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The null hypothesis is shown below:

$H_0:{\mathrm{σ}}^2≤(3{)}^2$

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

Against the alternative hypothesis as shown below:

$H_a:{\mathrm{σ}}^2>(15{)}^2$

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 ${\mathrm{χ}}_4^2$.

Calculate the mean and standard deviation as shown below:

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

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

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

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

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

$=\frac{4×10.1761}{9}$

$=4.52$

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 $>\mathrm{Î\pm }$

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