Chapter 9.2, Problem 9.22E

a)

To determine

To find the mean and standard deviation of the proportion $\widehat{p}$ .

Answer to Problem 9.22E

Mean = 0.8 and standard deviation = 0.0894

Explanation of Solution

Given:

Sample size = n = 20

Population proportion = p = 0.80

Formula:

Mean:

  ${\mu }_{\widehat{p}}=p$

Standard deviation:

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

Calculation:

The mean = ${\mu }_{\widehat{p}}=p=0.8$

The standard deviation:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{0.8(1-0.8)}{20}}\approx 0.0894$

b)

To determine

To find sample proportion $\widehat{p}$

Answer to Problem 9.22E

Sample proportion = 0.4

Explanation of Solution

Given:

  $x=8$

Formula:

  $\widehat{p}=\frac{x}{n}$

Calculation:

Using the formula,

  $\widehat{p}=\frac{8}{20}=0.4$

c)

To determine

To sketch the graph for part b)

Explanation of Solution

Given:

  $\widehat{p}=0.4$

Calculation:

Following is the graph which shows sampling distribution of sample proportion = 0.4:

d)

To determine

To explain how likely is it that you would get a value of $\widehat{p}$ .

Explanation of Solution

Given:

  $\widehat{p}=0.4$

Population proportion = p = 0.8

Formula:

  $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

Calculation:

First need to find z-score:

  $z=\frac{0.4-0.8}{\sqrt{\frac{0.8(1-0.8)}{20}}}=-4.47$

Therefore,

  $\begin{array}{l}P(\widehat{p}<0.4)=P(Z<-4.47)\\ P(\widehat{p}<0.4)=0\end{array}$

Using excel formula, =NORMSDIST(-4.47)

Hence, the sample proportion of 0.4 is unlikely to occur by chance as probability is less than 0.05. Thus, the player is not an 80% free-throw shooter.