Chapter 9, Problem 9.69RE

a)

To determine

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

Answer to Problem 9.69RE

There is not sufficient evidence to support the claim that the majority of U.S. teens would say that young people should wait to have sex until they are married.

Explanation of Solution

Given:

Sample size = n = 439

  $x$ = 246

Confidence level = 0.95

Claim: Proportion is higher than 50%.

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Test statistic:

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

Calculation:

The sample proportion is,

  $\widehat{p}=\frac{246}{439}=0.5604$

Null and alternative hypothesis:

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

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

Test statistic:

  $z=\frac{0.5604-0.5}{\sqrt{0.5(1-0.5)/439}}=2.53$

Test statistic z = 2.53

Using the P-value approach:

  $\begin{array}{l}P\left(Z>2.53\right)=1-P\left(Z<2.53\right)\\ P\left(Z>2.53\right)=1-0.9943=0.0057\end{array}$

The p-value is p = 0.0057

Using excel formula, =NORMSDIST(2.53)

And a = 0.05

Since p = 0.0057 <0.05, it is concluded that the Null Hypothesis is rejected.

Conclusion: There is not sufficient evidence to support the claim that the majority of U.S. teens would say that young people should wait to have sex until they are married.

b)

To determine

To find the confidence interval for population proportion.

Answer to Problem 9.69RE

The 95% confidence interval is (0.514,0.607)

Explanation of Solution

Given:

Sample proportion: $\widehat{p}=0.5604$

Confidence level = c = 95% = 0.95

Sample size = n = 439

Formula:

Margin of error:

  $E=Z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

Confidence interval:

  $(\widehat{p}-E,\widehat{p}+E)$

Confidence level = c = 95% = 0.95

So, level of significance = a = 1-0.95 = 0.05

Z _a/2 = 1.96 …Using excel function, =ABS(NORMSINV(1-0.05/2))

Margin of error is,

  $E=1.96\times \sqrt{\frac{0.5604(1-0.5604)}{439}}=0.046$

Therefore, the 95% confidence interval is,

  $(0.5604-0.046,0.5604+0.046)=(0.514,0.607)$

Hence, we are 95% confidence that the true population proportion of heads is between 0.514 and 0.607.

c)

To determine

To explain whether from inference a) and b), teens are actually telling the truth.

Answer to Problem 9.69RE

No.

Explanation of Solution

Before using inference, we always assume that the truth was told. Therefore, inference does not determine whether truth actually told or not. Its just try to give evidence for given claim.