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](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%5Cleq50%5C%5C%5C%5CH_a%3A%5Cmu%3E%2050)
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](https://tex.z-dn.net/?f=%5Csigma_M%3D%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B6%7D%7B%5Csqrt%7B55%7D%7D%3D0.809)
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](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7BM-%5Cmu%7D%7B%5Csigma_M%7D%3D%5Cdfrac%7B52.5-50%7D%7B0.809%7D%3D%5Cdfrac%7B2.5%7D%7B0.809%7D%3D3.09)
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](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7BM-%5Cmu%7D%7B%5Csigma_M%7D%3D%5Cdfrac%7B51-50%7D%7B0.809%7D%3D%5Cdfrac%7B1%7D%7B0.809%7D%3D1.24)
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](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7BM-%5Cmu%7D%7B%5Csigma_M%7D%3D%5Cdfrac%7B51.8-50%7D%7B0.809%7D%3D%5Cdfrac%7B1.8%7D%7B0.809%7D%3D2.22)
The null hypothesis is rejected, as z>zc and falls in the rejection region.