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43, With f= 8696.01 = Forze, there are significant significant difference among the exam means at the |
differences among the three treatment means. 205 level: - ’ |
The normal plot of residuals shows no reason to doubt b. Because f= 1.65 < 2.61 = Fos.449, there is no |
normality,.and'the plot’bf residuals against the fitted significant difference among the retention means at |
values shows no reason to doubt constant variance. the 208 level |
There is no significant difference between treatments B64, a, |
and C, but Treatment A differs (it is lower) significantly |
from the others at the .01 level. ieee of Sf USF |
Means: SS |
A29.49 B3131 31.40 Piet i e229: 282 a) |
Error 25 2.690 108 |
45. Because f = 8.87 > 7.01 = Foi.as, reject the hypothesis Total 29 36 |
that the variance for B is 0. |
Because f= 2.15 < 2.76 =Fosaas, there is no |
49. a. significant difference among the diet means at the .0S |
ee level. |
Source at 8s Ms F b. (~.59, 92) Yes, the interval includes 0. |
A 2 30763.0 15381.5 3.79 es) |
5 3 34185.6 11395.2 2.81 63, a, Test Mo: ti = 2 = fs versus Hy: the three means |
Interaction 6 43581.2 7263.5 1.79 are not ail the same. With f = 4.80 and Fo52.16 = |
Error 24 97436.8 4059.9 3.63 < 4.80 < 623 =Foi2is it follows that |
Total 35_205966.6 01 < P-value < .05 (more’ precisely, P = .023). |
an Reject Ho in favor of H, at the 5% level but not at |
b. Because 1.79 <2.04=Fige2% there is no the 1% level. |
significant interaction. b. Only the first and third means differ significantly at |
¢. Because 3.79 > 3.40 = F os2.24, there is a significant the 5% level. |
difference among the A means at the .05 level. A 3 % |
d. Because 2.81 < 3.01 = F.os2s, there is no 5 |
significant difference among the B means at the .05 Ga5F 2692) 28.07 |
level. |
e. Using w = 64.93, 65. Because f = 1123 > 4.07 = Fs.,s. there are significant |
differences among the means at the .05 level. |
3 1 2 For Tukey multiple comparisons, w = 7.12: |
3960.2 4010.88 4029.10 |
--- Trang 843 --- |
830 Chapter 12 |
¢. No, there is a wide range of y values for a given x; for |
ieee eM RM EID example when temperature is 18.2 the ratio ranges |
29.92 33.96 125.84 129.30 from 9 102.68. |
3. Yes. Yes. |
‘The means split into two groups of two. The means within |
each group do not differ significantly, but the means in 5. b. Yes |
the top group differ strongly from the means in the ¢, The relationship of y to x is roughly quadratic. |
bottom group. |
7. 0.5050 psi be L3 psi. 130 psi d. —130 psi |
67. The normal plot is reasonably straight, so there is no * m ‘ |
reason to doubt the normality assumption. Sean WS mimi bea smiiain 8S mufminy |
1,305 m/min d. 4207, .3446 e. 0036 |
01m —10n—b.3.08, 2.5 |
pounce: 20 Ss MS zs €..3653 4624 |
A 1 322.667 322.667 980.5 13. a. y = 63 + .652x |
B 3 35.623 11.874 36.1 b. 23.46, -2.46 |
AB 3 8.557 2.852 8.7 ©. 392, 5.72 |
Error 16 5.266 +329 d. 956 |
Total 23 372.113 e. y = 2.29 + .564x, 17 = 688 |
15. a. y = —15.2 + .0942x |
With f = 8.7 > 3.24 =Fossis there is significant b. 1.906 |
interaction at the .05 level. ¢. —1.006 , ~0.096, 0.034, 0.774 |
In the presence of significant interaction, main effects are d. 451 |
not very useful. |
17. a. Yes |
b. slope, .827; intercept, —1.13 |
¢. 40.22 |
Chapter 12 Gaon |
e975 |
1. a, Temperature |
aa 19. a. y= 75.2 — .209x 54.274 |
a b. The coefficient of determination is .791, meaning that |
th 3 the predictor accounts for 79.1% of the variation in y. |
17) 445 ¢. The value of s is 2.56, so typical deviations from the |
17| 67 regression line will be of this size. |
17 Stem: hundreds and tens . |
18] 0000011 Leaf: ones Aly be a 00d |
18) 2222 7.72 |
18) 445 | . 7 . |
1s] 6 25. fly = 1.8fy +32, Bi, = 1.8, |
iy 8 29. a. Subtracting ¥ from each x; shifts the plot ¥ units to |
‘The distribution is fairly symmetric and bell-shaped with the left. The slope is left unchanged, but the new |
a center around 180. y intercept is y, the height of the old line at x = x. |
Ratio b. fy =¥ = By + BX and Bi = By |
o | e890 31. a. 00189 |
b. 7101 |
; gone ¢. No, because here E(x; ~ ¥)°is 24,750, smaller than the |
value 70,000 in part (a), so V(B,) = o7/Z(x; — x)” is |
A | “dees higher here. |
A 66 |
1| 8889 Stem: ones; 33. a. (51, 1.40) |
3 | a4 Peske Leathe b. To test Ho: By =1 vs. Hy: fy <1, we compute |
5 1 = —2258 > ~1.383 = 1109, so there is no |
3 | % reason to reject the null hypothesis, even at the 10% |
level. There is no conflict between the data and the |
: & assertion that the slope is at least 1. |
3! 00 35. a. By = 1.536, and a 95% CT is (.632, 2.440) |
b. Yes, for the test of Ho: B, = O-vs. Hy: fy # 0, we find |
The distribution is concentrated between 1 and 2, with 1 = 3.62, with P-value .0025. At the .01 level |
some positive skewness. conclude that there is a useful linear relationship. |
b. No, x does not determine y: for a given x there may be ¢. Because 5 is beyond the range of the data, predicting |
more than one y. at a dose of 5 might involve too much extrapolation. |
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