SOLUTIONS TO FINAL EXAM
VERSION 2
1)
A) The intercept of the fitted model, ˆ  .502 , can be interpreted as the estimated
expected value of the number of months ahead for the adjusted score, given an
unadjusted score of zero (at the national average).
B) The right-tailed p-value corresponding to the estimated slope is .042/2 = .021, so
the estimated slope is significantly positive at the 5% level of significance, but not
at the 1% level of significance.
C) We first calculate the fitted value for California, yˆ  .502  .746( −6.4) = −4.2724.
Next, we calculate the residual for California,
ei  yi  yˆ i  6.3  (4.2724)  2.0276 , which is negative, so the data point
lies below the fitted line.
D) We can test the null hypothesis that the true intercept is 0 using the p-value
corresponding to the constant term, p=.620. Since this p-value is not small, we do
not reject the null hypothesis that the true intercept is 0.
2)
A) We need to compute the t-statistic by hand, t = (.746−1)/(.309) = −.822. We are
doing a two-tailed test at the 5% level of significance. Since we have 10−2=8
degrees of freedom, we would reject the null hypothesis if |t| > t.025 , where from
Table 6 we obtain t.025  2.306. Since |t| does not exceed 2.306 we do not reject
the null hypothesis.
B) The confidence interval for β is ˆ  t.025 SE  .746  2.306(.309)  (.0334,1.459).
3)
A) No, based on the multiple regression Population is not statistically significant and
for the simple regression of Adjusted vs. Population we have R2=.042, which is
low.
B) The R2 went up from 42.13% to 46.86% when Population was added as an
explanatory variable. This is not a very large increase, and we know that R2 always
increases when a new variable is introduced. Meanwhile, The F-statistic is not
statistically significant in the multiple regression model. The p-value for the F-test
in the multiple linear regression model is .109, so that the model as a whole is not
statistically significant even at the 10% level, whereas in the simple linear
regression model it was significant at the 5% level. So in this sense the introduction
of Population as an explanatory variable weakened the overall regression model.
C) For the simple linear regression model (n=10, k=1) we obtain
AICC=log(75.92)+2(1+2)/(10−1−3)=5.3297. For the multiple linear regression
model (n=10, k=2) we obtain AICC=log(69.712)+2(2+2)/(10−2−3)=5.844. So the
preferred model is the simple linear regression model, since it has the smallest value
of AICC.
4) Figure 3 perhaps shows an overall increasing pattern, except for the two points on
the right, which correspond to Ohio and Pennsylvania. But these two residuals are not
extremely far below zero, since we have s = 3.16, so they are only about 1 standard
deviation below zero.
5) The expected value of the sample mean equals the population mean, μ=$2,749. This
is due to the equation from sampling theory,  x   , in other words, E[ X ]  .
There is no requirement that the population distribution be symmetric. We even saw
in the sampling applet that I demonstrated in class that when the population
distribution is skewed, it still is the case that E[ X ]  .
6) No, it’s not possible. Since ˆ  0 , it is not possible for all of the fitted values ŷi to be
the same. Therefore, SSR  i 1 ( yˆ i  y ) 2  0. It follows that R 2  SSR / SST  0.
n
7) Consider the data point ( xi , yi ). Moving this data point one unit to the right
corresponds to increasing the x-coordinate by one, so that ( xi , yi ) is changed
to ( xi  1, yi ) . So in the new data set (with the one data point changed) none of the yvalues have changed from their original values. Therefore, y , the average of the yvalues, remains unchanged, and finally SST  i 1 ( yi  y ) 2 remains unchanged. The
n
change in SST is zero.
8) No, it is not possible. Since R2=0, it follows that ˆ  0. Therefore, the corresponding
t-statistic is t  ˆ / SE ( ˆ )  0. But since the t-statistic is zero its absolute value
cannot be greater than the critical value of t.025 , which is approximately 1.96. Indeed,
the p-value corresponding to ˆ is 1, which is not less than .05. So ˆ is not
statistically significantly different from zero at the 5% level of significance.
9) The standard deviation of a discrete random variable X is the square root of the
variance, Var ( X )   ( x   ) 2 p( x) where the x denote the possible values,
all x
p( x)  Pr ob{ X  x} and   E[ X ]. If X could take on two or more possible values,
then because of the square in the formula above we would have Var ( X )  0. Since
the question states that X has a standard deviation of zero, we have Var ( X )  0 and
therefore X can take on just one possible value, and the corresponding value of p(x)
must be 1. (In other words, X is a constant.) For example, we could have X=5 with
probability 1. In this case, we would have E[X]= 5 Prob{X=5} = 5. So the mean of X
is not necessarily zero.
10) Since the sample size is large and the alternative hypothesis is left-tailed, we assume
that the p-value is the tail area under a standard normal distribution to the left of the
observed t-statistic. It is also clear that the observed t-statistic must be negative, since
the p-value is less than .5. From Table 5, we see that the area under a standard normal
distribution to the left of −2.51 is .5−.4940 = .006. Thus, t=−2.51. We therefore have
x 1
so that x  2.51(3.2 / 250 )  1  .4920.
t  2.51 
3.2 / 250