Chapter 9.2, Problem 9.27E

a)

To determine

To find null and alternative hypotheses.

Answer to Problem 9.27E

  $\begin{array}{l}{H}_0:p=0.30\\ H_a:p\ne 0.30\end{array}$

Explanation of Solution

Given:

The claim is the probability that light turns red on any press of button is 0.30.

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

  $\begin{array}{l}{H}_0:p=0.30\\ H_a:p\ne 0.30\end{array}$

b)

To determine

To explain the sampling distribution of null hypothesis.

Explanation of Solution

Given:

Sample size = n = 100

Population proportion = p = 0.30

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 = ${\mu }_{\widehat{p}}=p=0.30$

The standard deviation:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{0.30(1-0.30)}{100}}\approx 0.0458$

c)

To determine

To find the p-value.

Answer to Problem 9.27E

P-value = 0.0292

Explanation of Solution

Given:

  $\widehat{p}=\frac{x}{n}=\frac{20}{100}=0.2$

Population proportion = p = 0.8

n = 100

Formula:

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

Calculation:

First need to find z-score:

  $z=\frac{0.2-0.3}{\sqrt{\frac{0.3(1-0.3)}{100}}}=-2.18$

Therefore,

  $2\times P(Z<-2.18)=2\times 0.0146=0.0292$

Using excel formula, =NORMSDIST(-2.18)

Therefore, p-value = 0.0292

d)

To determine

To explain the conclusion.

Answer to Problem 9.27E

There is insufficient evidence to support the claim that the probability that light turns red on any press of button is 0.30.

Explanation of Solution

Given:

P-value = 0.0292

As p-value = 0.0292 which is < 0.05, reject H0. There is insufficient evidence to support the claim that the probability that light turns red on any press of button is 0.30.