Chapter 9.2, Problem 9.28E

a)

To determine

To find null and alternative hypotheses.

Answer to Problem 9.28E

  $\begin{array}{l}{H}_0:p=0.5\\ H_a:p>0.5\end{array}$

Explanation of Solution

Given:

The claim is proportion is more than 0.5

The claim can be either in null or alternative hypothesis. Therefore, in this case, the null and alternative hypotheses:

  $\begin{array}{l}{H}_0:p=0.5\\ H_a:p>0.5\end{array}$

b)

To determine

To explain the sampling distribution.

Explanation of Solution

Given:

Sample size = n = 1000

Population proportion = p = 0.50

Formula:

Mean:

  ${\mu }_{\widehat{p}}=p$

Standard deviation:

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

Calculation:

The sampling distribution of the sample proportion is approximately normal as sample size is large.

Therefore, mean of the sampling distribution of sample proportion:

The mean = ${\mu }_{\widehat{p}}=p=0.5$

The standard deviation:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{0.50(1-0.50)}{1000}}=0.0158$

c)

To determine

To find the p-value.

Answer to Problem 9.28E

P-value = 0

Explanation of Solution

Given:

  $\widehat{p}=\frac{x}{n}=\frac{640}{1000}=0.64$

Population proportion = p = 0.5

n = 1000

Formula:

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

Calculation:

First need to find z-score:

  $z=\frac{0.64-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=8.85$

Therefore,

  $\begin{array}{l}P(Z>8.85)=1-P(Z<8.85)\\ P(Z>8.85)=1-1=0\end{array}$

Using excel formula, =NORMSDIST(8.85)

Therefore, p-value = 0

d)

To determine

To explain the conclusion.

Answer to Problem 9.28E

There is sufficient evidence to support the claim that the majority of adults have this view of alternative affirmative action.

Explanation of Solution

Given:

P-value = 0

As p-value = 0 which is < 0.05, reject H0. There is sufficient evidence to support the claim that the majority of adults have this view of alternative affirmative action.