Chapter 6, Problem 14P

a.

To determine

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

Answer to Problem 14P

The sampling distribution of $\overline{p}$ for a sample of 300 is normal with mean 0.42 and standard deviation 0.0285.

Explanation of Solution

It is given that the sample proportion of doctors who think their patients receive unnecessary medical care is $p=0.42$ and the sample size $n=300$

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

Verify the conditions:

$\begin{array}{c}np=300\times 0.42\\ =126\\ \ge 5\end{array}$

And $\begin{array}{c}n(1-p)=300\times (1-0.42)\\ =174\\ \ge 5\end{array}$

The condition is 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 doctors who think their patients receive unnecessary medical care.

The mean of $\overline{p}$ is $E(\overline{p})=p=0.42$

The standard deviation of $\overline{p}$ is ${\sigma }_{\overline{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.42\times 0.58}{300}}=0.0285$

Thus, the sampling distribution of the proportion $\overline{p}$ of doctors who think their patients receive unnecessary medical care is normal with mean $E(\overline{p})=0.42$ and standard deviation ${\sigma }_{\overline{p}}=0.0285$

b.

To determine

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

Answer to Problem 14P

The sampling distribution of $\overline{p}$ for a sample of 500 is normal with mean 0.42 and standard deviation 0.0221.

Explanation of Solution

It is given that the sample proportion of doctors who think their patients receive unnecessary medical care is $p=0.42$ and the sample size is given as $n=500$

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

Verify the conditions:

$\begin{array}{c}np=500\times 0.42\\ =210\\ \ge 5\end{array}$

And $\begin{array}{c}n(1-p)=500\times (1-0.42)\\ =290\\ \ge 5\end{array}$

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 computed as below.

$E(\overline{p})=p=0.42$

${\sigma }_{\overline{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.42\times 0.58}{500}}=0.0221$

Thus, for a random sample of 500 doctors, the sampling distribution of the proportion $\overline{p}$ of doctors who think their patients receive unnecessary medical care is normal with mean $E(\overline{p})=0.42$ and standard deviation ${\sigma }_{\overline{p}}=0.0221$

c.

To determine

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

Answer to Problem 14P

The sampling distribution of $\overline{p}$ for a sample of 1000 is normal with mean 0.42 and standard deviation 0.0156.

Explanation of Solution

It is given that the sample proportion of doctors who think their patients receive unnecessary medical care is $p=0.42$ and the sample size is given as $n=1000$

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

Verify the conditions:

$\begin{array}{c}np=1000\times 0.42\\ =420\\ \ge 5\end{array}$

And $\begin{array}{c}n(1-p)=1000\times (1-0.42)\\ =580\\ \ge 5\end{array}$

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 of $\overline{p}$ are computed as below.

$E(\overline{p})=p=0.42$

${\sigma }_{\overline{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.42\times 0.58}{1000}}=0.0156$

Thus, for a random sample of 1000 doctors, the sampling distribution of the proportion $\overline{p}$ of doctors who think their patients receive unnecessary medical care is normal with mean $E(\overline{p})=0.42$ and standard deviation ${\sigma }_{\overline{p}}=0.0156$

d.

To determine

Establish the smallest standard error of $\overline{p}$ obtained.

Explanation of Solution

The standard error of $\overline{p}$ for a sample of 300 doctors, computed in part (a) is 0.0285

The standard error of $\overline{p}$ for a sample of 500 doctors computed in part (b) is 0.0221

The standard error of $\overline{p}$ for a sample of 1000 doctors computed in part (c) is 0.0156

It can be noted that standard error of $\overline{p}$ is lowest in part (c) and it can also be observed that $\overline{p}$ is same in all the parts and the sample size is largest in part (c).