Question

Fuel Expenditures. In estimating the mean monthly fuel expenditure, $\mu$, per household vehicle, the Energy Information $\mathrm{Ad}-$ ministration takes a sample of size $6841$. Assuming that $\sigma =\$20.65$. determine the margin of error in estimating $\mu$ at the $95\%$ level of confidence.

Short answer

Margin of Error is $956\%$and confidence level is $\$0.49$.

Step-by-step solution

Given Information

It is given that

$n=6841$

$\sigma =\$20.65$

Calculations to be done at$95\%$confidence level.

Determination of Za2

Margin of Error at $95\%$confidence level is

$1-\alpha =0.95$

$\Rightarrow \alpha =0.05$

$\Rightarrow \frac{\alpha }{2}=0.025$

Hence, $Z_{\frac{a}{2}}=Z_{0.025}=1.96$

Calculation of Margin of Error

At$95\%$ confidence level, Margin of Error is
$E=Z_{0.025}\times \frac{\sigma }{\sqrt{n}}$

$E=1.96\times \frac{20.65}{\sqrt{6841}}=0.49$

Required Margin of Error is$\$0.49$