Statistics II
4 December 2015
Exercise 1 (7 points)
From a very large university, independent random samples of 120 students majoring in marketing and 90
students majoring in finance were selected. The mean grade point average (GPA) for the random sample of
marketing majors was found to be 3.08, and the mean GPA for the random sample of finance majors was
2.88. From similar past studies the population standard deviation for the marketing majors is assumed to be
0.42; similarly, the population standard deviation for the finance majors is 0.64. Denoting the population
mean for marketing majors by μX and the population mean for finance majors by μY, find a 95% confidence
interval for (μM– μF).
(From: Newbold, P., W. Carlson and B. Thorne. Statistics for Business and Economics. Pearson-Prentice Hall.)
Discussion: Since both sample sizes are large, the central limit theorem lets us know the asymptotic
distribution of the statistic to work with the means of two populations. The independence of the two
populations should be tested. The method of the pivot will be applied. Trustable estimates of the population
variances are available, which can be treated as the real population variances (though we will use a notation to
make it clear that they are estimates).
(a) Confidence interval
The variables are
M
≡
Grade of a student majoring in marketing (any student)
M∼N(μM,^
σM
2=0.422)
F
≡
Grade of a student majoring in finance (any student)
F∼N(μF,^
σF
2=0.642)
(a1) Pivot: We know that
•There are two independent populations
•We are interested in μM – μF
•Variances are known (trustable estimates of them, really)
Then, from a table of statistics (e.g. in [T]), we select
T(M,F;μM,μF)= (¯
M−¯
F)−(μM−μF)
√
^σM
2
nM
+^σF
2
nF
→
dN(0,1)
(a2) Event rewriting
1−α= P(lα / 2≤T(M,F;μMμF) ≤rα /2)=P
(
−rα/ 2≤(¯
M−¯
F)−(μM−μF)
√
^
σM
2
nM
+^
σF
2
nF
≤+rα / 2
)
=P
(
−rα/ 2
√
^
σM
2
nM
+^
σF
2
nF
≤ ( ¯
M−¯
F)−(μM−μF) ≤+rα/ 2
√
^
σM
2
nM
+^
σF
2
nF
)
Bachelor's Degree in Economics 1 Complutense University of Madrid
