Question

The city manager of Shinbone has received a complaint from the local union of firefighters to the effect that they are underpaid. Not having much time, the city manager gathers the records of a random sample of 27 firefighters and finds that their average salary is $38,073 with a standard deviation of $575. If she knows that the average salary nationally is $38,202, how can she respond to the complaint

Answer 1

Answer:

She can answer, after performing the hypothesis test, that there is not enough evidence to support the claim that the city firefighters salary is significantly lower than the national average.

Step-by-step explanation:

She can statistically test the claim of the firefighters to see if it has statistical evidence.

This is a hypothesis test for the population mean.

The claim is that the city firefighters salary is significantly lower than the national average.

Then, the null and alternative hypothesis are:

$H_0: \mu=38202\\\\H_a:\mu< 38202$

The significance level is 0.1.  Is less conservative than 0.05, for example, so if there is little evidence, the null hypothesis with be rejected.

The sample has a size n=27.

The sample mean is M=38073.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=575.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{575}{\sqrt{27}}=110.659$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{38073-38202}{110.659}=\dfrac{-129}{110.659}=-1.17$

The degrees of freedom for this sample size are:

df=n-1=27-1=26

This test is a left-tailed test, with 26 degrees of freedom and t=-1.17, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.127) is bigger than the significance level (0.1), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the city firefighters salary is significantly lower than the national average.