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828 Chapter 11
85. The difference is significant at the .05, 01, and 001 7. a. The Levene test gives f = 1.47, P-value .236, so there
levels. is no reason to doubt equal variances.
5 : b. Because f= 10.48 > 4.02 = Foi, there are
89. b. No, given that the 95% CI includes 0, the test at the uignificany ainverencevathong Wik mean
.05 level does not reject equality of means.
91. (—299.2, 1517.8) Source DE Ss) MS FP
Plate 4 43993 10998 10.48 0.000
93. (1020.2, 1339.9). Because 0 is not in the CI, we would
length
reject equality of means at the .01 level.
Error 30 31475 1049
95. Because = 2.61 and the one-tailed P-value is .007, the Ba tal 5a, SBS
difference is significant at the .05 level using either a a
one-tailed or a two-tailed test. wae At kh Bs
97. a. Because 1 = 3.04 and the two-tailed P-value is .008, Splitting the paints into two groups, (3, 1, 4), (2, 5),
the difference is significant at the .05 level. there are no significant differences within groups but the
b. No, the mean of the concentration distribution paints in the first group differ significantly (they are
depends on both the mean and standard deviation lower) from those in the second group.
of the log concentration distribution.
13.3 1 4 2 5
99, Because ¢ = 7.50 and the one-tailed P-value is 0000001, 4275 462.0 469.3 502.8 532.1
the difference is highly significant, assuming normality. SSS
101. The two-sample / is inappropriate for paired data, The 15. w = 5.92; At the 1% level the only significant
paired 1 gives a mean difference .3, 1 = 2.67, and the _differences are between formation 4 and the first two
two-tailed P-value is .045, so the means are significantly formations.
different at the .05 level. We are concluding tentatively 2 1 3 4
that the label understates the aleohol percentage. 24.69 26.08 29.95 33.84
103. Because paired ¢ = 3.88 and the two-tailed P-value is
.008, the difference is significant at the .05 and .01 17. (~.029, 379)
levels, but not at the .001 level. 19.436
105. Because z = 2.63 and the two-tailed P-value is .009.
a (00% 21. a. Because f = 22.60 > 3.26 = Forse, there are
there is s, sionificane ifteconce E the 01 level, sivnificant differences among the wean
suggesting better survival at the higher temperature. b. (299.1, 35.7), 29.4, 99.1)
107. .902, .826, .029, 00000003 23. The nonsignificant differences are indicated by the
109. Because z = 4.25 and the one-tailed P-value is .00001, lnderscares.
the difference is highly significant and companies 10 6 3 1
appear to discriminate. 45.5 50.85 55.40 58.28
UL. With Z=(X—Y)/V/X/n+Y¥/m, the result is |
z= —5.33, two-tailed P-value = 0000001, so one 25 @ Assume normality and equal variances.
should conclude that there is a significant difference in Baseball = Le Ls 2th F rosaaf velit — 18,
parameters, there are no significant differences among the means.
113. (i) not bioequivalent (ii) not bioequivalent (iii) 27+ @ Because f= 3.75, P-value = .028, there are
bioequivalent significant differences among the means.
b. Because the normal plot looks fairly straight and the
P-value for the Levene test is .68, there is no reason to
doubt the assumptions of normality and constant
Chapter 11 variance.
¢. The only significant pairwise difference is between
1. a, Reject Ho: fli = M2 = Hs = fa = Hs in favor of Hy: braids and
Hi Ha, Hs, Ha Ms not all the same, because 4 3 2 1
f= 5.57 2 2.69 = Foos.4.s0. 5.82 6.35 7.50 8.27
b. Using Table A.9, 001 < P-value < .01. (The P-value od :
is .0018)
31. 63
3. Because f= 643>2.95=Foss2, there are
significant differences among the means. 33. aresin( y/x/n)
5. Because f= 10.85 >4.38=Foi335, there are 35. a Because f= 1.55 < 3.26 = Fos4i2, there are no
significant differences among the means. significant differences among the means.
ip, Because f = 2.98 < 3.49 = Fos.3,12, there are no
Source DE ss MS F P significant differences among the means.
Formation 3 509.1 169.7 10.85 0.000 37, with = 5.49 > 4.56 = Foisis, there are significant
Error 36 563.1 15.6 differences among the stimulus means. Although not all
Total 39 1072.3 differences are significant in the multiple comparisons
analysis, the means for combined stimuli were higher.
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Chapter11 829
Differences among the subject means are not very SL. a. With f= 155 <281=F,o212, there is no
important here. The normal plot of residuals shows no significant interaction at the .10 level.
reason to doubt normality. However, the plot of residuals b. With f = 376.27 > 18.64 = Foo1212 there is a
against the fitted values shows some dependence of the significant difference between the formulation
variance on the mean. If logged response is used in place means at the .001 level.
of response, the plots look good and the F test result is With f= 19.27 > 12.97 = Foor there is a
similar but stronger. Furthermore, the logged response significant difference among the speed means at the
gives more significant differences in the multiple .001 level
comparisons analysis. ¢. Main effects Formulation: (1) 11.19, (2) -11.19
Speed: (60) 1.99, (70) ~5.03, (80) 3.04
Means:
_ DW oT we wer war 5S Beeiste ANOVA table
24.825 27.875 29.1 40.35 41.22 45.05 Source DE ss MSE P
Pen 3 1387.5 462.50 0.68 0.583
39. With f = 2.56 < 2.61 = F 103,12; there are no significant surface 2 2888.1 1444.04 2.11 0.164
differences among the angle means. Interaction 6 8100.3 1350.04 1.97 0.149
41. a. With f= 1.04 < 3.28 = Fos234, there are no Error 12 8216.0 684.67
significant differences among the treatment means. Total 23 20591.8
Source DE aS us F With f = 1.97 < 2.33 = F 106,12, there is no significant
Dp interaction at the .10 level.
Treatment 2 28.78 14.39 1.04 With f = .68 < 2.61 = F yo, there is no significant
Block 17 2977.67 175.16 12.68 difference among the pen means at the .10 level.
Error 34 469.56 13.81 With f = 2.11 < 2.81 = F102,12, there is no significant
Total 53 3476.00 difference among the surface means at the .10 level.
b. The very significant f for blocks, which shows that 57: @- F = MSAB/MSE .
blocks differ strongly, implies that blocking was b. A: F = MSA/MSAB__B: F = MSB/MSAB
suocesstul: 59. a, Because f= 343 > 2.61 =Fos.an there is a