91影视

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

If

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

If

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

If

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

If

If

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

Obtain a graph were $尾isonverticalaxisand渭isonhorizontalaxis.$

From a.) $尾$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 $渭=73is0.62$since $尾$is the probability of type II error.

The Power of the test at the specified value of $渭$

The formula for power is given below:

$Power=1-尾$

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

Comparison:

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

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