Chapter 9.3, Problem 101E

(a)

To determine

To Explain: the shape, centre and variability of the sampling distribution of $\widehat{p}$ in random samples of size 500.

Answer to Problem 101E

Approximately normal with mean 0.08 and standard deviation 0.01213

Explanation of Solution

Given:

  $\begin{array}{l}{H}_0:p=0.08\\ p=0.08\\ n=500\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

The hypothesis distribution of the sample proportions $\widehat{p}$ is about Normal is the large counts condition is satisfied, that is, when

  $np\ge 10\text{and n}\left(1-p\right)\ge 10$ .

  $\begin{array}{l}np=500\left(0.08\right)=40\ge 10\\ \text{n}\left(1-p\right)=500\left(1-0.08\right)=460\ge 10\end{array}$

The mean of the sampling distribution of the sample proportions $\widehat{p}$ is

  ${\mu }_{\widehat{p}}=p=0.08$

Then 10% condition requires that the sample is less than 10% of the population size. Assuming that the sample of 500 potatoes is less than 10% of the population of all potatoes, the 10% condition is satisfied.

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$ is

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.08\left(1-0.08\right)}{500}}\\ =0.01213\end{array}$

Therefore the sampling distribution of the sample proportion $\widehat{p}$ is about Normal with mean 0.08 and standard deviation 0.01213.

(b)

To determine

To find: the value of $\widehat{p}$ with an area of 0.05 to the right of it.

Answer to Problem 101E

  $\widehat{p}=0.09995$

Explanation of Solution

Given:

From the result part (a) is approximately Normal distribution

  $\begin{array}{c}\mu =0.08\\ \sigma =0.01213\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

If a z-score has probability of 0.05 to the right, so it has a probability of 1-0.05=0.95 to left.

It is observed that the probability of 0.95 lies exactly between 0.9495 and 0.9505. Which associate to the Z-score of 1.64 and 1.65 and then estimate the Z-score associating to 0.95 as the Z-score exactly in the middle between 1.64 and 1.65, which is 1.645

  $z=1.645$

The Z-score is

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

The two found expressions of the Z-score then

  $\frac{x-0.08}{0.01213}=1.645$

  $x-0.08=1.645\left(0.01213\right)$

  $x=0.08+1.645\left(0.01213\right)$

  $x=0.09995385=0.09995$

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

(c)

To determine

To Explain: the shape, centre and variability of the sampling distribution of $\widehat{p}$ in random samples of size 500.

Answer to Problem 101E

Approximately normal with the mean 0.11 and standard deviation 0.01399

Explanation of Solution

Given:

  $\begin{array}{c}p=0.11\\ n=500\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

The sampling distribution of the sample proportions $\widehat{p}$ is about Normal is the large counts condition is satisfied, that is, when $np\ge 10\text{and n}\left(1-p\right)\ge 10$ .

  $\begin{array}{c}np=500\left(0.11\right)=55\ge 10\\ \text{n}\left(1-p\right)=500\left(1-0.11\right)=445\ge 10\end{array}$

The mean of the sampling distribution of the sample proportions $\widehat{p}$ is

  ${\mu }_{\widehat{p}}=p=0.11$

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$ is

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.11\left(1-0.11\right)}{500}}\\ =0.01399\end{array}$

Therefore the sampling distribution of the sample proportion $\widehat{p}$ is about Normal with mean 0.11 and standard deviation 0.01399.

(d)

To determine

To find: the probability of getting a sample proportion from the part (c) greater than the value found in part (b).

Answer to Problem 101E

0.7642=76.42%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =0.11\\ \sigma =0.01399\\ x=0.09995\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

Z-score is

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

Probability is

  $\begin{array}{c}P\left(\widehat{p}>0.09995\right)=P\left(Z>-0.72\right)\\ =1-P\left(Z<-0.72\right)\\ =1-0.2358\\ =0.7642\\ =76.42\%\end{array}$

Therefore the power of the test is 0.7642 or 76.42%