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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. |
--- Trang 842 --- |
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 |
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