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.

Yes.
Because \(n_1 = n_2 = n_3\).
\(\bar{x}_{Heavy} = 4.55, s_{Heavy} = 1.00; \bar{x}_{Medium} = 4.525, s_{Medium} = 1.13; \bar{x}_{Light} = 2.60, s_{Light} = .70\)
- 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.