Chapter 7.2, Problem 36E

(a)

To determine

To Explain: the means of the sampling distribution of $\widehat{p}$

Answer to Problem 36E

0.90

Explanation of Solution

Given:

  $\begin{array}{c}p\text{=90\% =0}\text{.90}\\ n\text{ =15}\end{array}$

Calculation:

The mean of the sampling distribution of the sample proportions is same to the population proportion $p$ .

  ${\mu }_{\widehat{p}}=p=90\%=0.90$

(b)

To determine

To Calculate: and interpret the standard deviation of the sampling distribution of $\widehat{p}$

Answer to Problem 36E

0.0775

Explanation of Solution

Given:

  $\begin{array}{c}p\text{=90\% =0}\text{.90}\\ n\text{ =15}\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

The mean of the sampling distribution of the sample proportions is

  ${\mu }_{\widehat{p}}=p=90\%=0.90$

The standard deviation of the sampling distribution of the sample proportion is

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.90\left(1-0.90\right)}{15}}\\ =0.0775\end{array}$

The proportion of males among the $\text{15}$ randomly selected workers in the sample varies on average by $0.0775$ from the mean of $0.90$ .

(c)

To determine

To Explain: the shape of the sampling distribution of $\widehat{p}$ .

Answer to Problem 36E

Not about normal

Skewed to left

Explanation of Solution

Given:

  $\begin{array}{c}p\text{=90\% =0}\text{.90}\\ n\text{ =15}\end{array}$

Calculation:

The sampling distribution of the sample proportions $\widehat{p}$ is approximately Normal is the large counts condition is satisfied, that is , when $np\ge 10$ and $n\left(1-p\right)\ge 10$ .

  $\begin{array}{c}np=15\left(0.90\right)=13.5\ge 10\\ n\left(1-n\right)=15\left(1-0.90\right)=1.5<10\end{array}$

Since the population proportion of $0.90$ is closer to 1 than 0, the sample proportion is more likely to be close to 1 than 0 and therefore the sampling distribution of the sample proportions will be skewed to the left