Question

In order to determine the average price of hotel rooms in Atlanta, a sample of 38 hotels were selected. It was determined that the average price of the rooms in the sample was $109.3. The population standard deviation is known to be $18. We would like to test whether or not the average room price is significantly different from $110.Compute the test statistic.

Answer 1

Answer:

There is not enough evidence to support the claim that the average room price in Atlanta is significantly different from $110.

P-value = 0.81

Test statistic z=-0.24

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average room price in Atlanta is significantly different from $110.

Then, the null and alternative hypothesis are:

$H_0: \mu=110\\\\H_a:\mu\neq 110$

The significance level is assumed to be 0.05.

The sample has a size n=38.

The sample mean is M=109.3.

The standard deviation of the population is known and has a value of σ=18.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{18}{\sqrt{38}}=2.92$

Then, we can calculate the z-statistic as:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{109.3-110}{2.92}=\dfrac{-0.7}{2.92}=-0.24$

This test is a two-tailed test, so the P-value for this test is calculated as:

$P-value=2\cdot P(z<-0.24)=0.81$

As the P-value (0.81) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the average room price in Atlanta is significantly different from $110.