Supplemental Topic 3 Solutions
Page 1 of 3
SUPPLEMENTAL TOPIC 3
EXERCISE SOLUTIONS
S3.1
yˆ  116 .84  0.832 (65)  0.951(30)  2.34(15)  1.08(50)  123 .49
yˆ  116 .84  0.832 (65)  0.951(50)  2.34(15)  1.08(30)  126 .07
yˆ  116 .84  0.832 (65)  0.951(50)  2.34(25)  1.08(50)  127 .87
yˆ  116 .84  0.832 (65)  0.951(50)  2.34(35)  1.08(70)  129 .67
S3.2
y i   0  1 xi1   2 xi 2   3 xi3   i , where we assume that  i are normally distributed with mean
0 and standard deviation . Response variable is y = freshman GPA. Explanatory variables are
x1 = total SAT, x 2 = high school GPA, x3 = high school rank.
S3.3
Answers will vary.
S3.4
Yes. This is more or less a horizontal band of points with no obvious pattern, as it should be if the
conditions are met. While there are one or two points that are separated from the others, they are
not obviously outliers because the residuals for the points are not clearly greater than for other
points.
S3.5
The observed data were for Indians who had migrated, so the results should not be “extrapolated”
to Indians who had not migrated.
S3.6
a. yˆ i  20.8  0.415 (49)  1.24(56)  69.905 .
b. e i  y i  yˆ i  64  69.905  5.905 .
S3.7
a. b0 = 20.2 (or 20.248) b1 =0.193 (or 0.19294) b2 =0.539 (or 0.53897).
b. yˆ i  20.248  0.19294 (65)  0.53897 (74)  72.67 inches. (answer will be 72.63 if fewer
decimal places used for regression coefficients).
c. e i  y i  yˆ i  71  72.67  1.67 inches (or 1.63 if answer for part (b) was 72.63).
S3.8
It corresponds to the average height of males students whose mothers and fathers have height = 0.
This is impossible.
S3.9
0.539 inches (or 0.53897). This is the value of b2 .
S3.10
a. 0.031. We can reject then null hypothesis, and can conclude that the student height and
mother’s height are related variables.
0.19294
 2.20 .
b. t 
0.08777
S3.11
a. H0: 1   2  0 versus Ha: At least one of 1 and  2 is  0.
b. Reject the null hypothesis (F = 33.86, p-value = 0.000). Conclude that student height is related
to at least one of the parents’ heights.
S3.12
a. s = 2.591inches. Roughly, the average absolute size of the deviations between actual and
predicted student height is about 2.591 inches.
b. R2 = 48.1%. Mother’s and father’s height explain 48.1% of the variation among students’
heights.
Supplemental Topic 3 Solutions
S3.13
S3.14
Page 2 of 3
a. 69.976 to 71.365 inches. With 95% confidence, we estimate that the mean height is between
69.976 and 71.365 inches for male students with mother’s height = 63 inches and father’s height =
71 inches.
b. 65.460 to 75.881 inches. About 95% of students whose mother’s and father’s heights are 63
and 71 inches, respectively, are between 65.460 and 75.881 inches tall.
c. The “Fit” is the predicted height for a student with mother’s height= 63 and father’s height =
71. yˆ i  20.248  0.19294 (63)  0.53897 (71)  70.67 inches.
There’s an outlier - the point with residual  10. Otherwise, there are no violations of the
necessary conditions.
S3.15
a. H0: 1   2  0 versus Ha: At least one of 1 and  2 is  0.
b. Reject the null hypothesis (F = 15.19, p-value = 0.000). Conclude that left arm length is related
to at least one of left foot length and right foot length.
S3.16
a. In each case, cannot reject the null hypothesis. For left foot length, p-value = 0.240. For right
foot length, p-value = 0.514.
b. The tests in part (a) should be interpreted to mean that a particular foot length is not needed in
the equation given that the other foot length remains in the equation. One, but not both, of the foot
lengths will be a useful predictor of arm length. This occurs because the foot lengths are highly
correlated and, in essence, give the same information so it’s redundant to use both to predict arm
length.
S3.17
a. s = 1.796 cm. Roughly, the average absolute size of the deviations between actual and
predicted arm lengths is about 1.796 cm.
b. R2 = 36.9%. The foot length variables explain 36.9% of the variation among arm lengths.
S3.18
a. b0  119 .58 , b1  0.9636 , b2  1.0903 , b3  2.6295 , b4  1.2080 , b5  0.6413
b. There is not a statistically significant relationship between blood pressure and the calf skinfold
measurement, given that the other variables are present in the equation. The p-value is 0.120.
0.6413
 1.59 .
c. t 
0.4022
S3.19
In a multiple regression involving correlated explanatory variables, the individual regression
coefficients can’t be interpreted in the usual way that a slope coefficient is interpreted. More
specifically, the interpretation can’t be isolated from the other coefficients. Another possible
answer is that the relationship is not statistically significant (see exercise S3.18) so it may be that,
in the population, average blood pressure does not change when skinfold is changed.
S3.20
a. R2 = 1.5%. The three explanatory variables explain 1.5% of the variation among grade point
averages.
b. The p-value = 0.512 for the F-statistic in the analysis of variance. This supports the null
hypothesis that 1   2   2  0 , which means that the response variable is not related to any of
the explanatory variables.
S3.21
a. y i   0  1 xi1   2 xi 2   i , where we assume that  i are normally distributed with mean 0 and
standard deviation .
b. b2 represents the coefficient that multiplies x 2 in the equation for the sample. Its value can be
determined.  2 represents the coefficient that multiplies x 2 in the equation for the population. Its
value will not be known, but it will be estimated by b2 .
S3.22
A predicted value ŷ estimates the mean value of y for a population of individuals with the same
particular values of the explanatory variables, and it estimates the value of y for an individual with
those particular values of the explanatory variables.
Supplemental Topic 3 Solutions
S3.23
Page 3 of 3
a. The plots are shown below. There is an outlier in the plot of Weight versus Neck. Both plots
have more or less a linear pattern.
Figure for Exercise S3.23a
b. Regression equation is
Weight =  226.07 + 8.758 Neck + 6.3386 Chest
c. Both explanatory variables are statistically significant predictors. For Neck, p-value = 0.034,
and for Chest, p-value = 0.005. The Minitab output follows.
Output for Exercise S3.23c
Predictor
Constant
Neck
Chest
Coef
-226.07
8.758
6.386
SE Coef
20.61
3.745
1.949
T
-10.97
2.34
3.28
P
0.000
0.034
0.005
d. R2 = 96.4%. Neck and chest measurements explain 96.4 of the variation among weights.