Question

1. in a random sample of 192 college students, 129 had part - time jobs. find the margin of error for the 95% confidence interval used to estimate the population proportion.

a) 0.0664
b) 0.116
c) 0.0598
d) 0.00225

Explanation:

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

$\hat{p}=\frac{x}{n}=\frac{129}{192} = 0.6719$

Step2: Determine $z$-value for 95% confidence interval

For 95% confidence interval, $z = 1.96$

Step3: Calculate margin of error formula

The formula for margin of error $E$ for a proportion is $E=z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.6719$, $n = 192$ and $z=1.96$ into the formula:
$E=1.96\sqrt{\frac{0.6719\times(1 - 0.6719)}{192}}$
$E=1.96\sqrt{\frac{0.6719\times0.3281}{192}}$
$E=1.96\sqrt{\frac{0.2204}{192}}$
$E=1.96\sqrt{0.001148}$
$E=1.96\times0.0339$
$E = 0.0664$

Answer:

A. 0.0664