# Solved (Free): Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes

#### ByDr. Raju Chaudhari

Mar 14, 2021

Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes. Let $\alpha = 0.05$.

#### Solution

Given that the sample size $n = 25$ and sample standard deviation $s = 15$.

#### Step 1 Hypothesis Problem

The hypothesis testing problem is
$H_0 : \sigma^2 = 150$ against $H_1 : \sigma^2 > 150$ ($\text{right-tailed}$)

#### Step 2 Test Statistic

The test statistic for testing above hypothesis testing problem is

 $$\chi^2 =\frac{(n-1)s^2}{\sigma^2}$$

#### Step 3 Level of Significance

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value

As the alternative hypothesis is $\text{right-tailed}$, the critical value of $\chi^2$ $\text{ is }$ $\text{36.415}$ (from $\chi^2$ statistical table).

The rejection region (i.e. critical region) is $\chi^2 > 36.415$.

#### Step 5 Test Statistic

The test statistic under the null hypothesis is

 \begin{aligned} \chi^2 &=\frac{(n-1)s^2}{\sigma^2_0}\\ &= \frac{(25-1)*(15)^2}{(12.2474487)^2}\\ &= 36 \end{aligned}

#### Step 6 Decision

The test statistic is $\chi^2 =36$ which falls $outside$ the critical region, we $\text{fail to reject}$ 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 ($\chi^2=36$) is p-value = $0.0549$.

The p-value is $0.0549$ which is $\text{greater than}$ the significance level of $\alpha = 0.05$, we $\text{fail to reject}$ the null hypothesis.

We conclude that the average delay is so consistent that the variance is no more than 150 minutes.