Refer to Exercise 7.99.

a. Find b for each of the following values of the population mean: 74, 72, 70, 68, and 66.

b. Plot each value of b you obtained in part a against its associated population mean. Show b on the vertical axis and m on the horizontal axis. Draw a curve through the five points on your graph.

c. Use your graph of part b to find the approximate probability that the hypothesis test will lead to a Type II error when m = 73.

d. Convert each of the b values you calculated in part a to the power of the test at the specified value of m. Plot the power on the vertical axis against m on the horizontal axis. Compare the graph of part b with the power curve of this part.

e. Examine the graphs of parts b and d. Explain what they reveal about the relationships among the distance between the true mean m and the null hypothesized mean m0, the value of b, and the power.

We have to find the $\beta$ value for population mean: 74, 72, 70, 68, and 66

If

$$\begin{gathered} \mu =74 \\ \beta =P(\overline{x}\ge 72.275) \\ =P\left(z>\frac{72.275-74}{15/\sqrt{\sqrt{49}}}\right) \\ =P(z>-0.81) \\ =1-P(z<-0.81) \\ 1-0.20897 \\ =0.79103 \end{gathered}$$

If

role="math" localid="1664532990317" $$\begin{gathered} \mu =66 \\ \beta =P(\overline{x}\ge 72.275) \\ =P\left(z>\frac{72.275-66}{15/\sqrt{49}}\right) \\ =P(z>2.92) \\ =1-P(z<2.92) \\ =1-0.9982 \\ =0.0018 \end{gathered}$$

If

role="math" localid="1664533009209" $$\begin{gathered} \mu =66 \\ \beta =P(\overline{x}\ge 72.275) \\ =P\left(z>\frac{72.275-66}{15/\sqrt{49}}\right) \\ =P(z>2.92) \\ =1-P(z<2.92) \\ =1-0.9982 \\ =0.0018 \end{gathered}$$

If

If

$$\begin{gathered} \mu =66 \\ \beta =P(\overline{x}\ge 72.275) \\ =P\left(z>\frac{72.275-66}{15/\sqrt{49}}\right) \\ =P(z>2.92) \\ =1-P(z<2.92) \\ =1-0.9982 \\ =0.0018 \end{gathered}$$

Obtain a graph were $\beta isonverticalaxisand\mu isonhorizontalaxis.$

From a.) $\beta$value with respect to each population mean is given in the below table.

From Figure 1), the approximate probability that the hypothesis test will lead to a Type II error when $\mu =73is0.62$since $\beta$is the probability of type II error.

The Power of the test at the specified value of $\mu$

The formula for power is given below:

$Power=1-\beta$

The table below shows the power of the test at the specified value of $\mu$

Comparison:

From figure 1.), it is clear that the $\beta$value highest with the increment of the $\mu$worth. From figure 2.), it is clear that the $\mu$worth lowering with the increment of the $\mu$worth. Also, the power curve beginning from 1.

After examining figure (1) and figure (2), if the distance among the true mean $\mu$as well as the null hypothesized mean ${\mu }_0$raising, then $\beta$ will decrease. If the distance between the true mean $\mu$ and the hypothesized mean ${\mu }_0$raising, then power will be raising.