Question

Independent random samples from normal distributions with equal variance of the amount of insider

trading during mergers by employees of investment banking firms (1) and brokerage houses (2) gave the

results listed in the accompanying table.

Investment Banking Firm 10 5 5 5 8 8 8

Brokerage House 6 0 5 2 2 3 3

(a) Find the 95% confidence interval estimate for the difference between means µ1 − µ2..

(b) Would you conclude that employees of Investment banking firms cause more insider trading than

their counterparts in Brokerage House?

Answer 1

$n_1=7\\n_2=7\\\bar{x}_1=\frac{10+5+5+5+8+8+8}{7}=7\\\bar{x}_2=\frac{6+0+5+2+2+3+3}{7}=3\\{s_1}^2=\frac{\left( 10-7 \right) ^2+3\cdot \left( 5-7 \right) ^2+3\cdot \left( 8-7 \right) ^2}{7-1}=4\\{s_2}^2=\frac{\left( 6-3 \right) ^2+\left( 0-3 \right) ^2+\left( 5-3 \right) ^2+2\cdot \left( 2-3 \right) ^2+2\cdot \left( 3-3 \right) ^2}{7-1}=4\\s^2=\frac{\left( n_1-1 \right) {s_1}^2+\left( n_2-1 \right) {s_2}^2}{n_1+n_2-2}=\frac{\left( 7-1 \right) \cdot 4+\left( 7-1 \right) \cdot 4}{7+7-2}=4\\Confidence\,\,interval\\\left( \bar{x}_1-\bar{x}_2-\sqrt{s^2\left( \frac{1}{n_1}+\frac{1}{n_2} \right)}t_{\frac{1+\gamma}{2},n_1+n_2-2},\bar{x}_1-\bar{x}_2+\sqrt{s^2\left( \frac{1}{n_1}+\frac{1}{n_2} \right)}t_{\frac{1+\gamma}{2},n_1+n_2-2} \right) =\\=\left( 7-3-\sqrt{4\left( \frac{1}{7}+\frac{1}{7} \right)}\cdot 2.1788,7-3+\sqrt{4\left( \frac{1}{7}+\frac{1}{7} \right)}\cdot 2.1788 \right) =\\=\left( 1.67076,6.32924 \right)$

I can conclude that employees of Investment banking firms cause more insider trading than their counterparts in Brokerage House