It is desired to test H0: m = 75 against Ha: m 6 75 using a = .10. The population in question is uniformly distributed with standard deviation 15. A random sample of size 49 will be drawn from the population.

a. Describe the (approximate) sampling distribution of x under the assumption that H0 is true.

b. Describe the (approximate) sampling distribution of x under the assumption that the population mean is 70.

c. If m were really equal to 70, what is the probability that the hypothesis test would lead the investigator to commit a Type II error?

d. What is the power of this test for detecting the alternative Ha: m = 70?

Null hypothesis: $H_0:\mu =75$

Alternate hypothesis: $H_0:\mu <75$

Thus, we have to sample finding the distribution of $\overline{x}$according to the central limit theorem. The sample mean follows the normal distribution with mean $\mu$as well as standard deviation. $\frac{\sigma }{\sqrt{n}}$

$$\begin{gathered} E\left(\overline{x}\right)=\mu \\ =75 \\ {\sigma }_{\left(\overline{x}\right)}=\frac{\sigma }{\sqrt{n}} \\ =\frac{15}{\sqrt{49}} \\ =\frac{15}{7} \end{gathered}$$

Thus, the sample distribution of $\left(\overline{x}\right)$is N$\left(75,\frac{15}{7}\right)$

They consider the rejection region corresponding to $a=0.10$. To calculate to the rejection region is $z<-1.28$then the statistics test is

$$\begin{gathered} {\overline{x}}_{oe}={\mu }_0-1.28\times \frac{s}{\sqrt{n}} \\ =75-1.28\times 2.142 \\ =75-2.741 \\ {\overline{x}}_{oe}=72.259 \end{gathered}$$

$$\begin{gathered} {\overline{x}}_{oe}=\mu +1.28\times \frac{s}{\sqrt{n}} \\ =75+1.28\times 2.142 \\ =75+2.741 \\ {\overline{x}}_{oe}=77.741 \end{gathered}$$

They necessary to obtain sampling distribution of $\overline{x}$under the assumption that the population mean is 70. Thus, we have to sample finding the distribution of $\overline{x}.$according to the central limit theorem. The sample mean follows a normal distribution with mean $\mu$as well as standard deviation. $\frac{\sigma }{\sqrt{n}}$

$$\begin{gathered} E\left(\overline{x}\right)=\mu \\ =70 \\ {\sigma }_{\left(\overline{x}\right)}=\frac{\sigma }{\sqrt{n}} \\ =\frac{15}{\sqrt{49}} \\ =\frac{15}{7} \end{gathered}$$

Thus, the sample distribution of $\left(\overline{x}\right)$is N $\left(70,\frac{15}{7}\right)$

Following values in part (a) to z-value in the alternative distribution with $\mu =70$

$$\begin{gathered} zL=\frac{({\overline{)\overline{x}}}_{oe}-\overline{)\mu })}{\sigma \overline{x}} \\ =\frac{72.259-70}{2.142} \\ zL=1.054 \end{gathered}$$

Using Table IV in Appendix B, they see that the area among $z=0$as well as $z=1.054is0.3531,$while the area between $z=0$as well as $z=3.613$is about 0.5 (since $z=3.613$it is off the scale).

In this issue, we assume that $\mu$it is truly equal to 70. Then there is a chance that the hypothesis test will cause the investigator to make a type - II error $\left(\beta \right)$:

For this, we obtain the area between:

$$\begin{gathered} z=1.054 \\ z=3.613istreatedas \\ \beta =0.5-0.3531 \\ \beta =0.1469 \end{gathered}$$

The power is defined to be the probability of (correctly) the null hypothesis when the alternative is $H_a:\mu =70true$

Power of the test: $(1-\beta )$

$$\begin{gathered} Power=(1-\beta ) \\ =1-0.1469 \\ =0.8531 \end{gathered}$$