Chapter 8.2, Problem 42E

To determine

To construct confidence interval for the population proportion.

Answer to Problem 42E

The 95% confidence interval for population proportion is 0.7414 < p < 0.7786

Explanation of Solution

Given:

  $\begin{array}{l}n=2025\\ \widehat{p}=76\%=0.76\end{array}$

Confidence level = 0.95

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Margin of error:

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

The confidence interval:

  $(\widehat{p}-E,\widehat{p}+E)$

The condition of random selection is satisfied because the U.S. adults randomly selected.

The sample should be less than 10% of the population. Here, 2025 U.S. adults are less than 10% all the U.S. adults in population. Hence, this condition met.

The last condition is, $n\widehat{p}andn(1-\widehat{p})$ must be $\ge 10$ . Therefore, $n\widehat{p}=2025\times 0.76=1539$$\ge 10$ and $n(1-\widehat{p})=2025\times (1-0.76)=486$$\ge 10$ which is met. Hence, all the conditions of confidence interval are met.

The confidence level = 0.95

So, level of significance = a=0.05

The z_c=z _a/2 critical value = 1.96 …Using excel formula, =ABS(NORMSINV(0.05/2))

The margin of error is,

  $\begin{array}{l}E=1.96\times \sqrt{\frac{0.76(1-0.76)}{2025}}\\ E=0.0186\end{array}$

The confidence interval is,

  $\begin{array}{l}(0.76-0.0186,0.76+0.0186)\\ (0.7414,0.7786)\end{array}$

Hence, the 95% confidence interval for population proportion is 0.7414< p < 0.7786

We are 95% confident that the true population proportion of U.S. adults who would include the 9/11 attacks on their list of 10 historic events is 0.7414 and 0.7786