Question

Consider the following hypothesis test:

Answer 1

Answer:

a. z=3.09. Yes, it can be concluded that the population mean is greater than 50.

b. z=1.24. No, it can not be concluded that the population mean is greater than 50.

c. z=2.22. Yes, it can be concluded that the population mean is greater than 50.

Step-by-step explanation:

We have a hypothesis test for the mean, with the hypothesis:

$H_0: \mu\leq50\\\\H_a:\mu> 50$

The sample size is n=55 and the population standard deviation is 6.

The significance level is 0.05.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{55}}=0.809$

For a significance level of 0.05, the critical value for z is zc=1.644. If the test statistic is bigger than 1.644, the null hypothesis is rejected.

a. If the sample mean is M=52.5, the test statistic is:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{52.5-50}{0.809}=\dfrac{2.5}{0.809}=3.09$

The null hypothesis is rejected, as z>zc and falls in the rejection region.

b. If the sample mean is M=51, the test statistic is:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51-50}{0.809}=\dfrac{1}{0.809}=1.24$

The null hypothesis failed to be rejected, as z<zc and falls in the acceptance region.

c. If the sample mean is M=51.8, the test statistic is:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51.8-50}{0.809}=\dfrac{1.8}{0.809}=2.22$

The null hypothesis is rejected, as z>zc and falls in the rejection region.