Chapter 7.2, Problem 29E

(a)

To determine

To calculate: The mean of the sampling distribution of $\widehat{p}$ .

Answer to Problem 29E

The required mean of the sampling distribution $\widehat{p}$ is 0.45.

Explanation of Solution

Given information:

Percent of orange candies in the machine = 45%

Number of candies for SRS = 25

Formula used:

The mean of the sampling distribution of a sample proportion $\widehat{p}$ and the population proportion $p$ are equal that is, ${\mu }_{\widehat{p}}=p$ .

Calculation:

Consider that the sample distribution be p and sample size for SRS be n.

So, $\begin{array}{c}p=45\%\\ =0.45\end{array}$

And, $n=25$

Know that the mean of the sampling distribution of a sample proportion $\widehat{p}$ and the population proportion $p$ are equal that is, ${\mu }_{\widehat{p}}=p$ .

Substitute 0.45 for p in the above formula.

  ${\mu }_{\widehat{p}}=0.45$

Since, the sampling proportion is an unbiased estimator for the population proportion; so the mean is ${\mu }_{\widehat{p}}=0.45$ .

Hence, the required mean of the sampling distribution $\widehat{p}$ is 0.45.

(b)

To determine

To calculate: The standard deviation of the sampling distribution of $\widehat{p}$ .

Answer to Problem 29E

The required standard deviation is 0.099487.

Explanation of Solution

Given information:

Percent of orange candies in the machine = 45%

Number of candies for SRS = 25

Formula used:

The standard deviation of the sampling distribution of $\widehat{p}$ is, ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Calculation:

Consider that the sample distribution be p and sample size for SRS be n.

So, $\begin{array}{c}p=45\%\\ =0.45\end{array}$

And, $n=25$

Know that the standard deviation of the sampling distribution of $\widehat{p}$ is ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Substitute 0.45 for $p$ and 25 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.45\left(1-0.45\right)}{25}}\\ =\sqrt{\frac{0.45\times 0.55}{25}}\\ =\sqrt{0.0099}\\ \approx 0.0994987\end{array}$

Since, the candy machine is large; so it contains more than 250 candies that is the 10% condition is satisfied.

Hence, the required standard deviation is 0.099487.

(c)

To determine

Whether the sampling distribution of $\widehat{p}$ is approximately Normal or not.

Answer to Problem 29E

The sampling distribution of $\widehat{p}$ is approximately Normal.

Explanation of Solution

Given information:

Percent of orange candies in the machine = 45%

Number of candies for SRS = 25

Formula used:

When the product of sample size and the sampling proportion that is, $np$ and $n\left(1-p\right)$ both are less than at least 10 then the sampling distribution is approximately Normal.

Consider that the sample distribution be p and sample size for SRS be n.

So, $\begin{array}{c}p=45\%\\ =0.45\end{array}$

And, $n=25$

Know that the product of sample size and the sampling proportion that is, $np$ and $n\left(1-p\right)$ both are less than at least 10 then the sampling distribution is approximately Normal.

Substitute 0.45 for p and 25 for n in the expression $np$ .

  $25\times \left(0.45\right)=11.25$

Again, substitute 0.45 for p and 25 for n in the expression $n\left(1-p\right)$ .

  $\begin{array}{c}25\left(1-0.45\right)=25\times 0.55\\ =13.75\end{array}$

Since, both $np$ and $n\left(1-p\right)$ are at least 10; so the condition for Normal distribution is met.

Hence, the sampling distribution of $\widehat{p}$ is approximately Normal.

(d)

To determine

The change in the sampling distribution of $\widehat{p}$ if the sample size were 50 rather than 25.

Answer to Problem 29E

The standard deviation of the sampling distribution $\widehat{p}$ changes to 0.0703562 when the sample size is 50.

Explanation of Solution

Given information:

Percent of orange candies in the machine = 45%

Number of candies for SRS = 50

Formula used:

The standard deviation of the sampling distribution of $\widehat{p}$ is, ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Calculation:

Consider that the sample distribution be p and sample size for SRS be n.

So, $\begin{array}{c}p=45\%\\ =0.45\end{array}$

And, $n=50$

Know that the standard deviation of the sampling distribution of $\widehat{p}$ is ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Substitute 0.45 for $p$ and 50 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.45\left(1-0.45\right)}{50}}\\ =\sqrt{\frac{0.45\times 0.55}{50}}\\ =\sqrt{0.00495}\\ \approx 0.0703562\end{array}$

Thus, the standard deviation is 0.0703562.

Hence, the standard deviation of the sampling distribution $\widehat{p}$ changes to 0.0703562 when the sample size is 50.