Chapter 10
Homework Set A
10.6 What's wrong? For each of the following, explain what is wrong and why.
(a) The slope describes the change in x for a change in y.
I would not lose sleep over this question.
(b) The population regression line is y = b0 + b1 x.
The population regression line is written μ yˆ =β1x+β 0
(c) A 95% confidence interval for the mean response is the same width regardless of x.
Not so, if you look at the formula for the standard error of ŷ (which estimates μ ŷ )
2
s2 
 S.E.b1  x*  x  
n
It indicates that as x* gets further away from x-bar, the S.E.b1 increases.
Thus the width of the margin of error changes depending on the value of x*.
10.7 What's wrong? For each of the following, explain what is wrong and why.
(a) The parameters of the simple linear regression model are bo, b1, and s.
(b) To test Ho: b1 = 0, use a t test.
10.8 95% confidence intervals for the slope. Find a 95% confidence interval for the slope in each of the
following settings:
(a) n = 25, ŷ = 1.3 + 12.10x, and SEb1 = 6.31
Degrees of freedom 23. Using =tinv(0.05,23) we get t* = 2.069
12.10 ± 2.069(6.31)
(-0.96, 25.16) This indicates that we cannot rule out zero as the possible slope value.
(b) n = 25, ŷ = 13.0 + 6.10x, and SEb1 = 6.31
Degrees of freedom 23. Using =tinv(0.05,23) we get t* = 2.069
6.10 
This indicates that we cannot rule out zero as the possible slope value.
(c) n = 100, ŷ = 1.3 + 12.10x, and SEb1 = 6.31
Degrees of freedom 98. Using =tinv(0.05, 98) we get t* = 1.984
12.10 
This indicates that we cannot rule out zero as the possible slope value.
10.9 Significance test for the slope. For each of the settings in the previous exercise, test null hypothesis that
the slope is zero versus the two-sided alternative.
a. n = 25, ŷ = 1.3 + 12.10x, and SEb1 = 6.31
Ho:  1 = 0 Ha:  1 ≠ 0
t=
12.10 - 0
6.31
= 1.92
2P(t > 1.92) = 0.0673 degrees of freedom is 23. I used tdist(1.92,23,2).
b. n = 25, ŷ = 13.0 + 6.10x, and SEb1 = 6.31
Ho:  1 = 0 Ha:  1 ≠ 0
t=
6.10 - 0
6.31
= 0.97
2P(t > 0.97) = 0.3421 degrees of freedom is 23. I used tdist(0.97,23,2).
c. n = 100, ŷ = 1.3 + 12.10x, and SEb1 = 6.31
Ho:  1 = 0 Ha:  1 ≠ 0
t=
12.1 - 0
6.31
= 1.92
2P(t > 1.92) = 0.0578 degrees of freedom is 98. I used tdist(1.92, 98 ,2).
1. Is there a relationship between Verbal Memory Retention, and the right Hippocampal Volume on the brain?
Parameter
Intercept
Slope
Estimate Std. Err.
-29.653652 30.72434
0.0813201 0.02598
DF
19
19
Sample size: 21
R (correlation coefficient) = 0.5833
R-sq = 0.34022236
Estimate of error standard deviation (s) : 17.607368
x = 1173.33
The hippocampus is located in the
medial temporal lobe of the brain.
The location of one of the hippocampi
is indicated by the crosshairs.
a. Does the normal quantile plot of the residuals indicate that the model involving the infinite number of groups have a
normal distribution?
The points do line up in a relatively straight line, which supports
the idea that the distribution of Verbal Memory Retention is
Normal.
b. Answer the question posed “Is there a relationship between Verbal Memory Retention, and the right Hippocampal
Volume on the brain?” Use the volume as the x-variable. State the null and alternative hypothesis.
H0 :  1 = 0 Ha:  1 > 0
0.08132
= 3.13
0.02598
Test Statistic: t =
P(b1 > 0.08132) = P(t > 3.13)
= 0.002757
c. Create a 95% confidence interval for the slope.
0.08132 ± 2.0930(0.02598)
(0.02694, 0.1357)
d. What is the predicted verbal memory retention for a volume of 1200?
1200(0.0813201) – 29.654 = 67.93
e. I would like to create a 95% confidence interval to estimate the true mean verbal memory retention for a volume of
1200?
ŷ  t
*
2
s2 
*

 S.E.b1  x  x  
n 
67.93 ± 2.0930
where x* = 1200. t* = 2.0930
17.612
2
  0.02598(1200  1117.33)  = (58.75, 77.11) We are 95% certain that the population
21
mean verbal memory retention for a volume of 1200 is found in the given interval.