Question

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.

Short answer

a)$0.0018$

b) $\beta isontheverticalaxisand\mu isonhorizonalaxis$

c) $\mu =73is0.62$

d) 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.

e) If the distance between the true mean and the hypothesized mean ${\mu }_0$raising, then power will be raising.

Step-by-step solution

(a) Population mean

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

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

(b) Horizontal and vertical axis

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

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

$\mu$	$\beta$
74	0.7910
72	0.4522
70	0.1469
68	0.0233
66	0.0018

(c) Type error II

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.

(d) Power of the test specified value

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 $\mu$

$\mu$	$\beta$	Power
74	0.7910	0.209
72	0.4522	0.5478
70	0.1469	0.8531
68	0.0233	0.9767
66	0.0018	0.9982

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.

(e) They reveal about the link among the distance between true mean

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.