Question

A company sends a random sample of 16 of its sales- people to a course designed to increase their moti- vation and, hence, presumably their effectiveness. In the following year these people generated sales with an average value of $625,000 and a sample stan- dard deviation of $80,000. During the same period, an independently chosen random sample of 10 sales- people who had not attended the course obtained sales with an average value of $608,000 and a sample standard deviation of $73,000. Assume that the two population distributions are normal and have the same variance. Find a 90% confidence interval esti- mate for the difference between the population mean sales for salespeople who attended the motivational course and for those salespeople who did not attend the course.

Answer 1

$n_1=16\\n_2=10\\\bar{x}_1=625000\\\bar{x}_2=608000\\s_1=80000\\s_2=73000\\\nu =\frac{\left( \frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2} \right) ^2}{\frac{1}{n_1-1}\left( \frac{{s_1}^2}{n_1} \right) ^2+\frac{1}{n_2-1}\left( \frac{{s_2}^2}{n_2} \right) ^2}=\frac{\left( \frac{80000^2}{16}+\frac{73000^2}{10} \right) ^2}{\frac{1}{15}\left( \frac{80000^2}{16} \right) ^2+\frac{1}{9}\left( \frac{73000^2}{10} \right) ^2}=20.6134\approx 20\\Confidence\,\,interval:\\\left( \bar{x}_1-\bar{x}_2-\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}t_{\nu ,\frac{1+\gamma}{2}},\bar{x}_1-\bar{x}_2+\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}t_{\nu ,\frac{1+\gamma}{2}} \right) =\\=\left( 625000-608000-\sqrt{\frac{80000^2}{16}+\frac{73000^2}{10}}\cdot 1.7247,625000-608000+\sqrt{\frac{80000^2}{16}+\frac{73000^2}{10}}\cdot 1.7247 \right) =\\=\left( -35678.2,69678.2 \right)$