# 9.8: Chapter 9 Solutions

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3. $$SS_{Factor}=4,939.2$$

5. $$df_{num}=2$$

7. $$MS_{Factor}=2,469.6$$

9. $$F_{obs}=3.742$$

11. $$F_{\alpha=.05, 2, 12}=3.89$$. $$F_{obs}$$ does not exceed $$F_{critical}$$, so we fail to reject $$H_0$$.

13. $$H_{0} : \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}$$; $$H_a$$: At least two of the means ($$\mu_{1}, \mu_{2}, \mu_{3}, \mu_{4}$$) are not equal.

15. $$df_{num}=3$$

17. $$SS_{Within}=13.2$$

19. $$MS_{Within}=.825$$

21.

Source of variation $$SS$$ $$df$$ $$MS$$ $$F$$
Between 25.75 3 8.583

10.404

Within 13.2 16

.825

Total 38.95 19

Table $$\PageIndex{1}$$

23. The closest available $$F$$-value to our $$F_{obs}=10.404$$ at (3, 16) $$df$$ is 9.01, which corresponds to $$\alpha$$ of .01. Therefore, the $$p$$-value must be less than .01. Because this $$p$$-value is less than $$\alpha$$ = .05, we reject $$H_{0}$$.

27. $$df_{num}=4$$

29. $$SS_{Factor}=195.6$$; $$MS_{Factor}=48.9$$

31. $$F_{obs}=2.060$$

33. $$F_{\alpha=.05, 4, 10}=3.48$$

35. No, $$F_{obs}$$ does not exceed $$F_{critical}$$, and the approximate $$p$$-value exceeds $$\alpha$$, so we fail to reject $$H_0$$.

37. $$df_{num}=2$$

39. $$SS_{Between}=5,700.4$$; $$MS_{Between}=2,850.2$$

41. $$F_{obs}=3.610$$

43. $$F_{\alpha=.01, 2, 12}=6.93$$

45. No, $$F_{obs}$$ does not exceed $$F_{critical}$$, and the approximate $$p$$-value exceeds $$\alpha$$, so we fail to reject $$H_0$$.

47. $$SS_{Between}=.903$$

49. $$df_{num}=4$$

51. $$F_{obs}=4.220$$

53.

a. Decision: Reject $$H_0$$, because the approximate $$p$$-value (between .01 and .05) is less than $$\alpha$$.

b. Conclusion: There is a significant difference in at least one of these regions of the country for the average age at which teenagers obtain their driver's licenses.

55. $$H_{0} : \mu_{East}=\mu_{Central}=\mu_{West}$$; $$H_a$$: At least two of the means ($$\mu_{East}, \mu_{Central}, \mu_{West}$$) are not equal.

$$F_{obs}=\frac{\frac{344.164}{2}}{\frac{1,219.550}{11}}=\frac{172.082}{110.868}=1.552$$

$$F_{\alpha=.05, 2, 12}=3.98$$

$$F_{obs}$$ does not exceed $$F_{critical}$$, so we fail to reject $$H_0$$.

57. $$H_{0} : \mu_{Linda}=\mu_{Tuan}=\mu_{Javier}$$; $$H_a$$: At least two of the means ($$\mu_{Linda}, \mu_{Tuan}, \mu_{Javier}$$) are not equal.

$$F_{obs}=\frac{\frac{23.212}{2}}{\frac{208.324}{12}}=\frac{11.606}{17.360}=.669$$

$$F_{\alpha=.01, 2, 12}=6.93$$

$$F_{obs}$$ does not exceed $$F_{critical}$$, so we fail to reject $$H_0$$.

59. $$H_{0} : \mu_{Home}=\mu_{News}=\mu_{Health}=\mu_{Computer}$$; $$H_a$$: At least two of the means ($$\mu_{Home}, \mu_{News}, \mu_{Health}, \mu_{Computers}$$) are not equal.

Source of variation $$SS$$ $$df$$ $$MS$$ $$F$$
Between 34,288.6 3 11,429.533

8.689

Within 21,047.6 16

1,315.475

Total 55,336.2 19

Table $$\PageIndex{2}$$

$$F_{\alpha=.05, 3, 16}=3.24$$. $$F_{obs}$$ does exceed $$F_{critical}$$, so we can reject $$H_0$$.

61. $$H_{0} : \mu_{CNN}=\mu_{FOX}=\mu_{Local}$$; $$H_a$$: At least two of the means ($$\mu_{CNN}, \mu_{FOX}, \mu_{Local}$$) are not equal.

$$F_{obs}=\frac{\frac{1,967.925}{2}}{\frac{3,375.133}{14}}=\frac{983.963}{241.081}=4.081$$

$$F_{\alpha=.05, 2, 14}=3.74$$

$$F_{obs}$$ exceeds $$F_{critical}$$, so we reject $$H_0$$.

63. $$H_{0} : \mu_{White}=\mu_{Black}=\mu_{Hispanic}=\mu_{Asian}$$; $$H_a$$: At least two of the means ($$\mu_{White}, \mu_{Black}, \mu_{Hispanic}, \mu_{Asian}$$) are not equal.

Source of variation $$SS$$ $$df$$ $$MS$$ $$F$$
Between 13.032 3 4.344

.885

Within 73.600 15

4.907

Total 86.632 18

Table $$\PageIndex{3}$$

$$F_{\alpha=.01, 3, 15}=5.42$$. $$F_{obs}$$ doesn't exceed $$F_{critical}$$, so we cannot reject $$H_0$$.

65.

1. Yes.
2. Because $$n_1 = n_2 = n_3$$.
3. $$\bar{x}_{Heavy} = 4.55, s_{Heavy} = 1.00; \bar{x}_{Medium} = 4.525, s_{Medium} = 1.13; \bar{x}_{Light} = 2.60, s_{Light} = .70$$
• g = 3
• n = 12
• $$SS_{Between}$$ = 10.012
• $$MS_{Between}$$ = 5.006
• $$SS_{Within}$$ = 8.277
• $$MS_{Within}$$ = .920
• $$df_{num}$$ = 2, $$df_{denom}$$ = 9
• $$F$$ statistic = 5.443
• $$F$$-critical = 8.02
• Graph: Check student's solution.
• Decision: $$F_{obs}$$ doesn't exceed $$F_{critical}$$, so we cannot reject $$H_0$$.
• Conclusion: There is not sufficient evidence to conclude that the mean paper airplane flight distances differ based on paper weights.

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