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598 Part IV: Complete Solutions, Chapter 11

Chapter 11: Chi-Square and F Distributions

Section 11.1

1. Chi-square distributions are skewed right.

2. The skewness to the right decreases. Yes, chi-square distributions become more symmetric as the number of degrees of freedom increases.

3. Use a right-tailed test.

4. The null hypothesis for the test for independence claims that the random variables are independent. The null hypothesis for the test for homogeneity claims that the proportion of interest from each population is the same.

5. (a)



= 0.05

H

0

: Myers-Briggs preference and profession are independent.

(b)

H

1

: Myers-Briggs preference and profession are not independent.

 2   ( O



E )

2

E





49.02

2





57.98

2





74.22

2





87.78

2





62.76

2





74.24

2



8.649

All expected frequencies are greater than 5. Use the chi-square distribution.

Since there are 3 rows and 2 columns, d.f.

= (3



1)(2



1) = 2.

(c) In Table 7 in Appendix II, with d .

f . = 2, 8.649

falls between entries 7.38 and 9.21. Therefore,

0.010 < P value < 0.025. Using a TI-84, P value ≈ 0.0132.

(d) Since the P value is less than the level of significance



= 0.05, we reject the null hypothesis.

(e) At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.

6. (a)



= 0.01

H

0

: Myers-Briggs preference and profession are independent.

H

1

: Myers-Briggs preference and profession are not independent.

(b)

 2  



O



E

 2

E





71.86

2





76.14

2





77.20

2





81.80

2





102.94

2





109.06

2



10.26



= (3



1)(2



1) = 2.


f . = 2,

 2 

10.26







(e) At the 1% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and profession are not independent.

Copyright © Houghton Mifflin Company. All rights reserved.

Part IV: Complete Solutions, Chapter 11 599

7. (a)



= 0.01

H

0

: Site type and pottery type are independent.

H

1

: Site type and pottery type are not independent.

(b)

 2 







O



E

 2

E



75



74.64

 2









74.64

62



61.39

61.39

 2





 2





59.89

89.25

 2





 2





54.47

68



71.61

71.61

 2





 2





84.11

66



65.14

65.14

 2

70



67.5

67.5

 2



0.5552



= (3



1)(3



1) = 4.


f . = 4,

 2 

0.5552

falls between entries 0.484 and 0.711.

Therefore, 0.950 < P value < 0.975. Using a TI-84, P value ≈ 0.9679.

(d) Since the P value is greater than the level of significance



= 0.01, we do not reject the null hypothesis.

(e) At the 1% level of significance, there is insufficient evidence to conclude that site and pottery type are not independent.

8. (a)



= 0.05

H

0

: Ceremonial ranking and pottery type are independent.

H

1

: Ceremonial ranking and pottery type are not independent.

(b)

 2  



O



E

 2

E





79.94

2





55.06

2





85.86

2





59.14

2





91.19

2





62.81

2



6.198



= (3



1)(2



1) = 2.


f . = 2, 6.198







(e) At the 5% level of significance, there is sufficient evidence to conclude that ceremonial ranking and pottery type are not independent.

9. (a)



= 0.05

H

0

: Age distribution and location are independent.

H

1

: Age distribution and location are not independent.



(b)

 2 









O



E

 2

E





14.08

 2





11.33

 2





 2





12.84

34



31.59

31.59

 2





 2



14.08



28



28.82

28.82

 2





 2



11.33



30



31.59

31.59

 2



0.67

10.34

 2



= (3



1)(3



1) = 4.


f . = 4,

 2 

0.67







(e) At the 5% level of significance, there is insufficient evidence to conclude that age distribution and location are not independent.

10. (a)



= 0.05

H

0

: Myers-Briggs type and area of study are independent

H

1

: Myers-Briggs type and area of study are not independent.

(b)

 2 









O



E

 2

E

53.46





28.49





39.76



15.602





24.79



64.04

37



28.49

 2



28.49

17.76

29.69





85.75

21.27





39.76



85.75

2

 2



= (4



1)(3



1) = 6.


f . = 6, 15.602







(e) At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs type and area of study are not independent.

11. (a)



= 0.05

H

0

: Ages of young adult and movie preference are independent.

H

1

: Ages of young adult and movie preference are not independent.



(b)

 2 









O



E

 2

E



10.60



 2





10.00

 2





12.06

9



9.04

9.04

 2





 2







3.623

11.33

10.29

 2





12



9.35

9.35

 2





12



9.67

9.67

 2





 2

10.65

 2



= (3



1)(3



1) = 4.


f . = 4,

 2 

3.623







(e) At the 5% level of significance, there is insufficient evidence to conclude that age of young adult and movie preference are not independent.

12. (a)



= 0.01

H

0

: Contribution level and ethnic group are independent.

H

1

: Contribution level and ethnic group are not independent.

(b)

 2  



O



E

 2

E







80.61

2





65.60

2





48.14

2





34.48



22.17

2





84.78

2





69.00

2





50.63

2

2







36.27

2





23.31

2





85.42

2



(65



69.53)

2

69.53



51.02

2





36.54

2





23.49

2





109.19

2





13.35



88.87

2





65.21

2





46.71

2





30.03

2

All expected frequencies are greater than 5. Use the chi-square distribution. Since there are 4 rows and

5 columns, d.f.

= (4



1)(5



1) = 12.


f . = 12,

 2 

13.35







(e) At the 1% level of significance, there is insufficient evidence to conclude that contribution level and ethnic group are not independent.

13. (a)



= 0.05

H

0

: Stone tool construction material and site are independent.

H

1

: Stone tool construction material and site are not independent.



(b)

 2  



O



E

 2

E





689.36

2





625.64

2





102.22

2





92.78

2





542.58

2





492.42

2





93.84

2





85.16

2



11.15



= (4



1)(2



1) = 3.


f . = 3,

 2 

11.15







(e) At the 5% level of significance, there is sufficient evidence to conclude that stone tool construction material and site are not independent.

14. (i)

Spending Breakdown by Level and Political Party

60

50

40

30

20

10

0

De mo cr at

Re pu bli ca n

De mo cr at

Re pu bli ca n

De mo cr at

Re pu bli ca n

Le ss

th an

5

B illi on

5 to

10

B illi on

M or e th an

10

B illi on

(ii) α = 0.01

H

0

: The proportion of Democrats and Republicans within each spending level is the same.

H

1

: The proportion of Democrats and Republicans within each spending level is not the same.

 2 









O



E

 2

E



9.78



 2





 2

16.63





17.37

 2



19.41



 2



2.176

18.59

 2





10.22

 2


3 columns, d.f.

= (2



1)(3



1) = 2. In Table 7 in Appendix II, with d .

f . = 2,

 

2.176

falls between entries 0.211 and 4.61. Therefore, 0.100 < P value < 0.900. Using a TI-84, P value ≈ 0.3370.

Since the P value is greater than the level of significance






At the 1% level of significance, there is insufficient evidence to conclude that the proportion of spending for Democrats and Republicans within each level of spending is not the same.

15. (i)

Communication Breakdown by Age and Preference

20

10

40

30

60

50

80

70

0

15-24 25-34

Cell Phone

15-24 25-34

Instant Message

15-24 25-34

E-mail

15-24 25-34

Other

(ii) α = 0.05

H

0

: The proportion of age groups within each communication preference is the same.

H

1

: The proportion of age groups within each communication preference is not the same.

 2 



O



E

E

 2





44.5



44.5



35



35

 2



  

10 10 10.5

10.5

 2



9.312


4 columns, d.f.

= (2



1)(4



1) = 3. In Table 7 in Appendix II, with d .

f . = 3, χ

2

= 9.312 falls between entries 7.81 and 9.35. Therefore, 0.025 < P value < 0.050. Using a TI-84, P value ≈ 0.0254.

Since the P value is less than the level of significance




At the 5% level of significance, there is sufficient evidence to conclude that the proportion of age groups within each communication preference is not the same.

Section 11.2

1. The degrees of freedom are the number of categories minus one.

2. Consider the sample size of all the observed frequencies. The expected frequency for a category is computed by taking the proportion of the sample size designated by the proposed distribution.

3. The greater the differences between the observed frequencies and the expected frequencies, the larger the χ

2 value becomes. Larger χ 2

values lead to the conclusion that the differences are too large to be explained by chance alone.

4. No. When we reject the null hypothesis in a goodness-of-fit test, we say that the observed distribution is different from the expected distribution. The test does not tell us how the distributions differ within each category.

5. (a)



= 0.05

H

0

: The distributions are the same.

H

1

: The distributions are different.



(b)

 2 









O



E

 2

E

32.76



11.79



61.88



305.31



55.06

 2

All expected frequencies are greater than 5. Use the chi-square distribution. d .

f . = k



1 = 4



1 = 3.

(c) In Table 7, with d .

f . = 3,

 2 

11.79


Therefore, 0.005 < P -value < 0.010. Using a TI-84, P-value ≈ 0.0081.

(d) Since the P -value is less than the level of significance




(e) At the 5% level of significance, the evidence is sufficient to conclude that the Red Lake Village population does not fit the general Canadian population.

6. (a)



= 0.05

H

0


H

1


(b)

 2 









O



E

 2

E

106.86



13.017

 2





119.19

 2





36.99

 2





102.75

 2





68



45.21

 2

45.21


f . = k



1 = 5



1 = 4.


f . = 4, 13.017


Therefore, 0.010 < P -value < 0.025. Using a TI-84, P-value ≈ 0.0111.





(e) At the 5% level of significance, the evidence is sufficient to conclude that the Dove Creek household distribution does not fit the general U.S. household distribution.

7. (a)



= 0.01

H

0


H

1


(b)

 2  



O



E

 2

E





906



910.92

 2





910.92



0.1984

157.52

 2





169.40

 2





194.67

 2





53.50

 2


f . = k



1 = 5



1 = 4.


f . = 4, the entry 0.207 has 0.995 area in the right tail. Note that as

 2

decrease, the area in the right tail increases so a 0.207

means the corresponding

P -value > 0.995. Using a TI-84, P-value ≈ 0.9953.



(d) Since the P -value is greater than the level of significance




(e) At the 1% level of significance, the evidence is insufficient to conclude that the regional distribution of raw materials does not fit the distribution at the current excavation site.

8. (a)



= 0.05

H

0


H

1


(b)

 2 









O



E

 2

E

102.40



1.084

 2





123.84

 2





38.40

 2





27



29.76

29.76

 2





23



25.60

 2

25.60


f . = k



1 = 5



1 = 4.


f . = 4,

 2 

1.084


0.100 < P -value < 0.900. Using a TI-84, P-value ≈ 0.8968.





(e) At the 5% level of significance, the evidence is insufficient to conclude that the natural distribution of browse does not fit the feeding pattern.

9.

(i) Essay.

(ii) (a)



= 0.01

H

0


H

1


(b)

 2



O



E

 2

 

E







14.57





1.5693

83.70





83.70

14.57





2

210.80



210.80

 2

All expected frequencies are greater than 5. Use the chi-square distribution. d.f.

= k



1 = 6



1 = 5.


f . = 5, 1.5693







(e) At the 1% level of significance, the evidence is insufficient to conclude that the average daily July

temperature does not follow a normal distribution.

10.

(i) Essay.



(ii) (a)



= 0.01

H

0


H

1


(b)

 2  



O



E

 2

E





  

 2



14.57

83.70

210.80

210.80

 2

 

83.70

14.57



0.2562

All expected frequencies are greater than 5. Use the chi-square distribution. d.f.

= k



1 = 6



1 = 5.


f . = 5,

 2 

0.2562

is less than entry 0.412. Therefore, P -value > 0.995. Using a

TI-84, P-value ≈ 0.9984.





(e) At the 1% level of significance, the evidence is insufficient to conclude that the average daily

January temperature does not follow a normal distribution.

11. (a)



= 0.05

H

0


H

1


(b)

 2 









O



E

 2

E

150

 2





85



75

75

 2





220



200

 2





200

75



75

 2

75



9.333


f . = k



1 = 4



1 = 3.


f . = 3,

 2 

9.333







(e) At the 5% level of significance, the evidence is sufficient to conclude that current fish distribution is different than that of five years ago.

12. (a)



= 0.05

H

0


H

1




(b)

 2  



O



E

 2

E





11.92



284.16

2





222.0

2





177.6

2





133.2

2





71.04

2


f . = k



1 = 5



1 = 4.


f . = 4,

 2 

11.92







(e) At the 5% level of significance, the evidence is sufficient to conclude that the subject distribution of books in the library is different from that of books checked out by students.

13. (a)



= 0.01

H

0


H

1


(b)

 2 









O



E

 2

E





121.50

502



498.15

498.15

 2





40



36.45

 2





 2





36.45

56



72.90

72.90

480



461.70

 2





461.70

10



24.30



24.30

 2

2



13.7


f . = k



1 = 6



1 = 5.


f . = 5, 13.7







(e) At the 1% level of significance, the evidence is insufficient to conclude that census distribution and the ethnic origin distribution of city residents are different.

14. (a)



= 0.01

H

0


H

1


(b)

 2  



O



E

 2

E





88



62.28

 2







62.28



40



51.90

51.90

150.51

 2





76



72.66

72.66

 2





 2





52



57.09

 2

57.09

124.56



15.65

 2


f . = k



1 = 6



1 = 5.




f . = 5,

 2 

15.65







(e) At the 1% level of significance, the evidence is sufficient to conclude that the distribution of customer ages and the age distribution shown in the Snoop report are different.

15. (a)



= 0.01

H

0


H

1


(b)

 2  



O



E

 2

E





82.775

2





48.4

2





34.375

2





26.675

2





21.725

2





18.425

2





15.95

2





14.025

2





12.65

2



3.559


f . = k



1 = 9



1 = 8.


f . = 8,

 2 

3.559



(d) Since the P -value interval is greater than the level of significance




(e) At the 1% level of significance, the evidence is insufficient to conclude that the distribution of firstnon-zero digits in the accounting file do not follow Benford’s Law.

16. (a)



= 0.01

H

0


H

1


(b)

 2  



O



E

 2

E





50

2





50

2





50

2





50

2





50

2





50

2



13.44


f . = k



1 = 6



1 = 5.


f . = 5,

 2 

13.44



(d) Since the P -value interval is greater than the level of significance



= 0.01, we fail to reject the null hypothesis.

(e) At the 1% level of significance, the evidence is insufficient to conclude that distribution of observed outcomes for the die is different from the expected distribution of a fair die. In other words, at the 1% level of significance, we cannot conclude that the die is not balanced.



17. (a)

P (1)

P (2)



P (0)





 e

   r r !

e



1.72

(1.72)

0



0.179

0!

e



1.72

(1.72)

1

1!

e



1.72

(1.72)

2





2!

0.308

0.265

e

1.72

(1.72)

3

P (3)

 

0.152

3!

P (4 or more)

 

P (3 or less)

(b)

(c)

 2



0.096

r

0

1

2

3

4 or more

  ( O



E )

2

E

E = 90 P ( r )

90(0.179) = 16.11

90(0.308) = 27.72

90(0.265) = 23.85

90(0.152) = 13.68

90(0.096) = 8.64

(d)





16.11

2





27.72

2





23.85



 2





13.68

8.64



12.55

d .

f . = k



1 = 5



1 = 4



= 0.01

2

2

H

0

: The Poisson distribution fits.

H

1

: The Poisson distribution does not fit.

In Table 7 in Appendix II, with d .

f . = 4,

 2 

12.55


0.010 < P value < 0.025. Using a TI-84, P value ≈ 0.0137. Since the P value is greater than the level of significance



= 0.01, we do not reject the null hypothesis. At the 1% level of significance, we cannot say that the Poisson distribution does not fit.



18. (a)

P (1)



P (0)





P (2)

 e

   r r !

e



2.80

(2.80)

0

0!

e



2.80

(2.80)

1

1!

e



2.80

(2.80)

2





2!



0.061

0.170

0.238

e



2.80

(2.80)

3

P (3)

 

0.222

3!

e



2.80

(2.80)

4

P (4)

 

0.156

4!

P (5 or more)

 

P (4 or less)

(b)

(c)

 2



0.153

r

0

1

2

3

4

5 or more

 

( O



E )

2

E

E = 100 P ( r )

100(0.061) = 6.1

100(0.170) = 17.0

100(0.238) = 23.8

100(0.222) = 22.2

100(0.156) = 15.6

100(0.153) = 15.3

(d)





6.1

2





17.0

2





23.8

2





22.2

2





15.6

2





15.3

2



13.116

d .

f . = k



1 = 6



1 = 5



= 0.05

H

0

: The Poisson distribution fits.

H

1

: The Poisson distribution does not fit.


f . = 5, 13.116


0.010 < P value < 0.025. Using a TI-84, P value ≈ 0.0223. Since the P value is less than the level of significance



= 0.05, we reject the null hypothesis. At the 5% level of significance, the evidence is sufficient to say that the Poisson distribution does not fit.

Section 11.3

1.

Yes, it needs to be normal. No, the chi-square test of variance requires the x distribution to be exactly a normal distribution.

2. Use a histogram and a boxplot to check for symmetry and outliers. Look at a normal quantile plot using

Minitab, SPSS, or the TI-84 to check for a straight line. Also, use a goodness-of-fit test to see if the distribution follows the empirical rule.

3. (a)



= 0.05

H

0

:

H

1

:



2 = 42.3



2 > 42.3



(b)

 2 

 n





1

2

 s

2



 

42.3

d .

f . = 23



1 = 22



23.98

Assume a normal population distribution.

(c) Since > is in H

1

, a right-tailed test is used.

Using Table 7 in Appendix II and d .

f . = 22,

 2 

23.98







(e) At the 5% level of significance, there is insufficient evidence to conclude that the variance is greater in the new section.

(f) For . .



22 and

  

0.025,

 2

U



36.78.

2

For . .



22 and

  

0.975,



2

The 95% confidence interval for



2 is

2

L



10.98.

 n



1



 2

U s

2

 

  2 

 n



1

 s

2

 2

L

 

  2 

36.78

10.98

27.57

  2 

92.37

4. (a)

(b)



= 0.05

H

0

:

H

1

:



2 = 5.1



2 < 5.1

 2 

 n



1

 s

2



 2 d .

f . = 41



1 = 40



5.1





25.88


(c) Since < is in H

1

, a left-tailed test is used.


f . = 40,

 2 

25.88


Therefore,

0.975 < right-tail area < 0.950

1



0.975 < P value < 1



0.950

0.025 < P value < 0.050

Using a TI-84, P value ≈ 0.0411.





(e) At the 5% level of significance, there is sufficient evidence to conclude that the variance of age at first marriage is less than 5.1.



(f) For . .



40 and

  

0.05,

 2

U



55.76.

2

For . .



40 and

  

0.95,



2




2 is

2

L

 n



1

 s

2

  2 

 n



1

 s

2

 2

U

 

  2 

 2

L

 

55.76

26.51

2.37

  2 

4.98



26.51.

5. (a)

(b)



= 0.01

H

0

:

H

1

:



2 = 136.2



2 < 136.2

 2 

 n





1

2

 s

2



 

136.2

d .

f . = 8



1 = 7



5.92



1



f . = 7,

 2 

5.92



1



0.900 < P value < 1



0.100

0.100 < P value < 0.900


(d) Since the P value is greater than the level of significance, we do not reject the null hypothesis.

(e) At the 1% level of significance, there is insufficient evidence to conclude that the variance for number of mountain climber deaths is less than 136.2

(f) For . .



7 and

  

0.05,

 2

U



14.07.

2

For . .



7 and

  

0.95,



2




2 is

2

L



2.17.

 n



1

 s

2

 

 2

U

14.07

  2 

  2 

 n



1

 s

2

 2

L

 

2.17

57.26

  2 

371.29

6. (a)

(b)



= 0.05

H

0

:

H

1

:



2 = 47.1



2 > 47.1

 2 

 n



1 s

 2



2



 

47.1

d .

f . = 15



1 = 14



24.73





1



f . = 14, 24.73



(d) Since the P value is less than the level of significance, we reject the null hypothesis.

(e) At the 5% level of significance, there is sufficient evidence to conclude the variance of annual salaries is greater in Kansas.

(f) For . .



14 and

  

0.025,

 2

U



26.12.

2

For . .



14 and

  

0.975,

 2

L

2




2 is



5.63.

 n



 2

U

 

1

 s

2

  2 

  2 

 n



1

 s

2

 2

L

 

5.63

26.12

44.59

  2 

206.89

7. (a)

(b)



= 0.05

H

0

:

H

1

:



2 = 9



2 < 9

 2 

 n





1

2

 s

2



  

3

2

2 d .

f . = 23



1 = 22



8.82



1



f . = 22,

 2 

8.82



1



0.995 < P value < 1



0.990

0.005 < P value < 0.010



(e) At the 5% level of significance, there is sufficient evidence to conclude that the variance of protection times for the new typhoid shot is less than 9.

(f) For . .



22 and

  

0.05,

 2

U



33.92.

2

For . .



22 and

  

0.95,



2




is

2

L

 n



1

 s

2

 2

U

 n



1

 s

 2

L

2



12.34.

   2

33.92

  

12.34

2

1.53

2.54



8. (a)

(b)



= 0.01

H

0

:

H

1

:



2 = 225



2 > 225

 2 

 n





1

2

 s

2



  

  2

2 d .

f . = 10



1 = 9



23.04



1



f . = 9, 23.04




(e) At the 1% level of significance, there is sufficient evidence to conclude that the variance of the duration times of the tranquilizer is larger than stated in the journal.

(f) For . .



9 and

  

0.025,

 2

U



19.02.

2

For . .



9 and

  

0.975,

 2

L

2




is



2.70.

 n



1

 s

2

 2

U

 n



1

 s

2

 2

L

   2

19.02

   2

2.70

16.5

43.8

9. (a)

(b)



= 0.01

H

0

:

H

1

:



2 = 0.18



2 > 0.18

 2 

 n





1

2

 s

2

 d .

f . = 61



1 = 60

 

0.18



90



1



f . = 60,

 2 

90 falls between entries 88.38 and 91.95. Therefore,



(e) At the 1% level of significance, there is sufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is justified in claiming the blades must be replaced.



(f) For . .



60 and

  

0.05,

 2

U



79.08.

2

For . .



60 and

  

0.95,

2




is

 2

L



43.19.

 n



1

 s

2

 2

U

 

79.08

 n



1

 s

2

 2

L

 

43.19

0.45 mm 0.61 mm

10. (i) (a)



= 0.01

H

0

:

H

1

:



2 = 3,600



2



3,600

(b)

 2 

 n





1

2

 s

2

 d .

f . = 24



1 = 23

  

 

3, 600

2



33.12


(c) The area to the right of 33.12

is less than 50%, so we double the right-tail area to find the

P value for the two-tailed test. Right-tail area is between 0.050 and 0.100. Doubling these values for a two-tailed test gives 0.100 < P value < 0.200.





(e) At the 1% level of significance, there is insufficient evidence to conclude that the variance of test

scores on the preliminary exam is different from 3,600.

(ii) For . .



23 and

 

 n



1





2

U s

2

2



0.005,

 2

U



44.18.

For . .



23 and

  

0.995,

2




2 is

 2

L



9.26.

 n



1

 2

U

44.18

 s

   2

2

  2 

  2 

 n



1

 s

2

 2

L

   2

9.26

2698.78

  2 

12,876.03

2699

  2 

12,876

(iii) The 99% confidence interval for

 is

 n



1

 s

2



2

L

2698.78

51.95

12,876.03

113.47



11. (i) (a)



= 0.05

H

0

:

H

1

:



2 = 23



2



23

(b)

 2 

 n





1

2

 s

2

 d .

f . = 22



1 = 21

23



13.06


(c) The area to the left of

 2 

13.8

is less than 50%, so we double the left-tail area to find the P value for the two-tailed test. Right-tail area is between 0.950 and 0.900. Subtracting each value from 1, we find the left-tail area is between 0.050 and 0.100. Doubling the left-tail area for a twotailed test gives 0.100 < P value < 0.200.




= 0.05, we do not reject the null

hypothesis.

(e) At the 5% level of significance, there is insufficient evidence to conclude that the variance of

battery life is different from 23.

(ii) For . .



21 and

  

0.05,

 2

U



32.67.

2

For . .



21 and

  

0.95,

 2

L

2




2 is



11.59.

 n



1

 s

2

 2

U

  2

  

  2

32.67



 n



1

 s

2

 2

L



  

11.59

9.19

  2 

25.91

(iii) The 90% confidence interval for



is

 n



1



 2

U s

2

 n



1

 s

2

 2

L

9.19

25.91

Section 11.4

3.03

5.09

1. The populations should be independent.

2.

Yes, both populations must have a normal distribution. Distributions that are mound-shaped and more or less symmetric do not qualify for the F test.

3. F distributions are not symmetric. Values of the F distribution are always nonnegative.

4. You need to calculate the degrees of freedom for both the numerator and the denominator.

5.

(a)



= 0.01. Population I is annual production from the first plot.

H

0

:

 2 

1

 2

2

H

1

:



1

2   2

2



(b) Since s 2



0.332 is larger than s 2



0.089, we designate Population I as the first plot. s

1

2

F

  s

2 d.f.

N

0.332



3.73

0.089

2

= n

1



1 = 16



1 = 15, and d.f.

D

= n

2



1 = 16



1 = 15

The populations follow independent normal distributions. The samples are random samples from each population.


1


From Table 8 in Appendix II, F = 3.73 falls between entries 3.52 and 5.54. Therefore, 0.001 < P value

< 0.010. Using a TI-84, P value ≈ 0.0075.





(e) At the 1% level of significance, there is sufficient evidence to show that the variance in annual wheat production from the first plot is greater than that of the second plot.

6.

(a)



= 0.05. Population I is annual production from the second plot.

H

0

:



1

2   2

2

H

1

:



1

2   2

2

(b) Since s 2






0.318, we designate Population I as the second plot.

F s

1

2

  s

2

2

1.078

0.318



3.39

. .

N

 n

1

    d f

D n

2

   


(c) Since



is in H

1

, a two-tailed test is used.

From Table 8 in Appendix II, F = 3.39 falls between entries 3.14 and 3.95. Therefore,


0.050 < P value < 0.100






(e) At the 5% level of significance, there is insufficient evidence to show that the variance in annual wheat production from the first plot is different from that of the second plot.

7.

(a)



= 0.05. Population I has data from France.

H

0

:



1

2   2

2

H

1

:



1

2   2

2

(b) Since s 2






1.038, we designate Population I as France.

F s

1

2

  s

2

2

2.044

1.038



1.97

. .

N

 n

1

    d f

D n

2

   


(c) Since



is in H

1




0.100 < P value < 0.200








(e) At the 5% level of significance, there is insufficient evidence to show that the variance in corporate productivity of large companies in France and those in Germany differ. Volatility of corporate productivity does not appear to differ.

8.

(a)



= 0.05. Population I has data from South Korean companies.

H

0

:



1

2   2

2

H

1

:

 2

1

  2

2

(b) Since s 2






0.624, we designate Population I as the South Korean companies.

F s

1

2

  s

2

2

2.247

0.624



3.60

d.f.

N

= n

1



1 = 13



1 = 12, and d.f.

D

= n

2



1 = 9



1 = 8



1



< 0.050. Using a TI-84, P value ≈ 0.0388.





(e) At the 5% level of significance, there is sufficient evidence to show that the variance in productivity of

South Korean companies is larger than those of Sweden.

9.

(a)



= 0.05. Population I has data from aggressive growth companies.

H

0

:



1

2   2

2

H

1

:

 2

1

  2

2

(b) Since s 2






137.31, we designate Population I as aggressive growth. s

1

2

F

  s

2 d.f.

N

348.43



2.54

137.31

2

= n

1



1 = 21



1 = 20, and d.f.

D

= n

2



1 = 21



1 = 20



1



< 0.025. Using a TI-84, P value ≈ 0.0216.





(e) At the 5% level of significance, there is sufficient evidence to show that the variance in percentage annual returns for funds holding aggressive-growth small stocks is larger than that for funds holding value stocks.

10.

(a)



= 0.05. Population I refers to data from intermediate-term bonds.

H

0

:



1

2   2

2

H

1

:

 2

1

  2

2

(b) Since s 2






13.59, we designate Population I as intermediate-term bonds.

F s

1

2

  s

2

2

72.06

13.59



5.30

. .

N

 n

1

    d f n

2

1 13 1 12




(c) Since



is in H

1




0.002 < P value < 0.020






(e) At the 5% level of significance, there is sufficient evidence to show that the variance in annual percentage return of mutual funds holding short-term government bonds is different from that of funds holding intermediate-term corporate bonds.

11.

(a)



= 0.05. Population I has data from the new system.

H

0

:



1

2   2

2

H

1

:

 2

1

  2

2

(b) Since s 2






31.6, we designate Population I as the new system.

F s

1

2

  s

2

2

58.4

31.6



1.85

. .

N n

1

  d f

D n

2

   

24


(c) Since



is in H

1




0.100 < P value < 0.200






(e) At the 5% level of significance, there is insufficient evidence to show that the variance in gasoline consumption for the two injection systems is different.

12.

(a)



= 0.05. Population I refers to data from the old thermostat.

H

0

:



1

2   2

2

H

1

:

 2

1

  2

2

(b) Since s 2






5.1, we designate Population I as the old thermostat.

F

 d.f.

N s

1

2 s

2



12.8



2.51

5.1

2

= n

1



1 = 16



1 = 15, and d.f.

D

= n

2



1 = 21



1 = 20



1



< 0.050. Using a TI-84, P value ≈ 0.0281.





(e) At the 5% level of significance, there is sufficient evidence to show that the variance in temperatures with the old thermostat is greater than that with the new. The new thermostat appears to be more dependable.

Section 11.5

1.

(a)



= 0.01



H

0

:



1

 

2

 

3

H

1

: Not all the means are equal.

(b) Site I

 x n





1

7

286

 x

1

2 

15,312

SS

1



3626.857



 x

TOT x

2

TOT



286 164 176



626



Site II



 x n

2 x

2

2



4



164



8354

SS

2



1630

 

31,116

Site III n



6

 x

3



176

 x

2

3



7450

SS

3



2287.333

N 7 4 6 17

SS

TOT k



3

  x

2

TOT



  x

TOT

 2

N



31,116





626

 2

17



8, 064.470

SS

BET

 

All groups





  x i

 2



 n i





 x

TOT

 2

N





286

7

     

4



6

626

17

 2



520.280

SS

W



SS

1



SS

2



SS

3



3, 626.857 1, 630



2, 287.333



7, 544.190

Check that SS

TOT



SS

BET



SS

W

: 8, 064.470

 

BET

W

TOT k

N

1 k

3 1 2

17 3 14

N

MS

BET



SS

BET

BET



520.280



260.14

2

MS

W



SS

W d f

W



7, 544.190

14



538.87

For

F



MS

BET 

260.14

MS

W

538.87

. .

N



2 and d f

D



14.



0.48

(c) From Table 8 in Appendix II, F = 0.48 falls below the entry 2.73. Therefore, P value > 0.100. Using a

TI-84, P value ≈ 0.6270.




= 0.01, do not reject H

0

.

(e) At the 1% level of significance, there is insufficient evidence to conclude that the means are not all equal.

(f) Summary of ANOVA results

Source of

Variation

Sum of

Squares

Degrees of

Freedom MS

F

Ratio P Value

Test

Decision

Between groups 520.280

Within groups 7544.190

2

14

260.14 0.48 >0.100

538.87

Do not reject H

0

Total 8064.470 16



2.

(a)



= 0.05

H

0

:



1

 

2

 

3

 

4

(b)

H

1


 x

1

Site I n



5



76

 x

2 n

Site II



6



104

Site III n



4

 x

3



112

 x

1

2 

1336

SS

1



180.8

 x

2

2

SS

2



2374



571.333



 x

TOT x

2

TOT

N



76 104 112 107



399

   

9, 519

5 6 4 6 21 k



4

SS

TOT

  x

2

TOT



  x

TOT



N



9, 519





399

 2

21



1, 938

SS

BET

 

All groups





  x i

 2



 n i





 x

TOT

 2

N

 x

2

3

SS

3



3558



422

 x

4 n

Site IV



6



107

 x

2

4



2251

SS

4



342.833

SS

W





        

5 6

SS

1



SS

2



SS

3



SS

4

4



6

399

 2



421.033

21

  

1, 516.967

Check that SS

TOT



SS

BET



SS

W

:1,938

 

BET d f

W

TOT k 1 4 1 3

N k 21 4 17

N 1 21 1 20

MS

BET



SS

BET

BET



421.033

3



140.344

MS

W



SS

W

W



1, 516.967

17



89.233

For

F



MS

BET 

140.344



1.573

MS

W

89.233

. .

N



3 and d f

D



17.

(c) From Table 8 in Appendix II, F = 1.573 falls below the entry 2.44. Therefore, P value > 0.100. Using a

TI-84, P value ≈ 0.2327.




= 0.05, we do not reject H

0

.

(e) At the 5% level of significance, there is insufficient evidence to conclude the means are not all equal.




Source of

Variation

Sum of

Squares


Degrees of

Freedom

3

MS

F

Ratio P Value

140.344 1.573 >0.10

Test

Decision

Do not reject H

0

Within groups 1,516.967 17 89.233

Total 1,938.000 20

3.

(a)



= 0.05

H

0

:



1

 

2

 

3

 

4

H

1


(b) See Section 11.5 or solutions to Problems 1 and 2 for examples of calculations from formulas.

F



MS

BET 

29.879

35.325



0.846

MS

W

. .

N



3 and d f

D



18

(c) From Table 8 in Appendix II, F = 0.846 falls below entry 2.42. Therefore, P value > 0.100. Using a TI-

84, P value ≈ 0.4867.





0

.

(e) At the 5% level of significance, there is insufficient evidence to conclude the means are not all equal.


Source of

Variation

Sum of

Squares

Degrees of

Freedom MS

F

Ratio P Value

Test

Decision



3

18

29.879 0.846 >0.100 Do not reject H

0

35.324

Total 725.464 21

4.

(a)



= 0.01

H

0

:



1

 

2

 

3

H

1


(b) See Section 11.5 or solutions to Problems 1 and2 for examples of calculations from formulas.

F



MS

BET 

107.8

132.1



0.816

. .

N

MS

W d f



2 and d f

D



15.


84, P value ≈ 0.4608.





0

.



Source of

Variation

Sum of

Squares

Degrees of

Freedom MS

F

Ratio P Value

Test

Decision

Between groups 215.680 2 107.840 0.816 >0.100 Do not reject H

0

Within groups 1,981.725 15

Total 2 197.405

5.

(a)



= 0.05

H

0

:



1

 

2

 

3

H

1


17

132.115



(b) See Section 11.5 or solutions to Problems 1 and2 for examples of calculations from formulas.

F



MS

BET 

651.58



MS

W

130.19

. .

N



2 and d f

D



9

5.005

(c) From Table 8 in Appendix II, F = 5.005 falls between entries 4.26 and 5.71. Therefore, 0.025 < P value < 0.050. Using a TI-84, P value ≈ 0.0346.




= 0.05, we reject H

0

.

(e) At the 5% level of significance, there is sufficient evidence to conclude that the means are not all equal.


Source of

Variation

Sum of

Squares

Degrees of

Freedom MS

F

Ratio P Value

Test

Decision

Between groups 1,303.167

Within groups 1,171.750

2

9

651.58 5.005 0.025 < P value < 0.050 Reject H

0

130.19

Total 2,474.917

6.

(a)



= 0.05

H

0

:



1

 

2

 

3

H

1


11


F



MS

BET 

1.2208

0.4138



2.95

. .

N

MS

W d f



2 and d f

D



18

(c) From Table 8 in Appendix II, F = 2.95 falls between entries 2.62 and 3.55. Therefore, 0.050 < P value

< 0.100. Using a TI-84, P value ≈ 0.0779.




= 0.05, do not reject H

0

.

(e) At the 5% level of significance, there is insufficient evidence to conclude the that means are not all equal.


Source of

Variation

Sum of

Squares

Degrees of

Freedom MS

F

Ratio P Value

Test

Decision

Between roups 2.442

Within groups 7.448

2

18

1.2208 2.95 0.050 < P value < 0.100 Do not reject H

0

0.4138

Total 9.890 20

7.

(a)



= 0.01

H

0

:



1

 

2

 

3

H

1



F



MS

BET 

1.021

3.039



0.336

MS

W

. .

N

 d f

D



11


84, P value ≈ 0.7217.





0

.





Source of

Variation

Sum of

Squares

Degrees of

Freedom MS

F

Ratio P Value

Test

Decision

Between groups

Within groups

2.042

33.428

2

11

1.021 0.336

3.039

> 0.100 Do not reject H

0

Total 35.470 13

8.

(a)



= 0.05

H

0

:



1

 

2

 

3

 

4

H

1



F



MS

BET 

6.322

0.424



14.910

MS

W

. .

N



3 and d f

D



13

(c) From Table 8 in Appendix II, F = 14.910 falls above entry 10.21. Therefore, P value < 0.001. Using a

TI-84, P value ≈ 0.0002.





0

.



Source of

Variation

Sum of

Squares

Degrees of

Freedom MS

F

Ratio P Value

Test

Decision

Between groups 18.965 3 6.322 14.910 <0.001 Reject H

0

Within groups 5.517 13 0.424

Total 24.482

9.

(a)



= 0.05

H

0

:



1

 

2

 

3

 

4

H

1


16


F



MS

BET 

79.408

17.223



4.611

MS

W

. .

N



3 and d f

D



15

(c) From Table 8 in Appendix II, F = 4.611 falls between entries 4.15 and 5.42. Therefore, 0.010 < P value < 0.025. Using a TI-84, P value ≈ 0.0177.





0

.





Source of

Variation

Sum of

Squares



Total

Section 11.6

496.565

Degrees of

15

18

MS

F

Ratio P Value

Test

Decision

79.408 4.611 0.010 < P value < 0.025 Reject H

0

17.223

1.

There are two factors. One factor is walking device , with three levels, and the other factor is task , with two levels. The data table has six cells.

2.

There are two factors. One factor is rank , with four levels, and the other factor is institution , with two levels. The data table has eight cells.

3.

Since the P value is less than 0.01, there is a significant difference in mean cadence according to the factor walking device used.

The critical value is F

0.01

= 6.01. Since the sample F = 30.94 is greater than F

0.01

,

F lies in the critical region, and we reject H

0

.

4.

(a) There are two factors. One factor is education level , with four levels, and the other factor is media type , with five levels.

(b) For education,

H

0

: No difference in population mean index according to education level.

H

1

: At least two education levels have different mean indices.

F education



MS education

MS error

P value



0.075



320

108



2.96

At 5% level of significance, do not reject H

0

for education level.

The data do not indicate any differences in population mean index according to education level.

(c) For media,

H

0

: No difference in population mean index by media type.

H

1

: At least two types of media have different population mean indices.

F media



MS media 

1



0.01

MS error

108

P value



1.000 to three digits after the decimal.

At the 5% level of significance, do not reject H

0 for media.

The data do not indicate any differences in population mean index according to media type.

5.

(a) There are two factors. One factor is income level , with four levels, and the other factor is media type , with five levels.

(b) For income,

H

0

: There is no difference in population mean index based on income level.

H

1

: At least two income levels have different population mean indices.

F income



MS income

MS error

P value



0.088



359

130



2.77


0

.

The data do not indicate any differences in population mean index according to income level.



(c) For media,

H

0

: No difference in population mean index by media type.

H

1

: At least two media types have different population mean indices.

F media



MS media

MS error

P value



0.998



4

130



0.03


0

.

The data do not indicate any differences in population mean index according to media type.

6.

(a) There are two factors. One factor is class, with four levels, and the other factor is gender, with two levels.

(b) H

0

: No interaction between factors.

H

1

: Some interaction between factors.

F interaction



MS interaction

MS error

P value



0.736



0.095

0.224



0.43


0

.

The data do not indicate any interaction between factors.

(c) H

0

: No difference in population mean GPA based on class.

H

1

: At least two classes have different population mean GPAs.

F class



MS class 

0.742

0.224



3.32

MS error

P value



0.037. At the 5% level of significance, reject H

0

.

There is sufficient evidence to say that at least two classes have different population mean GPAs.

(d) H

0

: No difference in population mean GPA based on gender.

H

1

: Some difference in population mean GPA based on gender.

F gender



MS gender 

0.281

0.224



1.26

MS error

P value



0.273. At the 5% level of significance, do not reject H

0

.

There is insufficient evidence to say that males and females have different GPAs.


Part IV: Complete Solutions, Chapter 11

7.

627

Yes, the design fits the model for randomized block design.

Chapter 11 Review

1.

The chi-square and F distributions have only nonnegative values.

2. The normal and Student’s t distributions are always symmetric.

3. Since we are interested in proportions, the test for homogeneity is appropriate.

4. Since we are comparing to a fixed variance, a test of one variance is appropriate.



5.

One-way ANOVA

(i)



= 0.01

H

0

:



1

 

2

 

3

 

4

H

1


(ii) See Section 11.5 or solutions to Problems 1 and 2 for examples of calculations from formulas.

F



MS

BET 

2, 050

778



2.63

. .

N

MS n d f



3 and d f

D



16

(iii) From Table 8 in Appendix II, F = 2.63 falls between entries 2.46 and 3.24. Therefore, 0.050 < P value

< 0.100. Using a TI-84, P value ≈ 0.0857.

(iv) Since the P value is greater than the level of significance




0

.

(v) At the 1% level of significance, there is insufficient evidence to conclude that not all the packaging mean sales are equal.

(vi) Summary of ANOVA results

Source of

Variation

Sum of

Squares

Degrees of

Freedom MS

F

Ratio P Value

Test

Decision

Between groups 6,150

Within groups 12,455

3

16

2,050 2.63 0.050 < P value < 0.100

778

Do not reject H

0

Total 18,605 19

6.

Chi-square test of independence

(i)



= 0.01

H

0

: Time to take a test and test score are independent.

(ii)

H

1

: Time to take a test and test score are not independent.

 2 









O



E





E

2

18.93

17



21.07



42.60

48 47.40



71.00

85 79.00



 2

9.47

8 10.53



21.07

47.40

79.00

10.53



3.92


= (2



1)(4



1) = 3.

 2

(iii) In Table 7 in Appendix II, with d .

f . = 3,

 2 

3.92







(v) At the 1% level of significance, there is insufficient evidence to claim that time to do a test and test results are not independent.

7.

(a) Chi-square for testing



2

(i)



= 0.01

H

0

:



2 = 1,040,400

H

1

:



2 > 1,040,400



(ii)

 2 

 n





1

2

 s

2









1, 020

 2

 2 d.f.

= 30



1 = 29



51.03

(iii) Since > is in H

1



f . = 29,

 2 

51.03



(iv) Since the P value is less than the level of significance




(v) At the 1% level of significance, there is sufficient evidence to conclude that the variance is greater

than claimed.

(b) For . .



29 and

  

0.025,

 2

U



45.72.

2

For . .



29 and

  

0.975,

2




2 is

 2

L



16.05.

 n



1

 s

2

 2

U

  2 

 n



1

 s

2

 2

L

 

45.72

 2

  2 

   2

16.05

1,161,147.4

  2 

3,307,642.4 square foot-pounds

8.

One-way ANOVA

(i)



= 0.01

H

0

:



1

 

2

 

3

H

1


(ii) See Section 11.5 or solutions to Problems 1 and 2 for examples of calculations from formulas.

F



MS

BET 

0.501

1.129



0.44

. .

N

MS

W d f



2 and d f

D



9

(iii) From Table 8 in Appendix II, F = 0.44 falls below entry 3.01. Therefore, P value > 0.100. Using a TI-

84, P value ≈ 0.6572





0

.

(v) At the 1% level of significance, there is insufficient evidence to conclude that all the mean times to execute the programs are not equal.

(vi) Summary of ANOVA results

Source of

Variation

Sum of

Squares

Degrees of

Freedom MS

F

Ratio P Value

Test

Decision

Between groups 1.002 2 0.501 0.443 >0.100 Fail to reject H

0

Within groups 10.165 9 1.129

Total 11.167

9.

Chi-square test of independence

(i)



= 0.01

11



H

0

: Student grade and teacher rating are independent.

H

1

: Student grade and teacher rating are not independent.

(ii)

 2 









O



E

 2

E

10.00





40.00





43.33



9.8





13.33

65.00

10.00





2



21.67

15.00





5.00

20.00



 2



30.00

27



26.67

 2

26.67


= (3



1)(4



1) = 6.

(iii) In Table 7 in Appendix II, with d .

f . = 6,

 

9.8







(v) At the 1% level of significance, there is insufficient evidence to claim that student grade and teacher rating are not independent.

10.

Chi-square for testing



2

(i)



= 0.01

H

0

:



2 = 0.0625

H

1

:



2 > 0.0625

(ii)

 2 

 n





1

2

 s

2









0.25

 2

 2



25.41

d.f.

= 12



1 = 11


1



f . = 11,

 2 

25.41


0.005 < P value < 0.010. Using A TI-84, P value ≈ 0.0079.





(v) At the 1% level of significance, there is sufficient evidence to conclude that the variance has increased and the machine needs to be adjusted.

11.

Chi-square goodness of fit

(i)



= 0.01

H

0


H

1




(ii)

 2  



O



E

 2

E





42.0

2





31.5

2





63.0



 2



 2

52.5

21.0



11.93

d .

f . = k



1 = 5



1 = 4

2

(iii) In Table 7 in Appendix II, with 4 d .

f .,

 2 

11.93







0

.

(v) At the 1% level of significance, there is insufficient evidence to claim that the age distribution of the population of Blue Valley has changed.

12.

F test for two variances.

(i)



= 0.05

H

0

:



1

2   2

2

H

1

:

 2 

1



2

2

(ii) Since s 2 = 0.235 is larger than s 2 = 0.128, we designate Population I as the old process.

F

 d f

N s

1

2 s

2



0.235



1.84

0.128

2 n

1

1

 

20, and . .

D

 n

2

1 26 1 25

(iii) Since



is in H

1




0.100 < P value < 0.200






(v) At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is different.

13.

F test for the equality of two variances.

(i)



= 0.05

H

0

:



1

2   2

2

H

1

:



1

2   2

2

(ii) Since s 2 = 135.24 is larger than s 2 = 51.87, we designate Population I as the new process.

F d.f.

 s s

1

2

2

2

N

= n



135.24

51.87



2.61

1



1 = 16



1 = 15, and d.f.

D

= n

2



1 = 18



1 = 17


1



< 0.050. Using a TI-84, P value ≈ 0.0302.







(v) At the 5% level of significance, there is sufficient evidence to show that the variance for the lifetimes of bulbs manufactured using the new process is larger than that for bulbs made by the old process.

14.

Two-way ANOVA

(a) There are two factors. One factor is day with two levels, and the other factor is section with three

(b) levels.



= 0.01

H

0

: No interaction between day and section.

H

1

: Some interaction between day and section.

F interaction



MS interaction 

19.0

MS error

18.5

P value



0.371. Do not reject



1.03

H

0

; at the 1% level of significance, the data do not indicate any interaction between day and section.

(c) H

0

: No difference in population mean number of responses according to day.

H

1

: At least two population means are different among the days.

F day



MS day

MS error

P value



0.000



1024.0

18.5



55.28

Reject H

0

; at the 1% level of significance, some of the means are different among the days.

(d) H

0

: No difference in population mean number of responses according to section.

H

1

: At least two population means are different among the sections.

F section



MS section

MS error

P value



0.000



786.8



18.5

42.48

Reject H

0

; at the 1% level of significance, some of the means are different among the sections.


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