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

so we are given that margin of error is 2

Answer 2

yes

Answer 3

$$

\Large{
\text{confidence interval for sample mean is: } \\
\bar x \pm z^* \frac{\sigma }{\sqrt n}

\\ Margin~ of~ error = z_{.05/2} \cdot \frac{\sigma }{\sqrt n}
\\ \therefore \\
if ~ ~2 = 1.96 \cdot \frac{\sigma }{\sqrt n}
\\ then ~~ \frac {2}{1.96 } =
\frac{\sigma }{\sqrt n}
\\ \therefore
ME = z_{\frac{.02}{2}} \cdot \frac{\sigma }{\sqrt n} = z_{\frac{.02}{2}} \cdot \frac {2}{1.96 }
}
$$

Answer 4

ok! what is the final answer though i am not sure how to compute it

Answer 5

so you just need the critical score for 98% , two tailed

Answer 6

$$\Large{
z_{\alpha/2}\\~~\\
z_{.10/2} = 1.645\\
z_{.05/2} = 1.960\\
z_{.02/2} = 2.326

}
$$