Download Final Exam, Version 1, Solutions

SOLUTIONS TO FINAL EXAM
VERSION 1
1)
A) The plot shows an approximately linear relationship with no extreme
outliers, and reasonably constant variability. So based on the scatterplot
there is no apparent problem with fitting the simple linear regression
model to this data.
B) We cannot say exactly what will happen to log sales, but we estimate that
the expected value of log sales will increase by .16578.
C) Since the estimated coefficient of Housing Starts is positive, we can
calculate the right-tailed p-value as .004/2=.002. Since this p-value is less
than .003, there is indeed evidence of a positive linear relationship at the
.003 level of significance.
2)
A) The CI is ˆ1  t.025SE( ˆ1 ) . We have DF=22, so t.025  2.074, based on the
t-table. The CI is therefore .16578  (2.074)(.05147)  (.0590,.2725).
B) Such an interval would contain the true slope coefficient in 95% of all
samples which could be collected.
3) The left-tailed p-value is .252/2=.126, which is not less than 5%, so there is no
significant relationship between Mortgage Rate and log Sales, based on the
multiple regression output.
4) No. With Housing Starts alone, the R 2 was 32%, in the simple linear
regression. When we add in the Mortgage Rate, the multiple R 2 only goes up
to 36.3%, and the coefficient of Mortgage Rate is not statistically significant.
5) The p-value for the F-statistic is .009, which is the probability that an Fdistributed random variable with 2 and 21 degrees of freedom exceeds 5.97, so
the probability that it is less than 5.97 is .991.
6) The events {Straight} and {Flush} are not mutually exclusive, since a hand
can be simultaneously a straight and a flush. They are not complementary
events, since knowing that a hand is not a straight does not imply that it is a
flush. To see if these events are independent, we need to do some calculations.
First, get Prob{Straight}. Since there are 9 possibilities for the denomination
of the low card, and then 4 possible suits for each of the cards, we get
 52 
Pr ob{Straight}  (9)(4 5 ) /   =9216/2598960=.003546.
5
In class, we calculated Prob{Flush}, as
13  52 
Prob{Flush}= ( 4)   /   =.001981.
5  5 
Now, let's get Prob{Straight ∩ Flush}. If a hand is a straight and a flush (this
is called a straight-flush), once we know the denomination and suit of the
lowest card we know everything. The low card can have any of 9
denominations, and any of 4 suits, so there are 36 straight flushes, and we
 52 
get Pr ob{Straight  Flush}  36 /   =.00001385.
5
To see if the two events are independent, we now compute
Pr ob{Straight | Flush}  Pr ob{Straight  Flush} / Pr ob{Flush}  .00699
which is not the same as Prob{Straight}. So the events are not independent.
Answer is D.
7) Since the p-value is less then .01, it is also less than .05, so we can reject the
null hypothesis at level .05. The statement is true. Answer is A.
8) Under the null hypothesis, the t-statistic (since n is large) has a standard
normal distribution, so Prob{t>1.7}=.5.4554=.0446. Answer is C.
9) The t-statistic is t 
3.5  4
 1.77. The p-value is the probability that a
2 / 50
standard normal is less than 1.77, which is p=.5.4616=.0384. Answer is D.
10) Since x is normal with mean 0 and standard error 1 / n , we conclude that
n x has a standard normal distribution. Now, from the Normal chart, we have
.1587=.5.3413=Prob{Standard Normal > 1}. We want to find n such that
.1587=Prob{ x  15 / n} =Prob{ n x  15 / n }. So we set 15 / n  1 , so
15  n and n  15 2  225. Answer is D.
11) Assume that the null hypothesis is true. The significance level is Prob{Reject
the Null Hypothesis}=Prob{First p-value is less than .05 or Second p-value is
less than .05}. Since the two samples are independent of each other, so are the
two p-values, so the significance level becomes Prob{(First p-value is less
than .05)  (Second p-value is less than .05)}= .05 + .05  (.05)(.05) = .0975.
Answer is D.
12) The errors are independent and normally distributed with mean zero and
standard deviation 2. The probability that a particular error exceeds 1.2 is
Prob{Error>1.2}=Prob{Standard Normal>1.2/2}=Prob{Standard
Normal>.6}=.2743. The total number of errors that exceed 1.2 (of the 10
errors) has a binomial distribution with n  10 and p  .2743 . Thus, the
answer is np=2.743. Answer is B.
13) Since R 2  SSR / SST is less than 1, we must have SSR < SST. Answer is D.
14) Do a hypothesis test on the binomial proportion p, where p=Prob{Cliff
distinguishes correctly on a given trial}. We test H 0 : p  1 / 2 versus
69 / 93  .5
4.67.
H A : p  1 / 2 . Since p0  1 / 2 , the test statistic is Z 
(.5)(1  .5) / 93
From the t-table (last row) we can see that z.001 =3.090. Since we have
Z>3.090, Cliff's ability is statistically significant at a level that is smaller than
.001. Answer is E.
15) The two-tailed p-value is .478, which is greater than .05, so the difference
between Cliff's and Harjaap's abilities is not statistically significant at level
.05 (or at any smaller level, for that matter). Answer is D.