Question

If the margin of error in an estimate for the mean weight of a shipment is + or -2 pounds at a confidence level of 95 percent, what will be the margin of error at a confidence level of 98 percent? Be sure to show how you arrived at your answer

Answer 1

For a \((1-\alpha)\%\) confidence interval, the margin of error is given by
\[ME=Z_{\alpha/2}\frac{\sigma}{\sqrt n}\]
where \(\dfrac{\sigma}{\sqrt n}\) is the standard error (standard deviation divided by the square root of the sample size) and \(Z_{\alpha/2}\) is the critical value for the confidence level.

For an interval of any confidence level, the standard error will be the same if the standard deviation and sample size remain constant. You're given this equation:
\[2=Z_{.025}\frac{\sigma}{\sqrt n}~~\iff~~2=1.96\frac{\sigma}{\sqrt n}\]since 1.96 is the cutoff for a 95% CI. Solve for the standard error.

Once you have that, you have to find the margin of error at a 98% CI:
\[ME=Z_{.01}\frac{\sigma}{\sqrt n}~~\iff~~ME=2.33\frac{\sigma}{\sqrt n}\]
since 2.33 is the approximate cutoff for 98% confidence.