Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the $7.5$mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are $55$minutes, $61$minutes, $58$minutes, $63$ minutes, and $57$ minutes.

The null hypothesis is shown below:

$H_0:{\sigma }^2\le (3{)}^2$

That is, the standard deviation of $7.5$mile race is at most three minutes.

Against the alternative hypothesis as shown below:

$H_a:{\sigma }^2>(15{)}^2$

That is, the standard deviation of $7.5$mile race is greater than three minutes.

The degrees of freedom can be calculated as shown below:

$n-1=5-1$

$=4$

From the above calculation, it is clear that the distribution for the test is ${\chi }_4^2$.

Calculate the mean and standard deviation as shown below:

The test statistics can be calculated by using the formula shown below:

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

Here, $n$is sample size, $s^2$is sample variance and ${\sigma }^2$is population variance. Therefore, the calculation is shown below:

$\text{Test statistic}=\frac{(n-1)s^2}{{\sigma }^2}$

$=\frac{(5-1)(3.19{)}^2}{(3{)}^2}$

$=\frac{4\times 10.1761}{9}$

$=4.52$

Hence, the test statistic is $4.52$.

The $p$-value can be calculated in excel by using CHIDIST ( ) formula as shown below:

Hence the $p$- value is $0.340$.

The Chi-Square sketch is given below:

Alpha: $0.05$

Decision: Do not reject the null hypothesis$H_0$

Reason for decision: Because $p$-value $>\alpha$

Conclusion: There is insufficient evidence to ensure that the standard deviation of $7.5$mile race is greater than three minutes.