Answer to Question #224123 in Statistics and Probability for Carrie

Question

A financial analyst wished to establish whether the mean return on investment per cent (RoI%)

of the financial companies is greater than the mean RoI% of the manufacturing companies. The

financial analyst randomly sampled 28 financial companies and found their sample mean RoI%

to be 18:714 % with a standard deviation of 9:645%. For a random sample of 24 manufacturing

companies, the sample mean RoI% was 15:125 with a sample standard deviation of 8:823%.

This information is summarized below

Sample 1: Financial Sample 2: Manufacturing

n1 D 28 n2 D 24

x1 D 18:714 x2 D 15:125

S1 D 9:645 S2 D 8:823

Assuming that the population variances are equal, the 95% confidence interval for the difference

between two population means is

1. .5:987I 12:098/

2. .3:589I 5:1838/

3. .0:9678I 4:657/

4. .

Answer 1

$CI=\bar {X_1}-\bar{X_2}\pm t_{(1-\frac{\alpha}{2}),n_1+n_2-2}sp(\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

$sp=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}$

$\bar X_1=18.714$

$s_1=9.645$

$n_1=28$

$\bar X_2=15.125$

$s_2=8.823$

$n_2=24$

$sp=\sqrt{\frac{(27\times9.645^2)+(23\times 8.823^2)}{50}}$

$=9.276$

$t_{0.025,50}=2.009$

$CI=18.714-15.125\pm(2.009\times 9.276\times\sqrt{\frac{1}{28}+\frac{1}{24}})$

$=3.589\pm5.1827$

=[-1.594,8.772]