Chapter 9.1, Problem 9.18E

To determine

To find the 95% confidence interval using quick method and method used in chapter.

Explanation of Solution

Given:

Confidence level = C = 0.95

n = 600

  $x$ = 68

Formula:

Confidence interval for population proportion is,

  $CI=\widehat{p}\mp E$

Margin of error by quick method:

  $E=Zc\times \frac{1}{\sqrt{n}}$

Margin of error by method used in chapter:

  $E=Zc\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

Calculation:

Confidence interval by quick method:

Sample proportion:

  $\widehat{p}=\frac{x}{n}=\frac{68}{600}=0.1133$

Now need to find Zc

The level of significance is = a =0.05

Therefore, Zc = Z _a/2 = 1.96 ..Using standard normal table

Margin of error can be calculated as,

  $E=Zc\times \frac{1}{\sqrt{n}}=1.96\times \frac{1}{\sqrt{600}}=0.0800$

Therefore, the 95% confidence interval for population proportion is,

  $\begin{array}{l}\widehat{p}-E=0.1133-0.0800=0.0333\\ \widehat{p}+E=0.1133+0.0800=0.1933\end{array}$

Hence, confidence interval for population proportion is (0.0333, 0.1933)

Now we can find confidence interval using method used in chapter:

Margin of error:

  $E=1.96\times \sqrt{\frac{0.1133(1-0.1133)}{600}}=0.0254$

The 95% confidence interval is,

  $\begin{array}{l}\widehat{p}-E=0.1133-0.0254=0.0879\\ \widehat{p}+E=0.1133+0.0254=0.1387\end{array}$

Therefore, 95% confidence interval for population proportion is, (0.0879 , 0.1387)

Now, comparing two margin of errors:

0.0800 − 0.0254 = 0.0546

The quick method margin of error is 0.0546 larger than the margin of error used by method of this chapter.