Chapter 9.3, Problem 108E

To determine

To constructand explain a 95 percent confidence interval for p.

Answer to Problem 108E

  $0.5397<p<0.7063$

Explanation of Solution

Given:

  $\begin{array}{c}n=130\\ \widehat{p}=62.3\%=0.623\\ p=0.5\end{array}$

Formula used:

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

Calculation:

There are total three conditions when performing a significance test about a population proportion: Random, Normal and Independent.

Random: Satisfied, because of given random sample.

Normal: Satisfied, because the number of successes $np=130\left(0.5\right)=65$ and the number of failures $n\left(1-p\right)=130\left(1-0.5\right)=65$ are both greater than 10.

Independent: Satisfied, because the sample size of 130 is less than 10% of the population size.

Therefore it is observed that all three conditions have been met.

For confidence level $1-\alpha =0.95$ , determine $z_{\alpha /2}=z_{0.025}$ using table II (look up 0.025 in the table, the z-score is then they found z-score with opposite sign):

  $z_{\alpha /2}=1.96$

The margin of error is then:

  $\begin{array}{c}E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\\ =1.96\times \sqrt{\frac{0.623\left(1-0.623\right)}{130}}\\ =0.0833\end{array}$

The confidence interval then becomes:

  $0.5397=0.623-0.0833=\widehat{p}-E<p<\widehat{p}+E=0.623+0.0833=0.7063$