ch11 solutions

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252 The Practice of Statistics for AP*, 4/e Chapter 11 Section 11.1 Check Your Understanding, page 681: 1. 0 blue orange green yellow red brown : 0.23, 0.23, 0.15, 0.15, 0.12, 0.12 H p p p p p p = = = = = = : At least one of the 's is incorrect. a i H p 2. There were 12 7 13 4 8 2 46 + + + + + = candies in the bag. The expected count of both blue and orange candies is ( ) 46 0.23 10.58, = for green and yellow is ( ) 46 0.15 6.9, = and for red and brown is ( ) 46 0.12 5.52. = 3. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 12 10.58 7 10.58 13 6.9 4 6.9 8 5.52 2 5.52 10.58 10.58 6.9 6.9 5.52 5.52 χ = + + + + + 0.1906 1.2114 5.3928 1.2188 1.1142 2.2446 11.3724. = + + + + + = Check Your Understanding, page 684: 1. The expected counts calculated in Exercise 2 of the previous Check Your Understanding were all at least 5. We should use the chi-square distribution with 6 1 5 = df. 2. 3. From Table C 0.025 P-value 0.05. < < From the calculator the P-value is 0.0445. 4. Since the P-value is less than 0.05, we reject the null hypothesis and conclude that at least one of the proportions of M&M’s Peanut Chocolate Candies of a particular color is different than what the company reported. Check Your Understanding, page 689: 1. State: We want to perform a test at the 0.01 α = significance level of 0 red-straight red-curly white-straight white-curly 9 3 3 1 : , , , 16 16 16 16 H p p p p = = = = versus : At least one of the 's is incorrect. a i H p Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: A random sample was used. Large sample size: The expected counts in each category are red-straight: 9 200 112.5, 16 = red-curly and white straight: 3 200 37.5, 16 = and

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Page 1: Ch11 Solutions

252 The Practice of Statistics for AP*, 4/e

Chapter 11 Section 11.1 Check Your Understanding, page 681: 1. 0 blue orange green yellow red brown: 0.23, 0.23, 0.15, 0.15, 0.12, 0.12H p p p p p p= = = = = = : At least one of the 's is incorrect.a iH p 2. There were 12 7 13 4 8 2 46+ + + + + = candies in the bag. The expected count of both blue and orange candies is ( )46 0.23 10.58,= for green and yellow is ( )46 0.15 6.9,= and for red and brown is

( )46 0.12 5.52.=

3. ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2

2 12 10.58 7 10.58 13 6.9 4 6.9 8 5.52 2 5.5210.58 10.58 6.9 6.9 5.52 5.52

χ− − − − − −

= + + + + +

0.1906 1.2114 5.3928 1.2188 1.1142 2.2446 11.3724.= + + + + + = Check Your Understanding, page 684: 1. The expected counts calculated in Exercise 2 of the previous Check Your Understanding were all at least 5. We should use the chi-square distribution with 6 1 5− = df. 2.

3. From Table C 0.025 P-value 0.05.< < From the calculator the P-value is 0.0445. 4. Since the P-value is less than 0.05, we reject the null hypothesis and conclude that at least one of the proportions of M&M’s Peanut Chocolate Candies of a particular color is different than what the company reported. Check Your Understanding, page 689: 1. State: We want to perform a test at the 0.01α = significance level of

0 red-straight red-curly white-straight white-curly9 3 3 1: , , ,

16 16 16 16H p p p p= = = = versus

: At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: A random sample was used. Large sample size: The expected counts

in each category are red-straight: 9200 112.5,16 =

red-curly and white straight: 3200 37.5,16 =

and

Page 2: Ch11 Solutions

Chapter 11: Inference for Distributions of Categorical Data 253

white-curly: 1200 12.5,16 =

all of which are at least 5. Independent: It seems reasonable that there

would be more than 2000 pairs of fruit flies. The conditions are met. Do: The test statistic is ( ) ( ) ( ) ( )2 2 2 2

2 99 112.5 42 37.5 49 37.5 10 12.56.1867

112.5 37.5 37.5 12.5χ

− − − −= + + + = and the distribution has

4 1 3− = df. The P-value is 0.1029. Conclude: Since the P-value is greater than 0.01, we fail to reject the null hypothesis. We do not have enough evidence to reject the biologists predicted distribution of traits in offspring. Exercises, page 692: 11.1 (a) 0 cashews almonds macadamia brazil: 0.52, 0.27, 0.13, 0.08H p p p p= = = = versus

: At least one of the 's is incorrect.a iH p (b) There was a random sample of 150 nuts. The expected count for cashews is ( )150 0.52 78,= for almonds is ( )150 0.27 40.5,= for macadamia nuts is

( )150 0.13 19.5,= and for brazil nuts is ( )150 0.08 12.=

11.2 (a) 0 red black green18 18 2: , ,38 38 38

H p p p= = = versus : At least one of the 's is incorrect.a iH p (b) There

was a random sample of 200 spins. The expected count for red is 18200 94.74,38

=

for black is

18200 94.74,38

=

and for green is 2200 10.53.38

=

11.3 ( ) ( ) ( ) ( )2 2 2 22 83 78 29 40.5 20 19.5 18 12

0.321 3.265 0.013 3.000 6.599.78 40.5 19.5 12

χ− − − −

= + + + = + + + =

11.4 ( ) ( ) ( )2 2 2

2 85 94.74 99 94.74 16 10.531.001 0.192 2.841 4.034.

94.74 94.74 10.53χ

− − −= + + = + + =

11.5 (a) The expected counts calculated in Exercise 1 are all at least 5. Since there are 4 categories, use a chi-square distribution with 4 1 3− = df. (b)

Page 3: Ch11 Solutions

254 The Practice of Statistics for AP*, 4/e

(c) Using Table C, 0.05 P-value 0.10.< < Using the calculator the P-value is 0.0858. (d) Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to reject the company’s claim about the distribution of nuts. 11.6 (a) The expected counts calculated in Exercise 2 are all at least 5. Since there are 3 categories, use a chi-square distribution with 3 1 2− = df. (b)

(c) Using Table C, 0.10 P-value 0.15.< < Using the calculator, the P-value is 0.133. (d) Since the P-value is greater than 0.05, we fail to reject the null hypothesis and find that we do not have enough evidence to say that the roulette wheel probabilities are not what they should be. 11.7 State: We want to perform a test at the 0.05α = significance level of

0 firs pines other: 0.54, 0.40, 0.06H p p p= = = versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: A random sample was used. Large sample size: The expected counts in each category are firs: ( )156 0.54 84.24,= pines:

( )156 0.40 62.4,= and other: ( )156 0.06 9.36,= all of which are at least 5. Independent: It seems reasonable that there would be more than 1560 red-breasted nuthatches. The conditions are met. Do:

The test statistic is ( ) ( ) ( )2 2 22 70 84.24 79 62.4 7 9.36

7.41884.24 62.4 9.36

χ− − −

= + + = and the distribution has

3 1 2− = df. The P-value is 0.0245. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the nuthatches prefer particular types of trees when they are searching for seeds and insects. 11.8 State: We want to perform a test at the 0.05α = significance level of

0 sand mud rocks: 0.56, 0.29, 0.15H p p p= = = versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: This was a randomized experiment. Large sample size: The expected counts in each category are sand:

( )200 0.56 112,= mud: ( )200 0.29 58,= and rocks: ( )200 0.15 30,= all of which are at least 5. Independent: It seems reasonable that there would be more than 2000 seagulls. The conditions are met.

Do: The test statistic is ( ) ( ) ( )2 2 22 128 112 61 58 11 30

14.474112 58 30

χ− − −

= + + = and the distribution has

3 1 2− = df. The P-value is 0.0007. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the seagulls have a preference for the location where they stand.

Page 4: Ch11 Solutions

Chapter 11: Inference for Distributions of Categorical Data 255

11.9 A chi-square goodness-of-fit test would not be appropriate here because the data for each day were not counts but rather means, and because the same 50 students were used each day (so the observations were not independent of each other). 11.10 A chi-square goodness-of-fit test would not be appropriate here because many of these students likely had same teachers (so the observations were not independent of each other). 11.11 (a) State: We want to perform a test at the 0.05α = significance level of

0 1 2 3 4 5 6 7 8 9: 0.301, 0.176, 0.125, 0.097, 0.079, 0.067, 0.058, 0.051, 0.046H p p p p p p p p p= = = = = = = = = versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: A random sample was used. Large sample size: The expected counts in each category are first digit 1: ( )250 0.301 75.25,= first digit 2: ( )250 0.176 44,= first digit 3:

( )250 0.125 31.25,= first digit 4: ( )250 0.097 24.25,= first digit 5: ( )250 0.079 19.75,= first digit 6:

( )250 0.067 16.75,= first digit 7: ( )250 0.058 14.5,= first digit 8: ( )250 0.051 12.75,= and first digit 9:

( )250 0.046 11.5,= all of which are at least 5. Independent: It seems reasonable that there would be more than 2500 invoices at this company. The conditions are met. Do: The test statistic is

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

2 2 2 2 2 22

2 2 2

61 75.25 50 44 43 31.25 34 24.25 25 19.75 16 16.7575.25 44 31.25 24.25 19.75 16.75

7 14.5 8 12.75 6 11.521.563

14.5 12.75 11.5

χ− − − − − −

= + + + + + +

− − −+ + =

and the distribution has 9 1 8− = df. The P-value is 0.0058. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the invoices are inconsistent with Benford’s law. Follow-up analysis: The breakdown of the chi-square statistic is:

2 2.699 0.818 4.418 3.920 1.396 0.036 3.879 1.770 2.630.χ = + + + + + + + + From this we see that the largest contributors to the statistic are amounts with first digit 3, 4 and 7. There are too many amounts that start with 3 or 4 and not enough that start with 7. Similarly there are too few that start with 1 or with 9. (b) A Type I error would be to say that the company’s invoices did not follow Benford’s law (suggesting fraud) when in fact they were consistent with Benford’s law. A Type II error would be to say that the invoices were consistent with Benford’s law when in fact they were not. A Type I error would be more serious here – alleging that the company had committed fraud when it had not. 11.12 State: We want to perform a test at the 0.05α = significance level of

0 Hispanic Black White Asian Others: 0.28, 0.24, 0.35, 0.12, 0.01H p p p p p= = = = = versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the

conditions are satisfied. Random: A random sample was used. Large sample size: The expected counts in each category are Hispanic: ( )800 0.28 224,= Black: ( )800 0.24 192,= White: ( )800 0.35 280,= Asian: ( )800 0.12 96,=

and Others: ( )800 0.01 8,= all of which are at least 5. Independent: It seems

reasonable that there would be more than 8000 residents of a large housing complex. The conditions are met. Do: The test statistic is

( ) ( ) ( ) ( ) ( )2 2 2 2 22 212 224 202 192 270 280 94 96 22 8

26.0625224 192 280 96 8

χ− − − − −

= + + + + = and the distribution

has 5 1 4− = df. The P-value is 0.00003. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the residents of this housing complex do not follow the distribution by ethnic background of New York City as a whole. Follow-up analysis: The breakdown of the chi-square

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256 The Practice of Statistics for AP*, 4/e

statistic is: 2 0.6429 0.5208 0.3571 0.0417 24.5.χ = + + + + From this we see that the largest contributor to the statistic, by far, is the last amount. There were many more people in the “Other” category than we would have predicted. 11.13 (a) 0 lemon lime orange strawberry grape: 0.2, 0.2, 0.2, 0.2, 0.2H p p p p p= = = = = versus

: At least one of the 's is incorrect.a iH p (b) Since all 5 flavors have the same proportion, they all have the same expected count: ( )60 0.2 12.= (c) There are 5 1 4− = df for this chi-square statistic. Using Table C, the value for 0.05α = is 9.49 and for 0.01α = is 13.28. (d) Answers will vary. One possibility consists of 6 lemon, 6 lime, 16 orange, 16 strawberry and 16 grape. The chi-square statistic is

( ) ( ) ( ) ( ) ( )2 2 2 2 22 6 12 6 12 16 12 16 12 16 12

1012 12 12 12 12

χ− − − − −

= + + + + = with a P-value of 0.04.

11.14 (a) 0 1 2 3 4 5 6 7 8 9 0: 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1H p p p p p p p p p p= = = = = = = = = = versus : At least one of the 's is incorrect.a iH p (b) Answers will vary. We perform the test at the

0.05α = significance level. Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: A random sample was used. Large sample size: The expected count in each category is: ( )200 0.1 20,= which is at least 5. Independent: The calculator created these in an independent way. The conditions are met. Do: In one test of the calculator we got 18 0’s, 22 1’s, 23 2’s, 21 3’s, 21 4’s, 21 5’s, 17 6’s, 14 7’s, 21 8’s, and 22 9’s. The test statistic is

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

2 2 2 2 2 2 2 22

2 2

18 20 22 20 23 20 21 20 21 20 21 20 17 20 14 2020 20 20 20 20 20 20 20

21 20 22 203.5

20 20

χ− − − − − − − −

= + + + + + + +

− −+ + =

and the distribution has 10 1 9− = df. The P-value is 0.9411. Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We don’t have enough evidence to say that the calculator’s RandInt function is not working properly.

11.15 State: We want to perform a test at the 0.05α = significance level of 01: 0.083

12iH p = = for all

astrological signs versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: A random sample was used. Large sample

size: Since the total sample size is 4344, the expected count in each category is 14344 36212 =

which is

at least 5. Independent: There are more than 43,440 residents in the U.S. The conditions are met. Do:

The test statistic is ( ) ( ) ( )2 2 22 321 362 360 362 355 362

... 19.762362 362 362

χ− − −

= + + + = and the distribution has

12 1 11− = df. The P-value is 0.0487. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the 12 signs are not equally likely. Follow-up analysis: The breakdown of the chi-square statistic is:

2 4.34 0.00085 0.12 0.50 1.40 4.76 2.74 2.76 2.42 0.12 0.66 0.09.χ = + + + + + + + + + + + From this we see that the largest contributors to the statistic are Aries and Virgo. There are fewer Aries and more Virgos than we would expect. There are also more Libras and fewer Scorpios and Sagittarius than would be predicted. This suggests that there are more people born in the late summer and early fall months than we would predict if all time periods were equally likely.

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Chapter 11: Inference for Distributions of Categorical Data 257

11.16 State: We want to perform a test at the 0.05α = significance level of

0 : 0.167 for all flavorsiH p = versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: A random sample was used. Large

sample size: Since the total sample size is 120, the expected count in each category is 1120 206

=

which is at least 5. Independent: There are more than 1200 Froot Loops. The conditions are met. Do:

The test statistic is ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 22 28 20 21 20 16 20 25 20 14 20 16 20

7.920 20 20 20 20 20

χ− − − − − −

= + + + + + =

and the distribution has 6 1 5− = df. The P-value is 0.1618. Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to say that the six flavors are not equally likely. Follow-up analysis: Not necessary. 11.17 State: We want to perform a test at the 0.05α = significance level of

0 smooth wrinkled: 0.75, 0.25H p p= = versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: A random sample was used. Large sample size: He observed 556 peas so the expected counts are smooth: ( )556 0.75 417,= and wrinkled: ( )556 0.25 139,= both of which are at least 5. Independent: It seems reasonable that there would be more than 5560 peas. The conditions are met. Do: The test statistic is

( ) ( )2 22 423 417 133 139

0.3453417 139

χ− −

= + = and the distribution has 2 1 1− = df. The P-value is 0.5568.

Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to dispute Mendel’s beliefs. Follow-up analysis: Not necessary. 11.18 State: We want to perform a test at the 0.05α = significance level of

0 tall-cut tall-potato dwarf-cut dwarf-potato: 0.5625, 0.1875, 0.1875, 0.0625H p p p p= = = = versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the

conditions are satisfied. Random: A random sample was used. Large sample size: A total of 1611 plants were observed. The expected counts in each category are tall-cut: ( )1611 0.5625 906.1875,= tall-potato: ( )1611 0.1875 302.0625,= dwarf-cut: ( )1611 0.1875 302.0625,= and dwarf-potato: ( )1611 0.0625 100.6875,= all of which are at least 5. Independent: It seems reasonable that there would be more than 16,110 tomato plants. The conditions are met. Do: The test statistic is

( ) ( ) ( ) ( )2 2 2 22 926 906.1875 288 302.0625 293 302.0625 104 100.6875

1.469906.1875 302.0625 302.0625 100.6875

χ− − − −

= + + + = and the

distribution has 4 1 3− = df. The P-value is 0.6895. Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to dispute the genetic laws. Follow-up analysis: Not necessary. 11.19 d 11.20 a 11.21 c 11.22 d

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258 The Practice of Statistics for AP*, 4/e

11.23 The English grades for the heavy readers are centered around approximately 3.7 with a left skewed distribution. The IQR is approximately 0.4 and there is one low outlier at approximately 2.9. The English grades for light readers have a lower center at approximately 3.35. These grades are more symmetrically distributed with no outliers and the distribution has an IQR of approximately 0.7. In general, the grade distribution for light readers is has a lower and is more spread out than the grade distribution for heavy readers. 11.24 (a) The conditions required are Random: The sample was randomly selected. Normal: Both sample sizes were at least 30 and the only outlier was not a large outlier. Independent. Since the school was described as being a large high school, it seems reasonable to believe that there are more than 790 students in the school. All conditions are met. (b) State: Our parameters of interest are 1µ = the mean English grade of heavy readers and 2µ = the mean English grade of light readers. We want to estimate the difference 1 2µ µ− at a 95% confidence level. Plan: We should use a two-sample t interval for

1 2µ µ− if the conditions are satisfied. We checked the conditions in part (a). The conditions are met. Do: From the data, 1 47,n = 1 3.64,x =

10.324,xs =

2 32,n = 2 3.356,x = and

20.380.xs = We will use the

conservative degrees of freedom which is 31 in this case. So our 95% confidence interval is

( ) ( ) ( )2 20.324 0.38

3.64 3.356 2.040 0.284 0.1676 0.1164,0.4516 .47 32

− ± + = ± = Conclude: We are 95%

confident that the interval from 0.1164 to 0.4516 captures the true difference in the mean English grade of heavy and light readers. This suggests that the mean English grade is between 0.1164 and 0.4516 points higher for heavy readers than for light readers. (c) No. This was an observational study so no conclusion of cause and effect can be made. 11.25 (a) The y-intercept is 3.42. This means that we would predict an English grade of 3.42 for a student who had read no books. (b) The predicted English GPA for this student is

( )ˆ 3.42 0.024 17 3.828.y = + = This means that the residual is ˆ 2.85 3.828 0.978.y y− = − = − (c) The relationship between English grades and number of books read is not very strong. The value of 2r is only 0.083 which means that only 8.3% of the variation in English grades is accounted for by the linear relationship with the number of books read. 11.26 (a) What Luis has calculated is the probability of getting 5 of one particular number. What he has not taken into account is that there are 6 different possible ways of getting a Yahtzee – one for each number on the die. (b) No, Nassir should not be surprised. The probability of getting a Yahtzee is

516 0.000772.6

=

The probability of getting no Yahtzee is 1 0.000772 0.999228.− = Finally, the

probability of getting 25 rolls with no Yahtzee is ( )250.999228 0.9809.= Section 11.2 Check Your Understanding, page 698: 1. For the main campus the conditional distribution is: 6.0% use Facebook several times a month or less, 23.6% use it at least once a week, and 70.3% use it at least once a day. For the commonwealth campuses: 12.1% use it several times a month or less, 25.0% use it at least once a week, and 62.8% use it at least once a day.

Page 8: Ch11 Solutions

Chapter 11: Inference for Distributions of Categorical Data 259

2. It is important to compare proportions rather than counts in Question 1 because there was such a big difference in the sample size from the two different types of campuses. 3. The biggest difference between the two types of campuses is that students on the main campus are more likely to be everyday users of Facebook than students on the commonwealth campuses and those on the commonwealth campuses are more likely to use Facebook several times a month or less than those students on the main campus.

Check Your Understanding, page 703: 1.

0 : There is no difference in the distributions of Facebook use between students at Penn State's main campus and its commonwealth campusesH

: There is a difference in the distributions of Facebook use between students at Penn State's main campus and its commonwealth campuses.

aH

2. First note that there are 910 627 1537+ = students in the sample. Also, there are 55 76 131+ = total students who use Facebook several times a month or less. So, the expected count for the number of students at the main campus who use Facebook several times a month or less is computed as

( )131 910 77.56.1537

= The other expected counts are computed in a similar fashion and are given in the

table below. Use Facebook Main campus Commonwealth campus Several times a month 77.56 53.44 Once a week 220.25 151.75 Once a day 612.19 421.81

Page 9: Ch11 Solutions

260 The Practice of Statistics for AP*, 4/e

3.

( ) ( ) ( ) ( ) ( )

( )

2 2 2 2 22

2

55 77.56 76 53.44 215 220.25 157 151.75 640 612.1977.56 53.44 220.25 151.75 612.19

394 421.8119.489

421.81

χ− − − − −

= + + + +

−+ =

Check Your Understanding, page 705: 1. Using Table C with ( )( )df 3 1 2 1 2= − − = P-value 0.0005.< Using the calculator, the P-value is 0.000059. 2. Assuming that there is no difference in the distributions of Facebook use between students on Penn State’s main campus and students at Penn State’s commonwealth campuses, the probability of observing a sample that shows a difference in the distributions of Facebook use among students at the main campus and the commonwealth campuses as large or larger than the one found in this study is about 6 in 100,000. 3. Since the P-value was so small, reject the null hypothesis. It appears that the distribution of Facebook use is different among students at Penn State’s main campus and students at Penn State’s commonwealth campuses. Check Your Understanding, page 708: 1.

2. State: We want to perform a test of

0 : There is no difference in the distribution of quality of life in Canada and the US: There is a difference in the distribution of quality of life in Canada and the USa

HH

at the 0.01α = level. Plan: We should use a chi-square test for homogeneity if the conditions are satisfied. Random: The data came from separate random samples. Large sample size: We used technology to get the following expected counts:

Quality of life Canada US Much better 77.37 538.63 Somewhat better 71.47 497.53 About the same 109.91 765.09 Somewhat worse 41.70 290.30 Much worse 10.55 73.45

Page 10: Ch11 Solutions

Chapter 11: Inference for Distributions of Categorical Data 261

All of these counts are at least 5. Independent: We clearly have less than 10% of all heart attack victims in the U.S. and in Canada. All conditions have been met. Do: The test statistic is

( ) ( )2 22 75 77.37 65 73.45

... 11.725.77.37 73.45

χ− −

= + + = We use a chi-square distribution with

( )( )5 1 2 1 4− − = df and find a P-value of 0.0195. Conclude: Since the P-value is greater than 0.01, we fail to reject the null hypothesis. There is not enough evidence to conclude that there is a difference in the distribution of quality of life in Canada and the United States. Check Your Understanding, page 713: 1. This was an experiment. Each individual was exposed to a treatment that involved how they were contacted. 2. Contact method Answered “yes” Answered “no” Phone 168 632 Personal interview 200 600 Written response 224 576 3. State: We want to perform a test of

0 : There is no difference in the proportion who answer yes based on contact method: There is a difference in the proportion who answer yes based on contact methoda

HH

at the 0.05α = level. Plan: We should use a chi-square test for homogeneity if the conditions are satisfied. Random: The data came from a randomized experiment. Large sample size: We used technology to get the following expected counts:

Contact method Answered “yes” Answered “no” Phone 197.33 602.67 Personal interview 197.33 602.67 Written response 197.33 602.67

All of these counts are at least 5. Independent: Due to the random assignment, these three groups of people can be viewed as independent. Individual observations in each group should also be independent: knowing one person’s response gives no information about another person’s response. Do: The test

statistic is ( ) ( )2 22 168 197.33 576 602.67

... 10.619.197.33 602.67

χ− −

= + + = We use a chi-square distribution with

( )( )3 1 2 1 2− − = df and find a P-value of 0.0049. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that there is a convincing evidence of a difference in the proportion of people who answer yes based on how they are contacted. Check Your Understanding, page 718: 1. State: We want to perform a test of

0 : There is no association between territory type and success or not in the population of franchises: There is an association between territory type and success or not in the population of franchisesa

HH

at the 0.01α = level. Plan: We should use a chi-square test of independence/association if the conditions are satisfied. Random: This was a random sample of franchises. Large sample size: We used technology to get the following expected counts:

Exclusive territory Success Failure Yes 102.74 39.26 No 20.26 7.74

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All of these counts are at least 5. Independent: We sampled 170 franchises. There are more than 1700 franchises in the US. The conditions are met. Do: The test statistic is

( ) ( )2 22 108 102.74 13 7.74

... 5.911.102.74 7.74

χ− −

= + + = We use a chi-square distribution with

( )( )2 1 2 1 1− − = df and find a P-value of 0.0150. Conclude: Since the P-value is greater than 0.01, we fail to reject the null hypothesis. We do not have enough evidence to conclude that there is an association between whether franchises have an exclusive territory or not and whether they are successful or not. Exercises, page 724: 11.27 (a) There were 67 women and 67 men in the two samples so the conditional distributions for females and males are: Goal Female Male HSC-HM 14 0.209

67= 31 0.463

67=

HSC-LM 0.104 0.269 LSC-HM 0.313 0.075 LSC-LM 0.373 0.194 (b)

(c) In general it appears that females were classified mostly as low social comparison whereas males were classified mostly as high social comparison. Males were most likely to be in the high social comparison/high mastery group whereas females were most likely to be in the low social comparison/low social mastery group.

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11.28 (a) There were 202 in each of the black parents and Hispanic parents groups and 201 in the white parents group so the conditional distributions are: Survey result Black parents Hispanic parents White parents Excellent 12 0.059

202= 34 0.168

202= 22 0.109

201=

Good 0.342 0.272 0.403 Fair 0.371 0.302 0.299 Poor 0.119 0.119 0.119 Don’t know 0.109 0.139 0.070 (b)

(c) In general, Hispanic parents are more likely to think that the high schools are doing an excellent job. Whites are more likely than either of the other two groups to think that the high schools are doing either an excellent or a good job. Black parents are more likely to think that schools are only doing a fair job. 11.29 (a)

0 : There is no difference in the distribution of sports goals for female and male undergraduates: There is a difference in the distribution of sports goals for female and male undergraduatesa

HH

(b) First note that there are 67 67 134+ = students in the sample. Also, there are 14 31 45+ = total students who are classified as HSC-HM. So, the expected count for the number of female students

classified as HSC-HM is computed as ( )45 67 22.5.134

= The other expected counts are computed in a

similar fashion and are given in the table below. Goal Females Males HSC-HM 22.5 22.5 HSC-LM 12.5 12.5 LSC-HM 13 13 LSC-LM 19 19

(c) ( ) ( )2 22 14 22.5 13 19

... 24.898.22.5 19

χ− −

= + + =

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11.30 (a) 0 : There is no difference in the distribution of opinions about high school among black, Hispanic

and white parents: There is a difference in the distribution of opinions about high school among blaca

H

H k, Hispanic and white parents

(b) First note that there are 202 202 201 605+ + = parents in the sample. Also, there are 12 34 22 68+ + = total parents who thought the high schools are excellent. So, the expected count for the

number of black parents who think the high schools are excellent is computed as ( )68 202 22.70.605

= The

other expected counts are computed in a similar fashion and are given in the table below. Survey result Black parents Hispanic parents White parents Excellent 22.70 22.70 22.59 Good 68.45 68.45 68.11 Fair 65.44 65.44 65.12 Poor 24.04 24.04 23.92 Don’t know 21.37 21.37 21.26

(c) ( ) ( )2 22 12 22.70 14 21.26

... 22.426.22.70 21.26

χ− −

= + + =

11.31 (a) Random: The data came from random samples. Large sample size: The expected counts computed in exercise 29 were all at least 5. Independent: Clearly we have less than 10% of all male or female students possible. (b) Using Table C with df = 3,we get P-value 0.0005.< Using the calculator, the P-value is 0.00002. (c) Assuming that there is no difference in the distributions of goals for playing sports among males and females, the probability of observing a sample that shows a difference in the distributions of goals in playing sports among males and females as large or larger than the one found in this study is about 2 in 100,000. (d) Since the P-value is so small, we reject the null hypothesis. There is convincing evidence of a difference in the distributions of goals for playing sports among female and male undergraduates. 11.32 (a) Random: The data came from random samples. Large sample size: The expected counts computed in exercise 30 were all at least 5. Independent: Clearly we have less than 10% of all parents possible in each group. (b) Using Table C we get 0.0025 P-value 0.005.< < Using the calculator, the P-value is 0.0042. (c) Assuming that there is no difference in the distributions of opinions about high school among black, Hispanic, and white parents, the probability of observing a sample that shows a difference in the distributions of opinions about high school among black, Hispanic, and white parents as large or larger than the one found in this study is about 4 in 1000. (d) Since the P-value is so small, we reject the null hypothesis. There is convincing evidence of a difference in the distributions of opinions about high school among black, Hispanic, and white parents. 11.33 (a) The two-way table of the data is given below:

Environment Hatched Did not hatch Total Cold 16 11 27 Neutral 38 18 56 Hot 75 29 104 Total 129 58 187

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The proportions (conditional distributions) are given in the table below. Note that the conditional distributions go across rows in this case.

Environment Hatched Did not hatch Cold 16 0.593

27=

0.407

Neutral 38 0.67956

= 0.321

Hot 75 0.721104

= 0.279

This data does support the researchers’ belief to a certain extent. A smaller proportion of eggs hatched in the cold water. But a majority of the eggs in cold water did still hatch. (b) State: We want to perform a test of

0 : There is no difference in the proportion of eggs that hatch based on water temperature: There is a difference in the proportion of eggs that hatch based on water temperaturea

HH

at the 0.05α = level. Plan: We should use a chi-square test for homogeneity if the conditions are satisfied. Random: The data came from a randomized experiment. Large sample size: We used technology to get the following expected counts:

Environment Hatched Did not hatch Cold 18.63 8.37 Neutral 38.63 17.37 Hot 71.74 32.26

All of these counts are at least 5. Independent: Due to the random assignment, these three groups of eggs can be viewed as independent. Individual observations in each group should also be independent: knowing whether one egg hatches does not give information about whether another egg hatches or not.

The conditions are met. Do: The test statistic is ( ) ( )2 22 16 18.63 29 32.26

... 1.703.18.63 32.26

χ− −

= + + = We use

a chi-square distribution with ( )( )3 1 2 1 2− − = df and find a P-value of 0.4267. Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to say that different proportions of eggs hatch in different water temperatures. 11.34 (a) The two-way table of the data is given below:

Supplement Male Female Total PBM 9 11 20 NLCP 8 11 19 PL-LCP 8 11 19 TG-LCP 10 9 19 Total 35 42 77

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The proportions (conditional distributions) are given in the table below. Note that the conditional distributions go across rows in this case.

Supplement Male Female PBM 9 0.45

20=

0.55

NLCP 8 0.42119

= 0.579

PL-LCP 8 0.42119

= 0.579

TG-LCP 10 0.52619

= 0.474

It does appear that the groups are roughly balanced. (b) State: We want to perform a test of 0 : There is no significant difference in the proportions of females in different supplement groups

: There is a significant difference in the proportion of females in different supplement groupsa

HH

at the 0.05α = level. Plan: We should use a chi-square test for homogeneity if the conditions are satisfied. Random: The data came from a randomized experiment. Large sample size: We used technology to get the following expected counts:

Supplement Male Female PBM 9.09 10.91 NLCP 8.64 10.36 PL-LCP 8.64 10.36 TG-LCP 8.64 10.36

All of these counts are at least 5. Independent: Due to the random assignment, these three groups of infants can be viewed as independent. Individual observations in each group should also be independent: knowing whether one infant is female gives no information about any other infant’s gender. The

conditions are met. Do: The test statistic is ( ) ( )2 22 9 9.09 9 10.36

... 0.568.9.09 10.36

χ− −

= + + = We use a chi-

square distribution with ( )( )4 1 2 1 3− − = df and find a P-value of 0.9036. Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to say that the groups differ significantly based on gender. 11.35 We cannot use a chi-square test with this data because we do not have the actual counts of the travelers in each category. We also do not know if the sample was taken randomly, nor if the same people could have been counted more than once (traveling both for leisure one time and business another). 11.36 The observations in this table are not independent of each other. Data on each woman occurs in each row. 11.37 (a) The data are in the table below.

Treatment Success Failure Total Nicotine patch 40 204 244 Drug 74 170 244 Patch plus drug 87 158 245 Placebo 25 135 160 Total 226 667 893

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(b)

The best success rate is with both the patch and the drug. It also appears that the patch alone is not much better than the placebo. (c) The null hypothesis given says that each of the four treatments leads to the same probability of success. (d) There are a total of 893 subjects in the sample. Also, there are 226 subjects who were successful. So, the expected count for the number of successes among those who got

the nicotine patch is computed as ( )226 244 61.75.893

= The other expected counts are computed in a

similar fashion and are given in the table below. Treatment Success Failure Nicotine patch 61.75 182.25 Drug 61.75 182.25 Patch plus drug 62.00 183.00 Placebo 40.49 119.51

10.38 (a) The data are in the table below.

Treatment Stroke No stroke Total Placebo 250 1399 1649 Aspirin 206 1443 1649 Dipyridamole 211 1443 1654 Both 157 1493 1650 Total 824 5778 6602

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(b)

It appears that all 4 treatments have about the same success rate. (c) The null hypothesis given says that each of the four treatments leads to the same probability of success. (d) There are a total of 6602 subjects in the sample. Also, there are 824 subjects who had strokes. So, the expected count for the number of

who had strokes among those who got the placebo is computed as ( )824 1649 205.81.6602

= The other

expected counts are computed in a similar fashion and are given in the table below. Treatment Stroke No stroke Placebo 205.81 1443.19 Aspirin 205.81 1443.19 Dipyridamole 206.44 1447.56 Both 205.94 1444.06

11.39 State: We want to perform a test of

0 : There is no difference in the smoking cessation rates of the different treatments: There is a difference in the smoking cessation rates of the different treatmentsa

HH

at the 0.05α = level. Plan: We should use a chi-square test for homogeneity if the conditions are satisfied. Random: The data came from a randomized experiment. Large sample size: We computed the expected counts in Exercise 11.37. All of them are at least 5. Independent: Due to the random assignment, these four groups of subjects can be viewed as independent. Individual observations in each group should also be independent: knowing whether one person quits smoking gives no information about any other subject’s smoking status. The conditions are met. Do: The test statistic

is ( ) ( )2 22 40 61.75 135 119.51

... 34.937.61.75 119.51

χ− −

= + + = We use a chi-square distribution with

( )( )4 1 2 1 3− − = df and find a P-value of approximately 0. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the smoking cessation rate is different for at least one of the four treatments. 11.40 State: We want to perform a test of

0 : There is no difference in the stroke rates of the different treatments: There is a difference in the stroke rates of the different treatmentsa

HH

at the 0.05α = level. Plan: We should use a chi-square test for homogeneity if the conditions are satisfied. Random: The data came from a randomized experiment. Large sample size: We computed the

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expected counts in Exercise 11.38. All of them are at least 5. Independent: Due to the random assignment, these four groups of subjects can be viewed as independent. Individual observations in each group should also be independent: knowing whether one person has a stroke gives no information about whether any other subject had a stroke or not. The conditions are met. Do: The test statistic is

( ) ( )2 22 250 205.81 1493 1444.06

... 24.243.205.81 1444.06

χ− −

= + + = We use a chi-square distribution with

( )( )4 1 2 1 3− − = df and find a P-value of approximately 0. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the stroke rate is different for at least one of the four treatments. 11.41 The statistic breaks down into its individual components as follows:

2 7.662 2.596 2.430 0.823 10.076 3.414 5.928 2.008 34.937.χ = + + + + + + + = The largest component of this equation comes from those who had success using both the patch and the drug. Far more people fell in this group than would have been expected. The next largest component comes from those who had success using just the patch. Far fewer were in this group than would have been expected. This means that the patch alone really is not all that helpful, but the patch in combination with the drug is a good treatment. 11.42 The statistic breaks down into its individual components as follows:

2 9.487 1.353 0 0 0.101 0.014 11.629 1.658 24.243.χ = + + + + + + + = The largest component of this equation comes from those who had strokes on both aspirin and Dipyridamole. Far fewer were in this group than would have been expected. The next largest component comes from those who had strokes while on the placebo. Far more people were in this group than would have been expected. This means that the two drugs together far outperform the placebo and should probably be used if the side effects are not too great. 11.43 (a) The hypotheses are as follows:

0 : There is no difference in the improvement rates for the two treatments: There is a difference in the improvement rates for the two treatmentsa

HH

Assuming that there is no difference in the improvement rates between gastric freezing and the placebo, the probability of observing a sample that shows a difference in the improvement rates between gastric freezing and the placebo as large or larger than the one found in this study is about 57 in 100. These data do not provide convincing evidence of a difference in improvement rates for gastric freezing and placebo (b) The P-value for this test is identical to the P-value for the test in part (a). And since the hypotheses for the two-sample z test are the same (though usually stated in symbols rather than words), the conclusions to this test are the same as for the test in part (a). 11.44 (a) The hypotheses are as follows:

0 : There is no association between support for the death penalty and educational level: There is an association between support for the death penalty and educational levela

HH

Assuming that there is no association between support for the death penalty and education level, the probability of observing a difference in support of the death penalty among those with a high school degree and those with a Bachelor’s degree as large or larger than the one found in this study is approximately 0. This suggests that there is an association between support for the death penalty and educational level in the population. (b) The P-value for this test is identical to the P-value for the test in part (a). And since the hypotheses for the two-sample z test are the same (though usually stated in symbols rather than words), the conclusions to this test are the same as for the test in part (a).

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11.45 (a)

Buyers are much more likely to think the quality of recycled coffee filters is higher while nonbuyers are more likely to think the quality is lower. (b) State: We want to perform a test of

0 : Buying recycled products is independent of their perceived quality rating : Buying recycled products is not independent of their perceived quality rating a

HH

at the 0.05α = level. Plan: We should use a chi-square test for independence if the conditions are satisfied. Random: The data came from a random sample. Large sample size: The computer printout shows the expected counts. All of them are at least 5. Independent: The sample had 133 people in it. There are more than 1330 adults who could have been part of the sample. The conditions are met. Do: The test statistic is given by the output as being 7.638. We use a chi-square distribution with ( )( )3 1 2 1 2− − = df and find a P-value of 0.022. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that whether people buy recycled coffee filters or not is not independent of their opinion of the quality of recycled coffee filters. 11.46 (a)

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The higher the degree earned, the less likely people are to think that astrology is scientific. (b) State: We want to perform a test of

0 : Education level and belief about astrology are independent: Education level and belief about astrology are not independenta

HH

at the 0.05α = level. Plan: We should use a chi-square test for independence if the conditions are satisfied. Random: The data came from a random sample. Large sample size: The computer printout shows the expected counts. All of them are at least 5. Independent: The sample had 687 people in it. There are more than 6870 adults in the US who could have been part of the sample. The conditions are met. Do: The test statistic is given by the output as being 10.582. We use a chi-square distribution with ( )( )3 1 2 1 2− − = df and find a P-value of 0.005. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the education level and belief about the scientific nature of astrology are not independent. 11.47 (a) Of those who spend less than 2 hours on extracurricular activities per week, the proportion

who earn a C or better is 11 0.5520

= and the proportion who earn a D or worse is 0.45. Of those who spend

2 to 12 hours on extracurricular activities per week, the proportion who earn a C or better is 0.747 and the proportion who earn a D or worse is 0.253. Of those who spend more than 12 hours on extracurricular activities per week, the proportion who earn a C or better is 0.375 and the proportion who earn a D or worse is 0.625. A bar graph is given below. It appears that those who are moderately involved in extracurricular activities (2 to 12 hours per week) earn higher grades than those who are very active and those who are not active.

(b) A chi-square test should not be performed in this setting because there were only 8 students observed in the over 12 hours category. The expected count for at least one of these two cells will, necessarily, be less than 5. We also do not know how the students were chosen for this survey other than the fact that they were in a required chemical engineering course. We do not know if they were randomly selected or not, for instance.

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11.48 (a) The second row in the table shown is just a continuation of the first row – that is, the same people are still being measured. Also, the table does not show the people who did not get rid of the warts, so this table only shows the successes, not the failures. The table we should use is as follows:

Result Treatment Control Total Warts gone 25 3 28 Warts remain 7 20 27 Total 32 23 55

(b) It is still not appropriate to use a chi-square test because we do not know how the experiment was conducted. There is no mention of randomization so we have to wonder about that and independence. 11.49 (a)

The majority of people in most education levels seem to oppose such a law. The one education level where there is an even split is the group that has less than a high school education. (b) State: We want to perform a test of

0 : There is no association between education level and support of a gun law in the population of adults: There is an association between education level and support of a gun law in the population of a

HH adults

at the 0.05α = level. Plan: We should use a chi-square test for association/independence if the conditions are satisfied. Random: The data came from a random sample. Large sample size: We used technology to get the following expected counts:

Education Yes No Less than high school 46.94 69.06 High school grad 86.19 126.81 Some college 187.36 275.64 College grad 94.29 138.71 Postgraduate degree 71.22 104.78

All of these counts are at least 5. Independent: Our sample includes 1201 people. This is less than 10% of the population of the US. The conditions are met. Do: The test statistic is

( ) ( )2 22 58 46.94 99 104.78

... 8.525.46.94 104.78

χ− −

= + + = We use a chi-square distribution with

( )( )5 1 2 1 4− − = df and find a P-value of 0.0741. Conclude: Since the P-value is greater than 0.05, we

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fail to reject the null hypothesis. We do not have enough evidence to say that there is an association between educational level and support of a gun law in the population of adults.. 11.50 (a)

In soft water, the new product is slightly preferred over the standard product and there is not much difference between the warm water wash and the hot water wash. In hard water, the new product has a larger majority of the preference and is slightly more preferred among those using a warm water wash. (b) State: We want to perform a test of

0 : There is no association between type of wash and support for the new product among people who don't use the established brand

: There is an association between type of wash and support for the new a

H

H product among people whodon't use the established brand

at the 0.05α = level. Plan: We should use a chi-square test for association/independence if the conditions are satisfied. Random: The data came from a random sample. Large sample size: We used technology to get the following expected counts:

Product preference Soft, warm Soft, hot Hard, warm Hard, hot Standard 49.81 24.05 47.23 30.92 New 66.19 31.95 62.77 41.08

All of these counts are at least 5. Independent: Our sample includes 354 people. This is less than 10% of the population of the US who don’t currently use the established brand. The conditions are met. Do: The test statistic is

( ) ( )2 22 53 49.81 42 41.08

... 2.058.49.81 41.08

χ− −

= + + = We use a chi-square distribution with

( )( )4 1 2 1 3− − = df and find a P-value of 0.5605. Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to say that there is an association between type of wash and support for the new product among people who don’t use the established brand. 11.51 (a) Since this represents one sample and the subjects were then classified by their answer and their gender, this would be a chi-square test of association/independence. (b) The hypotheses are

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0 : Gender and where subjects live are independent: Gender and where subjects live are not independenta

HH

(c) Random: This was a random sample. Large sample size: The expected counts are given in the output and they are all at least 5. Independent: This sample of 4854 young adults is less than 10% of all young adults who could have been sampled. (d) If gender and place of living are independent, then we have a 1.2% chance of finding a sample with as much association or more than we found. This is fairly unlikely (less than a 5% chance) so we reject the null hypothesis and conclude that gender and where young adults live are not independent. 11.52 (a) Since there were two separate samples, one of American students and one of Asian students, we should perform a chi-square test for homogeneity. (b) The hypotheses are

0 : There is no difference in the reasons for shopping from catalogs between the two groups of students: There is a difference in the reasons for shopping from catalogs between the two groups of studea

HH nts

(c) Random: This data came from random samples. Large sample size: The expected counts are given in the output and they are all at least 5. Independent: There are at least 950 American students and 620 Asian students. (d) If the distribution of reasons for shopping from catalogs is the same for both Asian and American students, we have a 0.01% chance of finding as big a difference or bigger than we found between the two samples. This is quite unlikely, so we reject the null hypothesis and conclude that there are differences between American and Asian students in terms of why they shop from catalogs. 11.53 e 11.54 c 11.55 a 11.56 d 11.57 b 11.58 b 11.59 (a) One-sample t interval for the mean. (b) Two-sample z test for the difference between two proportions. 11.60 (a) Chi-square test for association/independence. (b) Two-sample t interval for the difference between two means. 11.61 This was an experiment. The subjects were randomly exposed to a treatment (the type of response they were allowed to make). 11.62 A chi-square goodness-of-fit test is only appropriate for testing whether a single categorical variable has a specified distribution.

11.63 (a) The mean for the 1-5 scale is ( ) ( ) ( ) ( ) ( )( )1 51 2 1 3 2 1 3 13 4 3 5 3.545.22

x − = + + + + = Using

technology, the standard deviation is 1.184. (b) The new mean for the 0-4 scale is ( )0 4 1 3.21 1 4.21.x − + = + = Adding one to all values does not change the standard deviation so it remains

0.568. (c) No, it would not be appropriate to compare the means from (a) and (b) using a two-sample t

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test because the original measurements are categorical not quantitative. Yes, they are recorded as numbers, but the “distance” between 1 and 2 may not be the same as the distance between 3 and 4 in people’s minds. 11.64 It is certainly possible that those who do not respond have similar views about the cafeteria food. For example, maybe people who really do not like the food feel uncomfortable answering this question. Chapter Review Exercises (page 731) R11.1 (a)

In this school the largest class consists of the freshmen and the class sizes decrease each year. But in the sample, the largest groups were the sophomores and the juniors. This suggests that the sample may not be representative of the school as a whole. (b) State: We want to perform a test at the

0.05α = significance level of 0 fresman sophomore junior senior: 0.29, 0.27, 0.25, 0.19H p p p p= = = = versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the

conditions are satisfied. Random: A random sample was used. Large sample size: The expected counts in each category are freshman: ( )206 0.29 59.74,= sophomore: ( )206 0.27 55.62,= junior:

( )206 0.25 51.5,= and senior: ( )206 0.19 39.14,= all of which are at least 5. Independent: It seems reasonable that there would be more than 2060 students in a large high school. The conditions are met.

Do: The test statistic is ( ) ( ) ( ) ( )2 2 2 2

2 54 59.74 66 55.62 56 51.5 30 39.145.016

59.74 55.62 51.5 39.14χ

− − − −= + + + = and

the distribution has 4 1 3− = df. The P-value, then, is 0.1706. Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to reject the idea that the sample is representative of the population of students.

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R11.2 The expected counts are listed in the table below. Teacher type 0 1 2 3 4 Knowledgeable 12.5 0.5 4.5 1.5 5 Ignorant 12.5 0.5 4.5 1.5 5

Note that 6 of the cells have expected counts that are less than 5. R11.3 (a)

Treatment Suffered cardiac event No cardiac event Total Stress management 3 30 33 Exercise 7 27 34 Usual care 12 28 40 Total 22 85 107

(b) The success rate for stress management is 30 0.909,33

= for exercise 0.794 and for the usual care 0.70.

(c) State: We want to perform a test of 0 : There is no difference in the actual proportion of cardiac events for the three treatments

: There is a difference in the actual proportion of cardiac events for the three treatments a

HH

at the 0.05α = level. Plan: We should use a chi-square test for homogeneity if the conditions are satisfied. Random: The data came from a randomized experiment. Large sample size: We used technology to get the following expected counts:

Treatment Cardiac event No Cardiac Stress management 6.79 26.21 Exercise 6.99 27.01 Usual care 8.22 31.78

All of these counts are at least 5. Independent: Due to the random assignment, these three groups of patients can be viewed as independent. Individual observations in each group should also be independent: knowing whether a person has a cardiac event gives no information about whether another person has a cardiac event or not. The conditions are met. Do: The test statistic is

( ) ( )2 22 3 6.79 28 31.78

... 4.840.6.79 31.78

χ− −

= + + = We use a chi-square distribution with ( )( )3 1 2 1 2− − = df

and find a P-value of 0.0889. Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to say that there is a difference in the actual proportion of cardiac events for the three treatments. R11.4 (a) This should be a chi-square test for association/independence because one random sample was used and people in the ads were classified both by gender of the target audience and whether the ad was sexual or not. (b) The hypotheses are

0 : Gender of target audience and whether or not the ad is sexual are independent: Gender of target audience and whether or not the ad is sexual are not independenta

HH

(c) The first number is the percentage of ads in women’s magazines which are found to be not sexy. This

is computed as 351 0.6094.576

= The second number is the expected count for the number of ads in

women’s magazines found to be not sexy and is computed as ( )1113 576 424.8.1509

= The third number is

the contribution to the chi-square statistic from this same cell and is computed as ( )2351 424.812.82.

424.8−

=

(The difference is due to rounding error.) (d) Since the P-value is so small (essentially 0), we would

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reject the null hypothesis and conclude that the gender of the target audience and whether magazine ads are sexual are not independent. R11.5 (a)

Both groups of children have the largest percentage reporting grades as the goal. But after that, boys were more likely to pick sports whereas girls were more likely to pick being popular. (b) State: We want to perform a test of

0 : There is no association between gender and goal at school for 4th, 5th, and 6th grade students: There is an association between gender and goal at school for 4th, 5th, and 6th grade studentsa

HH

at the 0.05α = level. Plan: We should use a chi-square test for association/independence if the conditions are satisfied. Random: The data came from a random sample. Large sample size: We used technology to get the following expected counts:

Gender Grades Popular Sports Female 129.70 74.04 47.26 Male 117.30 66.96 42.74

All of these counts are a least 5. Independent: Our sample includes 478 children. This is less than 10% of the population children in these grades in the US. The conditions are met. Do: The test statistic is

( ) ( )2 22 130 129.7 60 47.26

... 21.455.129.7 47.26

χ− −

= + + = We use a chi-square distribution with

( )( )3 1 2 1 2− − = df and find a P-value of 0.00002. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that there is an association between gender and goal at school for 4th, 5th, and 6th grade students. (c) The chi-square statistic, broken down, is

2 0.001 3.885 6.303 0.001 4.296 6.970 21.455.χ = + + + + + = The largest contributions are from the male/sports cell and the female/sports cell. More males than expected chose sports and fewer females than expected chose sports. R11.6 (a) The two-way table is

Type of subject Answer yes Answer no Total Student 22 39 61 Non-student 30 29 59 Total 52 68 120

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278 The Practice of Statistics for AP*, 4/e

State: We want to perform a test of 0 : There is no difference in the actual proportions who use unsafe passwords between

the students and nonstudents: There is a difference in the actual proportions who use unsafe passwords between

thea

H

H students and nonstudents

at the 0.05α = level. Plan: We should use a chi-square test for homogeneity if the conditions are satisfied. Random: The data came from random samples. Large sample size: We used technology to get the following expected counts:

Type of subject Yes No Student 26.43 34.57 Non-student 25.57 33.43

All of these counts are at least 5. Independent: We have sampled less than 10% of students and less than 10% of non-students. The conditions are met. Do: The test statistic is

( ) ( )2 22 22 26.43 29 33.43

... 2.669.26.43 33.43

χ− −

= + + = We use a chi-square distribution with

( )( )2 1 2 1 1− − = df and find a P-value of 0.1023. Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to say that students and non-students have different rates of unsafe password use. (b) State: We want to perform a test at the

0.05α = significance level of 0 1 2: 0H p p− = versus 1 2: 0aH p p− ≠ where 1p is the actual proportion of students who use unsafe passwords and 2p is the actual proportion of non-students who use unsafe passwords. Plan: We should use a two-sample z test for 1 2p p− if the conditions are satisfied. Random: We have random samples from each population. Normal: The number of successes and failures in both groups are at least 10 (see the two-way table in part (a)). Independent: Both samples are less than 10% of their respective populations (there are more than 610 students and 590 non-students who could have been sampled). The conditions are met. Do: The proportions of unsafe password users in the

two populations are 122ˆ 0.36161

p = = and 230ˆ 0.508.59

p = = The pooled proportion is 52ˆ 0.433.120Cp = =

The test statistic is ( )

( )( ) ( )( )0.361 0.508 0

1.62.0.433 0.567 0.433 0.567

61 59

z− −

= = −

+

Since this is a two-sided test the P-

value is ( ) ( )2 1.62 2 0.0526 0.1052.P z < − = = Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to conclude that there is a difference in the actual proportions of students and non-students who use unsafe passwords. Notice that the P-values from (a) and (b) are very close. They should be the same and are only different due to rounding issues.

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AP Statistics Practice Test (page 733) T11.1 b. There are 10 categories and the degrees of freedom is equal to the number of categories − 1. T11.2 c. This is a test for homogeneity since there were 4 different random samples used. T11.3 e. All three conditions hold. T11.4 d. 8% of the population is Hispanic and there was a sample size of 148. T11.5 c. There are 3 df and the P-value is 0.087. T11.6 c. The largest contribution is 3.20 and is from the Hispanic cell. T11.7 b. There are 3 rows and 2 columns so the df are ( )2 1 2.= T11.8 a. Multiply the row total by the column total and divide by the overall total. T11.9 d. This was not an experiment so no conclusion of causation can be made. Additionally, the P-value was quite high. T11.10 d. A Type I error is when we reject the null hypothesis when it is really true. T11.11 State: We want to perform a test at the 0.05α = significance level of

0 regular premium supreme: 0.6, 0.2, 0.2H p p p= = = versus : At least one of the 's is incorrect.a iH p Plan: We should use a chi-square goodness-of-fit test if the conditions are satisfied. Random: A random sample was used. Large sample size: The expected counts in each category are regular: ( )400 0.6 240,= and premium and supreme each: ( )400 0.2 80,= all of which are at least 5. Independent: It seems reasonable that there would be more than 400 people who stop at this gas station over a long period of time. The conditions are met. Do: The test statistic is

( ) ( ) ( )2 2 22 261 240 51 80 88 80

13.15240 80 80

χ− − −

= + + = and the distribution has 3 1 2− = df. The P-value,

then, is 0.0014. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that we the distribution of type of gas purchased is not what the distributor claims. T11.12 (a) Random assignment was used to make sure that the responding officers did not bias the assignment of treatments. It might have been the case that they would more likely use one technique in certain situations than another technique.

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280 The Practice of Statistics for AP*, 4/e

(b)

(c) The alternative hypothesis is that at least one ip is different from the others. (d) If the proportion of subsequent arrests is the same for all three techniques, we have a 7.96% chance of getting a difference between the three groups as large as or larger than the one observed ( 2χ = 5.063) in this study due to random assignment. This leads us to fail to reject the null hypothesis. We do not have enough evidence to say that the techniques lead to different subsequent arrest rates. T11.13 (a) State: We want to perform a test of

0 :Smoking status and education are independent:Smoking status and education are not independenta

HH

at the 0.05α = level. Plan: We should use a chi-square test for association/independence if the conditions are satisfied. Random: The data came from a random sample. Large sample size: We used technology to get the following expected counts:

Education Nonsmoker Former Moderate Heavy Primary school 59.48 50.93 42.37 34.22 Secondary school 44.21 37.85 31.49 25.44 University 42.31 36.22 30.14 24.34

All of these counts are at least 5. Independent: Our sample includes 459 French men. This is less than 10% of the population men (aged 20 to 60) in France. The conditions are met. Do: The test statistic is

( ) ( )2 22 56 59.48 16 24.34

... 13.305.59.48 24.34

χ− −

= + + = We use a chi-square distribution with

( )( )4 1 3 1 6− − = df and find a P-value of 0.0384. Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that education and smoking status among French men aged 20 to 60 are not independent. (b) The chi-square statistic, broken down, is

2 0.204 0.186 0.044 0.092 1.177 0.700 0.641 1.693 2.704 1.866 1.141 2.858 13.305.χ = + + + + + + + + + + + = The largest contributions are from the university educated males. There are more non-smokers and moderate smokers and fewer former smokers and heavy smokers among this group than we would have expected.