# Chapter 2 solutions

```Stat 250.3
December 10, 2003
HOMEWORK 7 – SOLUTIONS
14.10
a. This is given by the sample slope b1 = 0.269. (Or, with more decimal places, b1= 0.26917.)
b. Approximately b1  2 s.e. (b1 ) . This is 0.26917  (2  0.6616 ) or about 0.401 to 0.137. With 95% confidence,
we can say that the slope of the regression line in the population is between 0.401 and 0.137. Note: The exact
multiplier in this situation is t *  1.981 which can be found by using software like Minitab or Excel to find the value
of t * for which the probability is 0.95 in a t-distribution that has df = n2 = 1162=114.
c. H0: 1 = 0 versus Ha: 1  0.
b1
 0.26917
d. t 

 4.07 , df = n  2 = 116  2 = 114.
s.e.(b1 )
0.06616
e. The relationship is statistically significant. The p-value is 0.000, given in the output under “P” in the row labeled
Study.
14.11
a. s =1.509 hours. This is roughly the average deviation of individual y-values from the regression line.
b. yˆ  7.56  0.269 (4)  6.48 hours.
c. 6.48  (2 1.509 ) which is 6.48  3.018, or about 5.46 hour to 9.50 hours. Remember that about 95% of individuals
will be within two standard deviations of the mean.
d. Hours of study explains 12.7% of the observed variation in hours of sleep.
14.15
a. The notation is b1 because 7 is the slope for the sample regression line.
b. The approximate 95% C.I. is 7  (21.581) which is 7  3.162, or about 3.84 to 10.16. With 95% confidence, we
can say that in the population of college men, the mean increase in weight per 1-inch increase in height is somewhere
between about 3.84 and 10.16 pounds.
c. The confidence interval for the slope contains 5, so the statement about the population slope is reasonable.
d. The intercept is the mean value of weight when height is 0. This is meaningless because height=0 is impossible.
14.40
a. yˆ  29.981  0.57568 (65)  67.4 inches. With the more rounded off version of the equation given in the second line
of output, yˆ  30  0.576 (65)  67 .44 which is about 67.4 inches.
b. The interval is 64.946 inches to 75.612 inches. The two interpretations are: (1) In the population of college males
for whom father’s height is 70 inches, about 95% have a height in the given interval. (2) For a randomly selected
college male whose father is 70 inches tall, the probability is 0.95 that his height is in the given interval.
c. For the population of male college students whose fathers are 74 inches tall, we can say with 95% confidence that
the average height is between 71.596 and 73.566 inches.
d. The prediction intervals are wider because they include variation among the individual heights of the men whose
fathers are all the same height. The confidence intervals are more narrow because they are estimating the center
(mean) of the population of students heights at each value of father's height. The confidence intervals do not
incorporate variability among individuals.
e. Condition 2 appears to be violated. There are two outliers that may be unduly influencing the results. The two
outliers correspond to father/son heights of 66&quot;/57&quot; and 55&quot;/65&quot;. In both cases, one of the heights is less than 5 feet,
which is unlikely to be correct. The original surveys should be checked for accuracy. In the case of dad's height = 55&quot;,
the point could be dropped as long as the intention is to make inferences only for students whose fathers are within the
usual range, perhaps 60&quot; to 78&quot; or so. It is not as easy to justify dropping the case with son's height = 57&quot; because the
range of inference cannot be restricted for the y-variable.
```
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