Respuesta :
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:
[tex]H_0: \mu\leq50\\\\H_a:\mu> 50[/tex]
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:
[tex]\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{55}}=0.809[/tex]
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:
[tex]z=\dfrac{M-\mu}{\sigma_M}=\dfrac{52.5-50}{0.809}=\dfrac{2.5}{0.809}=3.09[/tex]
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:
[tex]z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51-50}{0.809}=\dfrac{1}{0.809}=1.24[/tex]
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:
[tex]z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51.8-50}{0.809}=\dfrac{1.8}{0.809}=2.22[/tex]
The null hypothesis is rejected, as z>zc and falls in the rejection region.
