Chapter 6, Problem 12P

a.

To determine

Obtain the sampling distribution of $\overline{p}$

Answer to Problem 12P

The sampling distribution of $\overline{p}$ is normal with mean 0.55 and standard deviation 0.0352.

Explanation of Solution

It is given that the proportion of entrepreneurs whose first startup was at 29 years or less is $p=0.55$ and the sample size $n=200$

The sampling distribution of the proportion is approximately normal if $np\ge 5$  and $n(1-p)\ge 5.$

Verify the conditions:

$\begin{array}{c}np=200\times 0.55\\ =110\\ \ge 5\end{array}$

And $\begin{array}{c}n(1-p)=200\times (1-0.55)\\ =90\\ \ge 5\end{array}$

The conditions are satisfied. Therefore, the sampling distribution of the proportion is normal.

The mean of the $\overline{p}$ is $E(\overline{p})=p$  and standard deviation of $\overline{p}$ is ${\sigma }_{\overline{p}}=\sqrt{\frac{p(1-p)}{n}}$

In this context, $\overline{p}$ is the sample proportion of entrepreneurs whose first startup was at 29 years or less

The mean of $\overline{p}$ is

$\begin{array}{c}E(\overline{p})=p\\ =0.55\end{array}$

The standard deviation of $\overline{p}$ is

$\begin{array}{c}{\sigma }_{\overline{p}}=\sqrt{\frac{p(1-p)}{n}}\\ =\sqrt{\frac{0.55\times 0.45}{200}}\\ =0.0352\end{array}$

Thus, the sampling distribution of the proportion $\overline{p}$ proportion of entrepreneurs whose first startup was at 29 years or less is normal with mean $E(\overline{p})=0.55$ and standard deviation ${\sigma }_{\overline{p}}=0.0352$

b.

To determine

Obtain the sampling distribution of $\overline{p}$

Answer to Problem 12P

The sampling distribution of $\overline{p}$ is normal with mean 0.45 and standard deviation 0.0352.

Explanation of Solution

It is given that is the sample proportion of entrepreneurs whose first startup was at 30 years or more is $p=0.45$ and the sample size $n=200$

The sampling distribution of the proportion is approximately normal if $np\ge 5$  and $n(1-p)\ge 5.$

Verify the conditions:

$\begin{array}{c}np=200\times 0.45\\ =90\\ \ge 5\end{array}$

And $\begin{array}{c}n(1-p)=200\times (1-0.45)\\ =110\\ \ge 5\end{array}$

The conditions are satisfied. Therefore, the sampling distribution of the proportion is normal.

The mean of the $\overline{p}$ is $E(\overline{p})=p$ and standard deviation of $\overline{p}$ is ${\sigma }_{\overline{p}}=\sqrt{\frac{p(1-p)}{n}}$

Mean and standard deviation are calculated as below

$\begin{array}{c}E(\overline{p})=p\\ =0.45\end{array}$

$\begin{array}{c}{\sigma }_{\overline{p}}=\sqrt{\frac{p(1-p)}{n}}\\ =\sqrt{\frac{0.45\times 0.55}{200}}\\ =0.0352\end{array}$

Thus, the sampling distribution of the proportion $\overline{p}$ of entrepreneurs whose first startup was at 30 years or more is normal with mean $E(\overline{p})=0.45$ and standard deviation ${\sigma }_{\overline{p}}=0.0352$

c.

To determine

Check whether the standard error of the sampling distributions of $\overline{p}$ obtained in part (a) and part (b) are different.

Answer to Problem 12P

The standard error of the sampling distributions of $\overline{p}$ obtained in part (a) and part (b) are precisely the same.

Explanation of Solution

The standard error of $\overline{p}$ computed in part (a) is 0.0352

The standard error of $\overline{p}$ computed in part (b) is 0.0352

It can be seen that standard error of $\overline{p}$ is exactly the same in part (a) and part (b). Since the numerator of the formula of standard error of $\overline{p}$ is $p(1-p)$ and whenever the value of $p(1-p)$ is same and the sample size is equal, the value of standard error will be the same.