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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.