91Ó°ÊÓ

$$\begin{gathered} H_0:p=0.08 \\ p=0.08 \\ n=500 \end{gathered}$$

${\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1−p)}{n}}$

When the large counts requirement is met, the hypothesis distribution of the sample proportions is roughly Normal.

$$\begin{gathered} npâ‰\yen 10andn(1−p)â‰\yen 10 \\ np=500(0.08)=40â‰\yen 10 \\ n(1−p)=500(1−0.08)=460â‰\yen 10 \end{gathered}$$

The sampling distribution of the sample proportions $\hat{p}$has a mean of

${\mathrm{μ}}_{\hat{p}}=p=0.08$

Then the $10\%$requirement states that the sample size must be smaller than $10\%$of the total population size. The $10\%$requirement is satisfied if the sample of $500$potatoes represents less than $10\%$of the total population of potatoes.

The sampling distribution of the sample proportion $\hat{p}$has a standard deviation of

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1−p)}{n}} \\ {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{0.08(1−0.08)}{500}} \\ =0.01213 \end{gathered}$$

As a result, the sample proportion $\hat{p}$ sampling distribution is roughly Normal, with a mean of $0.08$ and a standard deviation of $0.01213$

$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}$

If a $z-$score has a $0.05$ probability to the right, it has a $1-0.05=0.95$ probability to the left. The probability of $0.95$ is found to be exactly between $0.9495$ and $0.9505$ Which correspond to $Z-$scores of $1.64$ and $1.65$ respectively, and then estimate the $Z-$score corresponding to $0.95$ as the $Z-$score exactly in the middle of $1.64$ and $1.65,1.645$

$z=1.645$

The $Z-$score is

$$\begin{gathered} z=\frac{x−\mathrm{μ}}{\mathrm{σ}} \\ =\frac{x−0.08}{0.01213} \end{gathered}$$

The two found expressions of the $Z-$score then

$$\begin{gathered} \frac{x−0.08}{0.01213}=1.645 \\ x−0.08=1.645(0.01213)x \\ =0.08+1.645(0.01213)x \\ =0.09995385=0.09995 \end{gathered}$$

Therefore the sample proportion $\hat{p}=0.09995$ has a probability of $0.05$ to its right.

${\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1−p)}{n}}$

If the big count criterion is met, the sampling distribution of the sample proportions $\hat{p}$is nearly Normal, when $npâ‰\yen 10andn(1−p)â‰\yen 10$

$$\begin{gathered} np=500(0.11)=55â‰\yen 10 \\ n(1−p)=500(1−0.11)=445â‰\yen 10 \end{gathered}$$

The mean of the sampling distribution of the sample proportions p is

${\mathrm{μ}}_{\hat{p}}=p=0.11$

The standard deviation of the sampling distribution of the sample proportion p

is $$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1−p)}{n}} \\ {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{0.11(1−0.11)}{500}} \\ =0.01399 \end{gathered}$$

As a result, the sample proportion $\hat{p}$ sampling distribution is roughly Normal, with a mean of $0.11$ and a standard deviation of $0.01399$

$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}$

$Z-$score is

$$\begin{gathered} z=\frac{x−\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0.09995−0.11}{0.01399} \\ =−0.72 \end{gathered}$$

Probability is

$$\begin{gathered} P(\hat{p}>0.09995)=P(Z>−0.72) \\ =1−P(Z<−0.72) \\ =1−0.2358 \\ =0.7642=76.42\% \end{gathered}$$

Therefore the power of the test is $0.7642$ or $76.42\%$