Chapter 9, Problem 9.70RE

a)

To determine

To find the null and alternative hypotheses.

Answer to Problem 9.70RE

  $\begin{array}{l}{H}_o:p=\frac{1}{3}\\ H_a:p>\frac{1}{3}\end{array}$

Explanation of Solution

Given:

Sample size = n = 1500

  $x$ = 660

Confidence level = 0.95

Claim: Proportion is more than $\frac{1}{3}$

Null and alternative hypothesis:

  $\begin{array}{l}{H}_o:p=\frac{1}{3}\\ H_a:p>\frac{1}{3}\end{array}$

This corresponding to a right-tailed test, for which a z-test for one population proportion needs to be used.

b)

To determine

To find the sampling distribution of proportion and draw a Normal curve.

Answer to Problem 9.70RE

  ${\mu }_{\widehat{p}}=0.333$

  ${\sigma }_{\widehat{p}}=0.012$

Explanation of Solution

Given:

Sample size = n = 1500

  $x$ = 660

Confidence level = 0.95

Formula:

Mean:

  ${\mu }_{\widehat{p}}=p$

Standard deviation:

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

2 standard deviation from the mean:

  ${\mu }_{\widehat{p}}\mp 2{\sigma }_{\widehat{p}}$

Calculation:

The mean = ${\mu }_{\widehat{p}}=p=\frac{1}{3}=0.333$

The standard deviation:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{0.333(1-0.333)}{1500}}\approx 0.012$

Therefore,

  $\begin{array}{l}{\mu }_{\widehat{p}}-2{\sigma }_{\widehat{p}}=0.333-2\times 0.012=0.309\\ {\mu }_{\widehat{p}}+2{\sigma }_{\widehat{p}}=0.333+2\times 0.012=0.357\end{array}$

So, standard normal curve is,

  

c)

To determine

To draw a actual value of $\widehat{p}$ Normal curve.

Explanation of Solution

Given:

  ${\mu }_{\widehat{p}}=0.333$

  ${\sigma }_{\widehat{p}}=0.012$

Formula:

  $\widehat{p}=\frac{x}{n}$

Calculation:

  $\widehat{p}=\frac{660}{1500}=0.44$

The graph becomes,

  

d)

To determine

To perform hypothesis test for one sample proportion $\widehat{p}$ .

Answer to Problem 9.70RE

There is sufficient evidence to support the claim that more than 1/3 of all adults would use alternative medicine if traditional medicine did not produce the results they wanted.

Explanation of Solution

Given:

  $\widehat{p}=0.44$

n=1500

Confidence level = 0.95

Formula:

Test statistic:

  $z=\frac{\widehat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}$

Calculation:

Null and alternative hypothesis:

  $\begin{array}{l}{H}_o:p=\frac{1}{3}\\ H_a:p>\frac{1}{3}\end{array}$

This corresponding to a one-tailed test, for which a z-test for one population proportion needs to be used.

Test statistic:

  $z=\frac{0.44-0.333}{\sqrt{0.333(1-0.333)/1500}}=8.76$

Test statistic z = 8.76

Using the P-value approach:

  $\begin{array}{l}P\left(Z>8.76\right)=1-P\left(Z<8.76\right)\\ P\left(Z>8.76\right)=1-1=0\end{array}$

The p-value is p = 0

Using excel formula, =NORMSDIST(8.76)

And a = 0.05

Since p = 0<0.05, it is concluded that the Null Hypothesis isrejected.

Conclusion: There is sufficient evidence to support the claim that more than 1/3 of all adults would use alternative medicine if traditional medicine did not produce the results they wanted.