CC0294 Semester 1 – 2016 Answers
MCQ.
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15
C E A D B C C B B A E C E D B
Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30
E B A B E C C C E B D E D D A
LQ1.
(a) = √2 = 0.9877.
(b) 0: = 0 , : ≠ 0 where is the population correlation between vehicles and life
expectancy. The test statistic is = √−2
√1−2
= 38.98 > 38,0.005∗ = 2.712
(c) Reject 0 and conclude there is a statistically significant positive correlation between vehicles
produced and life expectancy.
(d) Each extra 1,000 vehicles produced over the period 1970-2009 is associated with an increase in
life expectancy of 0.05015 years, roughly 18 days.
(e) 0: = 0 , : ≠ 0 where is the population slope. The test statistic is = 38.99 >
38,0.005∗ = 2.712. Thus we reject 0at 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 (0.046,0.054).
(h) They are all testing the same thing: the association between vehicle production and life
expectancy.
(i) A classic case of confounding variables. Regression is valid only as a description of the data;
nothing can be said about causation between vehicle production and life expectancy.
LQ2.
(a) is a binomial random variable with = 10 and = 0.25.
(b) = = 2.5
(c) Using the binomial probability formula ( ≥ 1) = 1 − ( = 0) = 1 − 0.7510 = 1 −0.0563 = 0.9436 or just over 94%.
(d) ( ≥ 5) = 1 − ( < 5) = 1 − ( = 0) − ( = 1) − ( = 2) − ( = 3) −
( = 4) = 1 − 0.0563 − 0.1877 − 0.2816 − 0.2503 − 0.1460 = 0.0781
(e) 0: 2 − 1 = 0 , : 2 − 1 > 0 where 1is the class (population) mean on the first quiz and
2 the class mean on the second quiz.
(f) Under 0 the test statistic for a two-sample difference in means
= ̅2 − ̅1 − 0
�1
2
1
+ 222 =
8.3 − 7.7
�1.3375210 + 0.8233210 = 1.21
has an approximate t-distribution with 9 degrees of freedom. Since = 1.21 < 9,0.05∗ = 1.83
we do not reject the null at the 5% level and conclude that there has been no improvement.
(g) Let be the population paired difference in marks for students on the first and second quizzes.
An improvement in marks may be represented by the hypotheses 0: = 0 ,: > 0.
(h) Under 0 the test statistic for a paired difference in means is = ̅−0/√ = 0.60.8433/10 = 2.25 >
9,0.05∗ = 1.83, so we reject the null at the 5% level and conclude that there has been an
improvement.
(i) The paired difference test reduces the presence 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.