Question

use the sample data and confidence level given below to complete parts (a) through (d). a research institute poll asked respondents if they felt vulnerable to identity theft. in the poll, n = 978 and x = 525 who said \yes.\ use a 99% confidence level. click the icon to view a table of z - scores. a) find the best point estimate of the population proportion p. 0.537 (round to three decimal places as needed.) b) identify the value of the margin of error e. e = (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion $\hat{p}$

$\hat{p}=\frac{x}{n}=\frac{525}{978}\approx0.537$ (already given for part a)

Step2: Determine $z -$score for 99% confidence level

For a 99% confidence level, the significance level $\alpha = 1- 0.99=0.01$, and $\alpha/2=0.005$. Looking up in the $z -$score table, $z_{\alpha/2}=z_{0.005} = 2.576$.

Step3: Calculate margin of error $E$

The formula for the margin of error for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.537$, $n = 978$ and $z_{\alpha/2}=2.576$ into the formula. First, calculate $1-\hat{p}=1 - 0.537 = 0.463$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.537\times0.463}{978}=\frac{0.248631}{978}\approx0.000254$. $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\approx\sqrt{0.000254}\approx0.0159$. And $E=2.576\times0.0159\approx0.041$

Answer:

$E = 0.041$