with df = 6, the critical value of χ2 is 12.592. Step 4: χ2 = (n –1)s2/σ2 = (7 – 1)(25.12571428)/20 = 7.538 Step 5: Do not reject H0 since 7.538 < 12.592. Conclude that the population variance is not larger than 20 square pounds.

11.73 a. ∑x = 4897, ∑x2 = 3,010,461

( )2 22

2

(4897)3,010, 461

8 1840.6964291 8 1

xx

nsn

− −= = =

− −

∑∑

b. df = n – 1 = 8 – 1 = 7 χ2 for 7 df and .025 area in the right tail = 16.013; χ2 for 7 df and .975 area in the right tail = 1.690 The 95% confidence interval for σ2 is

2 2

2 2/2 1 /2

( 1) ( 1) (8 1)(1840.696429) (8 1)(1840.696459) to to 804.6509 to 7624.1864

16.013 1.690

n s n s

α αχ χ −

− − − −= =



c. Step 1: H0: σ2 = 750, Hl: σ2 ≠ 750 Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: For α = .05 with df = 7, the critical values of χ2 are 1.690 and 16.013. Step 4: χ2 = (n –1)s2/σ2 = (8 – 1)(1840.696429)/750 = 17.180 Step 5: Reject H0 since 17.180 > 16.013. Conclude that the population variance is different from 750 square dollars.

11.74 a. The counts for the categories Mom’s spouse and The guests bring food would have to be 57 and 27,

respectively, for the value of the test statistic to be as small as possible since 300(.19) = 57 and 300(.09) = 27.

b. Since 300(.38) = 114 and 300(.34) = 102, (O – E)2/E = 0 for each of these categories. For α = .10 with df = 3, the critical value of χ2 is 6.251. The squared difference between the observed and expected values for the two remaining categories will be equal as whatever value we add to one category will be subtracted from the other. Let d2 represent this common squared difference. Then we want the

minimum value of d such that 2 2

6.251.57 27

d d+ > Thus, d = 11, and the count for Mom’s spouse is

57 + 11 = 68. Then 2 2 2 2 2

2 ( ) (114 114) (102 102) (68 57) (16 27)

114 102 57 27 0 0 2.123 4.482 6.605 6.251

O E

Eχ − − − − −= = + + +

= + + + = >∑

c. The count for Mom’s spouse would be 57 – 11 = 46, and the count for The guests bring food is 38. 2 2 2 2 2

2 ( ) (114 114) (102 102) (46 57) (38 27)

114 102 57 270 0 2.123 4.482 6.605

O E

Eχ − − − − −= = + + +

= + + + =∑

The value is the same as in part b. 11.75 Step 1: H0: Opinions on disposal site are independent of gender.

Hl: Opinions on disposal site are dependent on gender. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(3 – 1) = 2 For α = .05, the critical value of χ2 is 5.991. Step 4: From the given data we can calculate the following: Total number opposed = 200(.60) = 120 Total number in favor = 200(.32) = 64 Total number undecided = 200(.08) = 16 Number of women opposed = 120(.65) = 78 Number of men opposed = 120 – 78 = 42 Number of men in favor = 64(.625) = 40 Number of women in favor = 64 – 40 = 24 Number of women undecided = 110 – 78 – 24 = 8 Number of men undecided = 16 – 8 = 8 Using these results, we may construct the following contingency table of observations and expected values.

Opposed In Favor Undecided Total Women 78 24 8 110 (66.0) (35.2) (8.8) Men 42 40 8 90 (54.0) (28.8) (7.2) Total 120 64 16 200



(continued)

2

2

2 2 2 2 2 2

( )

(78 66.0) (24 35.2) (8 8.8) (42 54.0) (40 28.8) (8 7.2)66.0 35.2 8.8 54.0 28.8 7.2

2.182 3.564 .073 2.667 4.356 .089 12.931

O E

Eχ −=

− − − − − −= + + + + +

= + + + + + =

∑

Step 5: Reject H0 since 12.931 > 5.991. Conclude that opinions on disposal site are dependent on gender.

11.76 a. Step 1: H0: 1 2p p− = 0, H1: 1 2p p− ≠ 0

Step 2: Since 1 1ˆn p , 1 1ˆn q , 2 2ˆn p , and 2 2ˆn q are all greater than 5, use the normal distribution.

Step 3: 1p̂ = 48/52 = .923 and 2p̂ = 44/54 = .815

1 2

1 2

x xp

n n

+=+

48 44

52 54

+=+

= .868 and q = 1 – p = 1 – .868 = .132

1 2ˆ ˆp ps − =

1 2

1 1p q

n n

⎛ ⎞+⎜ ⎟⎝ ⎠

=1 1

(.868)(.132)52 54⎛ ⎞+⎜ ⎟⎝ ⎠ = .06576597

z = 1 2

1 2 1 2

ˆ ˆ

ˆ ˆ( ) ( ) (.923 .815) 0

.06576597p p

p p p p

s −

− − − − −= = 1.64

From the normal distribution table, area to the right of z = 1.64 is 1 – .9495 = .0505. p-value = 2(.0505) = .1010

b. Step 1: H0: The distributions of final grades for the two instructors are independent. Η1: The distributions of final grades for the two instructors are not independent. Step 2: Since this is a test of independence, use the chi-square distribution.

Pass Did Not Total Pass Instructor A 48 4 52 (45.13) (6.87) Instructor B 44 10 54 (46.87) (7.13) Total 92 14 106

2 2 2 2 2

2 ( ) (48 45.13) (4 6.87) (44 46.87) (10 7.13)

45.13 6.87 46.87 7.13.183 1.199 .176 1.155 2.713

O E

Eχ − − − − −= = + + +

= + + + =∑

For df = 1 and χ2 = 2.713, the area in the right tail is about .1000. Hence, p – value = .1000

c. z2 = (1.64)2 = 2.6896 is approximately equal to χ2 = 2.713. The difference is due to rounding. The p-values are the same (the difference is due to rounding), which is always true since the tests in parts a and b are equivalent.

11.77 Step 1: H0: The proportions of red and green marbles are the same in all five boxes.

Η1: The proportions of red and green marbles are not the same in all five boxes. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(5 – 1) = 4 For α = .05, the critical value of χ2 is 9.488.



(continued)


Box 1 2 3 4 5 Total Red 20 14 23 30 18 105 (21) (21) (21) (21) (21) Green 30 36 27 20 32 145 (29) (29) (29) (29) (29) Total 50 50 50 50 50 250

22

2 2 2 2 2

2 2 2 2 2

( )

(20 21) (14 21) (23 21) (30 21) (18 21)

21 21 21 21 21(30 29) (36 29) (27 29) (20 29) (32 29)

29 29 29 29 29.048 2.333 .190 3.857 .429 .034 1.690 .138 2.793 .31011.822

O E

Eχ −=

− − − − −= + + + +

− − − − −+ + + + +

= + + + + + + + + +=

∑

Step 5: Reject H0 since 11.822 > 9.488. Conclude that the proportions of red and green marbles are not the same in all five boxes.

11.78 The following table displays the expected counts for the cells under the assumption that the two variables

are independent: Variable 1 A B C Total

X 39.60 33.60 46.80 120 Y 67.65 57.40 79.95 205 Variable 2 Z 57.75 49.00 68.25 175

Total 165 140 195 500

Since the expected counts are non-integer values, the value of 2( )O E

E

− for each cell will not be an

integer. Note that this implies that none of these values will be equal to zero. Moreover, since each of these values is positive, the sum will be a positive value. Hence, the value of the chi-square test statistic for this table cannot be equal to zero. Alternatively, the only way the χ2 value can be 0 is if all the expected counts and observed counts are equal. This cannot happen for any cell where the expected count is not an integer.

11.79 Step 1: H0: A normal distribution is an appropriate model for these data.

Η1: A normal distribution is not an appropriate model for these data. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 10, df = k – 1 = 10 – 1 = 9 For α = .05, the critical value of χ2 is 16.919. Step 4: We will find the probability values necessary to fill in the table used to calculate the chi-square test statistic using the standard normal distribution table. For example, for the second row, we will find p = P(−2 < z < −1.5) = P(z < −1.5) − P(z < −2) = .0668 − .0228 = .0440. Note the symmetry in the standard normal distribution corresponds to the symmetries in the probabilities, and, therefore, in the expected counts.



(continued)

Category O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε z score below −2 48 .0228 22.8 25.2 635.04 27.853 z score from −2 to −1.5 67 .0440 44.0 23.0 529.00 12.023 z score from −1.5 to −1 146 .0919 91.9 54.1 2926.81 31.848 z score from −1 to −0.5 248 .1498 149.8 98.2 9643.24 64.374 z score from −0.5 to 0 187 .1915 191.5 −4.5 20.25 .106 z score from 0 to 0.5 125 .1915 191.5 −66.5 4422.25 23.093 z score from 0.5 to 1 88 .1498 149.8 −61.8 3819.24 25.496 z score from 1 to 1.5 47 .0919 91.9 −44.9 2016.01 21.937 z score from 1.5 to 2 25 .0440 44.0 −19.0 361.00 8.205 z score of 2 or above 19 .0228 22.8 −3.8 14.44 .633 n = 1000 Sum = 215.568

χ2 = Σ (O – E)2/E = 215.568 Step 5: Reject H0 since 215.568 > 16.919. Conclude that a normal distribution is not an appropriate model for these data.

11.80 The hypothesis test in Exercise 11.61 is a test of homogeneity because we want to determine if the

populations (the four regions) have similar distributions with respect to opinions about a farm subsidy program. To conduct the test of homogeneity, a random sample of size 100 was selected from each region, and the opinions of these 400 persons were recorded. To change to a test used to determine if region and opinion are independent, we would take a random sample of 400 people from across the nation and record both the region in which they reside and their opinion.

11.81 a. Step 1: H0: All categories are equally likely.

Η1: All categories are not equally likely. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 4, df = k – 1 = 4 – 1 = 3 For α = .10, the largest reasonable significance level, the critical value of χ2 is 6.251. Step 4:

Category O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε A 21 .25 25 −4 16 .64 B 26 .25 25 1 1 .04 C 31 .25 25 6 36 1.44 D 22 .25 25 −3 9 .36 n = 100 Sum = 2.48

χ2 = Σ (O – E)2/E = 2.48 Step 5: Do not reject H0 since 2.48 < 6.251. Conclude that all categories are equally likely. Note that for any α < .10, χ2 > 6.251. Hence, we would not reject H0 for any reasonable significance level.

b. We must have ∑ | O – E | = 14 and ∑(O – E) = 0. Under these constraints, the value of the chi-square test statistic is maximized when the four values for O – E are 0, 0, 7, and –7. In such a case, calculating

the chi-square test statistic yields 2 2 2 2

2 (0) (0) (7) (7)3.92

25 25 25 25χ = + + + = .

Since 3.92 < 6.251, the p-value > .10.

Chapter 11 Self-Review Test 305

Self-Review Test 1. b 2. a 3. c 4. a 5. b 6. b 7. c 8. b 9. a 10. Step 1: H0: The current distribution of the source of health insurance is the same as the 2010 distribution.

Η1: The current distribution of the source of health insurance differs from the 2010 distribution. Step 2: Since this is a multinomial experiment, use the chi-square distribution. Step 3: k = 6, df = k – 1 = 6 – 1 = 5 For α = .05, the critical value of χ2 is 11.070. Step 4:

Source O p Ε = np Ο – Ε (Ο – Ε)2 (Ο – Ε) 2/Ε Employer 7286 0.4912 7368.00 −82.00 6724.000 0.913 Individual 698 0.0489 733.50 −35.50 1260.250 1.718 Medicaid 2402 0.1586 2379.00 23.00 529.000 0.222 Medicare 1927 0.1249 1873.50 53.50 2862.250 1.528 Other Public 171 0.0129 193.50 −22.50 506.250 2.616 Uninsured 2516 0.1635 2452.50 63.50 4032.250 1.644 n = 15,000 Sum = 8.641

χ2 = Σ (O – E)2/E = 8.641 Step 5: Do not reject H0 since 8.641 < 11.070. Conclude that the current distribution of the source of health insurance is the same as the 2010 distribution.

11. Step 1: H0: Educational level and ever being divorced are independent.

Η1: Educational level and ever being divorced are dependent. Step 2: Since this is a test of independence, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (2 – 1)(4 – 1) = 3 For α = .01, the critical value of χ2 is 11.345. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Educational Level Less than High school Some College High school degree college degree Total Divorced 173 158 95 53 479 (160.47) (136.04) (98.20) (84.30) Never 162 126 110 123 521 Divorced (174.54) (147.96) (106.81) (91.70) Total 335 284 205 176 1000

2

2

2 2 2 2

2 2 2 2

( )

(173 160.47) (158 136.04) (95 98.20) (53 84.30)160.47 136.04 98.20 84.30

(162 174.54) (126 147.96) (110 106.81) (123 91.70)

174.54 147.96 106.81 91.70.978 3.545 .104 11.621 .901 3.259 .0

O E

Eχ −=

− − − −= + + +

− − − −+ + + +

= + + + + + +

∑

95 10.684 31.187+ =

Step 5: Reject H0 since 31.187 > 11.345. Conclude that educational level and ever being divorced are dependent.


12. Step 1: H0: The percentages of people who play the lottery often, sometimes, and never are the same for each income group. Η1: The percentages of people who play the lottery often, sometimes, and never are not the same for each income group. Step 2: Since this is a test of homogeneity, use the chi-square distribution. Step 3: df = (R – 1)(C – 1) = (3 – 1)(3 – 1) = 4 For α = .05, the critical value of χ2 is 9.488. Step 4: The expected frequencies are given in parentheses below the observed frequencies in the table.

Income Group Low Middle High Total Play often 174 163 90 427 (170.80) (142.33) (113.87) Play sometimes 286 217 120 623 (249.20) (207.67) (166.13) Never Play 140 120 190 450 (180.00) (150.00) (120.00) Total 600 500 400 1500

2

2

2 2 2 2 2

2 2 2 2

( )

(174 170.80) (163 142.33) (90 113.87) (286 249.20) (217 207.67)

170.80 142.33 113.87 249.20 207.67(120 166.13) (140 180.00) (120 150.00) (190 120.00)

166.13 180.00 150.00 120.00.060 3.00

O E

Eχ −=

− − − − −= + + + +

− − − −+ + + +

= +

∑

2 5.004 5.434 .419 12.809 8.889 6.000 40.83382.450

+ + + + + + +=

Step 5: Reject H0 since 82.450 > 9.488. Conclude that the percentages of people who play the lottery often, sometimes, and never are not the same for each income group.

13. a. df = n – 1 = 20 – 1 = 19


2

2/2

( 1)n s

αχ−

to 2

21 /2

( 1)n s

αχ −

− =

(20 1)(.48)

38.582

− to

(20 1)(.48)

6.844

− = .2364 to 1.3326


b. Step 1: H0: σ2 = .25, Hl: σ2 > .25 Step 2: Since this is a test about a population variance, use the chi-square distribution. Step 3: For α = .01 with df = 19, the critical value of χ2 is 36.191. Step 4: χ2 = (n –1)s2/σ2 = (20 – 1)(.48)/.25 = 36.480 Step 5: Reject H0 since 36.480 > 36.191. Conclude that the population variance exceeds .25 square ounce.

chapter 11 chi-square tests

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