Solved-Independent random samples taken on two university campuses revealed the following information concerning the average

ByDr. Raju Chaudhari

Oct 7, 2020

Independent random samples taken on two university campuses revealed the following information concerning the average amount of money spent on textbooks during the fall semester.

. University A University B
Sample Size 50 40
Average Purchase \$260 \$250
Standard Deviation \$20 \$23

We want to determine if, on the average, students at University A spent more on textbooks then the students at University B.
a. Compute the test statistic.
b. Compute the p-value.
c. What is your conclusion? Let $\alpha = .05$.

Solution

a. Let $\mu_1$ be the average spending by students of University A, and $\mu_2$ be the average spending by students of University B.

Given that the sample size $n_1 = 50$, $n_2 = 40$, sample mean $\overline{x}_1= 260$, $\overline{x}_2= 250$, sample standard deviation $s_1 = 20$ and $s_2 = 23$.

The hypothesis testing problem is
$H_0 : \mu_1 = \mu_2$ against $H_1 : \mu_1 > \mu_2$ ($\textit{right-tailed}$)

As the population standard deviations are unknown, we use t-test for testing above hypothesis.

The test statistic is

 $$\begin{equation*} t=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 - \mu_2)}{sp\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \end{equation*}$$

where

 $$\begin{eqnarray*} s_p & = &\sqrt{\frac{(n_1-1)s_1^2 +(n_2-1)s_2^2}{n_1+n_2-2}}\\ & = & \sqrt{\frac{(50-1)202 +(40-1)23^2}{50+40-2}}\\ & = & 21.3815. \end{eqnarray*}$$

The test statistic under the null hypothesis is

 $$\begin{eqnarray*} t&=&\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1-\mu_2)}{sp\sqrt{\big(\frac{1}{n_1}+\frac{1}{n_2}\big)}}\\ &=& \frac{(260-250)-0}{21.3815\sqrt{\big(\frac{1}{50}+\frac{1}{40}\big)}}\\ &=& 2.2047 \end{eqnarray*}$$

b. $p$-value

This is a $\textit{right-tailed}$ test, so the p-value is the area to the right of the test statistic ($t=2.2047$).

$p$-value = $P(t_{88}> 2.2047) = 0.015$.

The p-value is $0.015$ which is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis.

c. Thus, on the average, students at University A spent more on textbooks then the students at University B.