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(ISOM111)[2002](f)final~2106^_10360.pdf
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ISMT111 Business Statistics
2002 Final Examination

Suggested solution
Question 1: [14 Marks]
(a) Let 1 denote mean salary for females and 2 denote mean salary for males
H0: 1-2 >=0 vs Ha: 1-2<0 Using a two sample t-test and compute the pooled variance estimator-S2p = (n1 - 1)S12 + (n2 - 1)S22/(n1 + n2 - 2) = {17(8000)2+11(11000)2}/28 Sp =9294.78. The observed test statistic is
x1 . x2 43000 . 55000
tobs =
= =.3.46
11 11

Sp + 9294.78 +
n1 n2 18 12

With 12+18-2=28 degrees of freedom and = .05, rejection region: t < -1.701.
As tobs falls in the rejection region, reject H0 at 5% significance level. The monthly
salary for females is lower that for males.

(b)
The two assumptions are: the salaries of males and females follow a normal distribution and the variances for both populations are equal.

(c)
Assuming that yearly salary equates to 12(monthly salary) we can convert this question into one about monthly salary. That is we can test the hypothesis that the average monthly salary is 600,000/12=50,000 H0: 2=50000 vs Ha: 2 50000


With 12 C 1 = 11 degrees of freedom 11, and = 0.10, rejection regions: t < -1.796 or t > 1.796 The observed test statistic is
x2 . 50000 55000 . 50000
== = 1.575
tobs
S2 11000

n2
12
As tobs does not fall in the rejection regions, do not reject H0 at 10% significance level. The average yearly salary for males is not different from $600,000.
(d) Given the true standard deviation =$8000. We need to calculate the sample mean
of the combined sample of n=30. The sample mean based on the combined data is nx + nx 18(43000) +12(55000)
11 22
x == = 478000
n1 + n2 18 +12 Use the Normal table to get Z0.025=1.96 A 95% C.I. for the mean monthly salary is 8000
478000 1.96
= [44937.24, 50662.76]
30

Question 2: [13 Marks]
(a) Test H0: p1-p2=0 vs Ha: p1-p2 0
x + x 46 +13
Common p. = 12 == 0.367 n1 + n2 109 + 52 The observed test statistic p. . p. 0.422 . 0.25
Zobs =
12 == 2.123 11 11
p.q. + 0.367(0.633) +
n1 n2 109 52 With = 0.10, rejection regions: Z < -1.645 or Z > 1.645 As Zobs falls in the upper rejection region, reject H0 at 10% significance level. There is a difference between males and females preferences.
(b)
Test H0: p1-p2=0 vs Ha: p1-p2 > 0 p-value =P(Z>Zobs)=P(Z>2.123)=1-.9830=.017

(c)
Based on the previous sample we can use p2 =0.25, B=0.1 and Z.025=1.96. The required sample size is n>= 72.03=73.



Question 3: [13 Marks]
(a) Let p1 be the proportion of students who think that the verdict is too lenient. Similarly, p2 is the proportion of the general public. We want to perform the test
H: p . p = 0 versus H: p . p < 0 . The observed test statistic is
01 2 a 12
p.1 . p.2 46/70 . 0.687
zobs =
= =.0.498, implying the 11 11
p.(1. p.)( + ) 0.683(1. 0.683)( + )
n1 n2 434 70 p-value of 0.31. Therefore, we do not have sufficient evidence (e.g. at = 10%) to say that the students are more forgiv