MCQ.
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15
C A C D B B B D C C E D A C B
Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30
D E D C D E A B C B E D C E B
LQ1.
(a) = √2 = 0.5245.
(b) 0: = 0, : ≠ 0 where is the population correlation between PPM and price. The test
statistic is = √−2
√1−2
= 2.889 > 22,0.005∗ = 2.819
(c) Reject 0 and conclude there is a statistically significant and positive correlation between
printer speed and price.
(d) On average, each extra PPM increase in printer speed raises printer price by \$6.08.
(e) 0: = 0 , : ≠ 0 where is the population slope. The test statistic is = 2.889 >
22,0.005∗ = 2.819. Thus we reject 0 at the 1% level.
(f) The purpose is to construct a confidence interval for the regression slope, which can also be
used to test a hypothesis about the slope.
(g) Given 1,000 samples, 1% is 10 samples, so for a 99% CI we count 5 observations in from the left
and right tails of the distribution, giving a CI of approximately (2.1, 10.6).
(h) They are all testing the same thing: the statistical significance of the association between PPM
and price.
(i) The study appears reasonable but there might be omitted variables, such as quality (printer
resolution). The sample size is also rather small ( < 30).
LQ2.
(a) is a binomial random variable with = 50 and = 0.06.
(b) = = 3
(c) ( = 3) = 50!
3!47! 0.0630.9447 = 0.231.
(d) ( > 3) = 1 − ( = 0) − ( = 1) − ( = 2) − ( = 3) = 1 − 0.04533 − 0.14467 −0.22624 − 0.23105 = 0.353
(e) The mean of the treatment group will be lower if the program reduces the length of
unemployment spells. Thus we have 0: − = 0 , : − > 0 where is the
(population) control group mean and (population) treatment group mean.
(f) Under 0 the test statistic for a two-sample difference in means
= ̅ − ̅ − 0

2

+ 2 =
81.143 − 77.143
�5.21027 + 7.38127 = 1.17
has an approximate t-distribution with 6 degrees of freedom. Since = 1.17 < 6,0.05∗ = 1.943
we do not reject the null at the 5% level and conclude that the program is not effective.
(g) Let be the population paired difference (control minus treatment) in unemployment spells
for the two groups. Shorter spells due to the program may be represented by the hypotheses
0: = 0 ,: > 0.
(h) Under 0 the test statistic for a paired difference in means is = ̅−0/√ = 42.646/√7 = 4.00 >
6,0.05∗ = 1.943, so we reject the null at the 5% level and conclude that the program reduces
unemployment spells.
(i) The paired difference test reduces the influence of confounding variables. Hence it reduces the
, implying the t statistic is larger (for a given difference) and increasing the possibility that the
hypothesis is rejected.