# 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.