# Solved (Free): A math instructor has written a new placement exam. He is concerned that the exam may be too long. The instructor claims that the average length of time required to finish

#### ByDr. Raju Chaudhari

Apr 4, 2021

A math instructor has written a new placement exam. He is concerned that the exam may be too long. The instructor claims that the average length of time required to finish the exam for all students who take it is at most 50 minutes. The exam is given to 20 randomly selected students, and their average length of time required to finish the exam was 54 minutes, with a standard deviation of 5.99 minutes. Test the instructor's claim at the 0.05 level of significance.

#### Solution

Given that the sample size $n = 20$, sample mean $\overline{x}= 54$ and sample standard deviation $s = 5.99$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \mu \leq 50$ against $H_1 : \mu > 50$ ($\text{right-tailed}$)

#### Step 2 Test Statistic

The test statistic is

 \begin{aligned} t& =\frac{\overline{x} -\mu}{s/\sqrt{n}} \end{aligned}
which follows $t$ distribution with $n-1$ degrees of freedom.

#### Step 3 Significance Level

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value(s)

As the alternative hypothesis is $\text{right-tailed}$, the critical value of $t$ $\text{is}$ $1.729$.

The rejection region (i.e. critical region) is $\text{t > 1.729}$.

#### Step 5 Computation

The test statistic under the null hypothesis is
 \begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\\ &= \frac{54-50}{5.99/ \sqrt{20 }}\\ &= 2.986 \end{aligned}

#### Step 6 Decision

The test statistic is $t =2.986$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

$p$-value Approach:

This is a $\text{right-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=2.986$) is p-value = $0.0038$.

The p-value is $0.0038$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

The claims that the average length of time required to finish the exam for all students who take it is at most 50 minutes is rejected. So the average length of time required to finish the exam for all students who take it is more than 50 minutes.