Question

The administrator at your local hospital states that on weekends the average wait time for emergency room visits is 11 minutes. Based on discussions you have had with friends who have complained about how long they wait to be seen in the ER over a weekend, you dispute the administrator's claim. You decide to test your hypothesis. Over the course of a few weekends, you record the wait time for 28 randomly selected patients. The average wait time for these selected patients is 12 minutes with a standard deviation of 2.5 minutes. Do you have enough evidence to support your hypothesis that the average ER wait time is longer than 11 minutes? Conduct your test with a 5% level of significance.

Answer 1

This is a hypothesis test for the population mean.

The claim is that on weekends the average wait time for emergency room visits is more than 11 minutes.

Then, the null and alternative hypothesis are:

`\begin{gathered} H_0\colon\mu=11 \\ H_a\colon\mu>11 \end{gathered}`

The significance level is 0.05.

The sample has a size n=28.

The sample mean is M=12.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.5.

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

`s_M=(s)/(√(n))=(2.5)/(√(28))=0.4725`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(12-11)/(0.4725)=(1)/(0.4725)=2.117`

The degrees of freedom for this sample size are:

`df=n-1=28-1=27`

This test is a right-tailed test, with 27 degrees of freedom and t=2.117, so the P-value for this test is calculated as (using a t-table):

`\text{P-value}=P(t>2.117)=0.0218`

As the P-value (0.0218) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

Conclusion: at a significance level of 0.05, there is enough evidence to support the claim that, on weekends, the average wait time for emergency room visits is more than 11 minutes.