Chapter 7.2, Problem 36E

a.

To determine

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

Answer to Problem 36E

The mean is 0.40.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Sample proportion}\left(\widehat{p}\right)=0.44\\ \text{Population proportion}\left(p\right)=0.40\\ \text{Sample size}\left(n\right)=1785\end{array}$

The mean of the sampling distribution can be calculated as:

  $\begin{array}{c}{\mu }_{\widehat{p}}=p\\ =0.40\end{array}$

Here, mean is 0.40 because of the unbiased estimator property.

b.

To determine

To find: The standard deviation of the sample proportion.

Answer to Problem 36E

The standard deviation is 0.0115954.

Explanation of Solution

Calculation:

The standard deviation of the sample proportion is computed as:

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.40\left(1-0.40\right)}{1785}}\\ =0.0115954\end{array}$

Thus, the standard deviation is 0.0115954.

c.

To determine

To find: Whether the normality assumptions are satisfied or not.

Explanation of Solution

Here,

  $\begin{array}{l}np=1785\left(0.40\right)=714>10\\ n\left(1-p\right)=1785\left(1-0.40\right)=1071>10\end{array}$

Thus, normal approximation could be used here.

d.

To determine

To find: The probability that 44% or more saying that they attended church or synagogue. Also, comment on the result of the poll.

Explanation of Solution

Calculation:

The probability that 44% or more saying that they attended church or synagogue. Also, comment on the result of the poll is computed as:

  $\begin{array}{c}P\left(\widehat{p}\ge 0.44\right)=P\left(Z\ge \frac{0.44-0.40}{0.0115954}\right)\\ =P\left(Z\ge 3.45\right)\\ =0.0003\end{array}$

Here, probability is less than 0.05. Thus, there is doubt on the occurrence of the event.