Answer to Question #311394 in Statistics and Probability for Steve

Question

Is the average amount spent on textbooks per semester by accounting majors significantly different from the average amount spent on text books per semester by management majors? Answer this question with a 90% confidence interval using the following data from random samples of students majoring in accounting or management. Discuss the assumptions

Accounting majors Management major

Mean $340 $285

Standard deviation 20 30

Sample size 40 50

Answer 1

$n_1=40\\n_2=50\\\bar{x}_1=340\\\bar{x}_2=285\\s_1=20\\s_2=30\\\nu =\frac{\left( \frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2} \right) ^2}{\frac{1}{n_1-1}\left( \frac{{s_1}^2}{n_1} \right) ^2+\frac{1}{n_2-1}\left( \frac{{s_2}^2}{n_2} \right) ^2}=\frac{\left( \frac{20^2}{40}+\frac{30^2}{50} \right) ^2}{\frac{1}{39}\left( \frac{20^2}{40} \right) ^2+\frac{1}{49}\left( \frac{30^2}{50} \right) ^2}=85.437\approx 85\\$

Then the confidence interval for $\mu _1-\mu _2$ is

$\left( \bar{x}_1-\bar{x}_2-\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}t_{\nu ,\frac{1+\gamma}{2}},\bar{x}_1-\bar{x}_2-\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}t_{\nu ,\frac{1+\gamma}{2}} \right) =\\=\left( 340-285-\sqrt{\frac{20^2}{40}+\frac{30^2}{50}}\cdot 1.663,340-285+\sqrt{\frac{20^2}{40}+\frac{30^2}{50}}\cdot 1.663 \right) =\\=\left( 46.2002,63.7998 \right)$

Since the confidence interval doesn’t contain 0, the null hypothesis is rejected, the mean values differ.

We assume that two samples are independent, with normal distribution.