Chapter 7.2, Problem 30E

(a)

To determine

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

Answer to Problem 30E

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

Explanation of Solution

Given information:

Percent of orange candies in the machine = 15%

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, $\begin{array}{c}p=15\%\\ =0.15\end{array}$

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 0.15 for p in the above formula.

  ${\mu }_{\widehat{p}}=0.15$

Since, the sampling proportion is an unbiased estimator for the population proportion; so the mean is ${\mu }_{\widehat{p}}=0.15$ .

Hence, the required mean of the sampling distribution $\widehat{p}$ is 0.15.

(b)

To determine

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

Answer to Problem 30E

The required standard deviation is 0.0714143.

Explanation of Solution

Given information:

Percent of orange candies in the machine = 15%

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, $\begin{array}{c}p=15\%\\ =0.15\end{array}$

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

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.15\left(1-0.15\right)}{25}}\\ =\sqrt{\frac{0.15\times 0.85}{25}}\\ =\sqrt{0.0051}\\ \approx 0.0714143\end{array}$

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 0.0714143.

(c)

To determine

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

Answer to Problem 30E

The sampling distribution of $\widehat{p}$ is not approximately Normal.

Explanation of Solution

Given information:

Percent of orange candies in the machine = 15%

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, $\begin{array}{c}p=15\%\\ =0.15\end{array}$

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 0.15 for p and 25 for n in the expression $np$ .

  $25\times \left(0.15\right)=3.75$

Again, substitute 0.15 for p and 25 for n in the expression $n\left(1-p\right)$ .

  $\begin{array}{c}25\left(1-0.15\right)=25\times 0.85\\ =21.75\end{array}$

It is seen that both $np$ and $n\left(1-p\right)$ are not at least 10 that is, one is less than 10; so the condition for Normal distribution has not met.

Hence, the sampling distribution of $\widehat{p}$ is not approximately Normal.

(d)

To determine

The change in the sampling distribution of $\widehat{p}$ if the sample size were 75 rather than 25.

Answer to Problem 30E

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 = 15%

Number of candies for SRS = 75

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, $\begin{array}{c}p=15\%\\ =0.15\end{array}$

And, $n=75$

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 0.15 for $p$ and 75 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.15\left(1-0.15\right)}{75}}\\ =\sqrt{\frac{0.45\times 0.55}{75}}\\ =\sqrt{0.0033}\\ \approx 0.057445\end{array}$

Thus, the standard deviation is 0.057445.

Hence, the standard deviation of the sampling distribution $\widehat{p}$ changes to 0.057445 when the sample size is 75.