Question

Power calculation: potatoes Refer to Exercise 85.

a. Suppose that $H_0:p=0.08$is true. Describe the shape, center, and variability of the sampling distribution of $\hat{p}$ in random samples of size $500$

b. Use the sampling distribution from part (a) to find the value of $\hat{p}$with an area of $0.05$to the right of it. If the supervisor obtains a random sample of $500$potatoes with a sample proportion of defective potatoes greater than this value of p^, he will reject $H_0:p=0.08$at the $\alpha =0.05$significance level.

c. Now suppose that $p=0.11$Describe the shape, center, and variability of the sampling distribution of $\hat{p}$in random samples of size $500$

d. Use the sampling distribution from part (c) to find the probability of getting a sample proportion greater than the value you found in part (b). This result is the power of the test to detect $p=0.11$

Short answer

Part (a) Approximately normal with mean $0.08$and standard deviation $0.01213$

Part (b)$\hat{p}=0.09995$

Part (c) Approximately normal with the mean $0.11$ and standard deviation $0.01399$

Part (d)$0.7642=76.42\%$

Step-by-step solution

Part (a) Step 1: Given information

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

Part (a) Step 2: Concept

${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

Part (a) Step 3: Explanation

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

$$\begin{gathered} np\ge 10andn(1-p)\ge 10 \\ np=500(0.08)=40\ge 10 \\ n(1-p)=500(1-0.08)=460\ge 10 \end{gathered}$$

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

${\mu }_{\hat{p}}=p=0.08$

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

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

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

Part (b) Step 1: Concept

$z=\frac{x-\mu }{\sigma }$

Part (b) Step 2: Explanation

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

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{x-0.08}{0.01213} \end{gathered}$$

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.

Part (c) Step 1: Concept

${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

Part (c) Step 2: Explanation

If the big count criterion is met, the sampling distribution of the sample proportions $\hat{p}$is nearly Normal, when $np\ge 10andn(1-p)\ge 10$

$$\begin{gathered} np=500(0.11)=55\ge 10 \\ n(1-p)=500(1-0.11)=445\ge 10 \end{gathered}$$

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

${\mu }_{\hat{p}}=p=0.11$

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

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

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

Part (d) Step 1: Concept

$z=\frac{x-\mu }{\sigma }$

Part (d) Step 2: Explanation

$Z-$score is

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.09995-0.11}{0.01399} \\ =-0.72 \end{gathered}$$

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 $76.42\%$