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Chapter 12 831
d. p, = 1.683, and a 95% CI is (531, 2.835). 59 a. For the test of Ho: p =0 vs. Hy: p > 0, we find
Eliminating the point causes only moderate change, r = .760, t = 4.05, with P-value < .001. At the .001
so the point is not extremely influential. level conclude that there is positive correlation.
_ b. Because? = .578 we say that the regression accounts
37, a, Yes; forthe test of Ha: By= 0 vs: Ha: B. # 0, we find for 57.8 % of the variation in endurance. This also
= —6.73, with P-value .00002. At the .01 level applies to prediction of lactate level from endurance.
conclude that there is a useful linear relationship.
b. (—2.77,-1.42) 61. For the test of Ho: p = Ovs. Hy: p #0, we find r = .773,
1 = 2.44, with P-value .072. At the .05 level conclude
M3: Nosece=: 7S -and the F-value'is 46; so there:isno evidence that there is not a significant correlation, With such a
for a significant impact of age on kyphosis. small sample size, a high r is needed for significance.
45. a. sy increases as the distance of x from x increases 63. a. Reject the null hypothesis in favor of the alternative.
b. (2.26, 3.19) b. No, with a large sample size a small r can be
eet 34 ATT) significant.
d. At least 90% ¢. Because t = 2.200 > 1.96 = to25,900g the correlation
47. a, The regression equation is y= ~1.58 +2.59x and ig ‘Statistically ((but. not inecessanly practically)
R= 838. significant at the .05 level.
b. A 95% confidence interval for the slope is (2.16, 67, a, 184,238, 426
3.01). In repetitions of the whole process of data b. The mean that is subtracted is not the mean ¥,,_; of
collection and calculation of the interval, roughly js Sy ans a, esthesmectens OF 26a, no he
95% of the intervals will contain the true slope. ‘Also the” denominaior” Of ry Ge Hoe
¢. When tannin = .6 the estimated mean astringency is (S50, nt, Sea
—0.0335 and the 95% confidence interval is (-0.125, VOT (i — Bn) 03 (41 — Fan)”. However, if
0.058) nis large then r is approximately the same as the
d. When tannin =.6 the predicted astringency is correlation. A similar relationship applies to r2.
—0,0335 and the 95% prediction interval is c. No
(0.5582, 0.4912) d. After performing one test at the .05 level, doing more
e, Our null hypothesis is that true average astringency tests raises the probability of at least one type I error to
is 0 when tannin is .7, and the alternative is that the more than .05.
true average is positive. The f for this test is 4.61, with
P-value = 000035, so yes there is compelling 69. The plot shows no reasons for concern about using the
evidences, simple linear regression model.
49. (431.2, 628.6) 71. a. The simple linear regression model may not be a
perfect fit because the plot shows some curvature,
51. a. Yes, for the test of Ho: B; =0 vs. Ha: Bi #0, b. The plot of standardized residuals is very similar to
we find ¢= 10.62, with P-value .000014. At the the residual plot. The normal probability plot gives
.001 level conclude that there is a useful linear no reason to doubt normality.
relationship.
b. (8.24, 12.96) With 95% confidence, when the flow 73: a. For the test of Ho: f, = 0 vs. Hy: Bi: #0, we find
rate is increased by 1 SCCM, the associated expected 1 = 10.97, with P-value .0004. At the .001 level
change invetch rate is in'the interval, conclude that there is a useful linear relationship.
€. (36.10, 40.41) This is fairly precise. b. The residual plot shows curvature, so the linear
d. (31.86, 44.65) This is much less precise than the relationship of part (a) is questionable.
imervalin (6) ¢. There are no extreme standardized residuals , and the
e. Because 2.5 is closer to the mean, the intervals will be plot of standardized residuals is similar to the plot of
narrower. ordinary residuals.
f Because 6 is outside the range of the data, it is 75. The first data set seems appropriate for a straight-line
unknown whether the regression will apply there. snodsl,, “The secoed. dath det shows i quadiatis
g. Use a 99% CI at each value: (23.88, 31.43), (29.93, gelationihip, 50 the sttuighidine relationship. is
35.98), O20 A145) inappropriate. The third data set is linear except for an
53, a. Yes outlier, and removal of the outlier will allow a line to be
b. Yes, for the test of Ho: By = Ovs. Ha By # 0, we find fit. The fourth data set has only two values of x, so there is
1 = —4.39, with P-value < .001. At the .001 level no way to tell if the relationship is linear.
conclude that there is a useful linear relationship. 77. a. To test for lack of fit, we find f= 3.30, with 3
&.(403.6, 468.2) numerator df and 10 denominator df, so the P-value
87. a. r= .923, sox and y are strongly correlated. is .079. At the .05 level we cannot conclude that the
by uadetedted relationship is poor.
c. unaffected b. The scatter plot shows that the relationship is not
d. The normal plots seem consistent with normality, but linear, in spite of (a). In this case, the plot is more
the scatter plot shows a slight curvature. sensitive than the test.
e. For the test of Ho: p =0 vs. Ha: p #0, we find 79 77.3
1= 7.59, with P-value .00002. At the .001 level b404
conclude that there is a useful linear relationship. ©. ‘The cvetictenty 8 thataleneretice ft aalas eaudéa by
the window, all other things being equal.