An economist would like to estimate the average household income in St. Lucie county. The mean household income of a random sample of 193 families in St. Lucie county was found to be $\$ 45,267$. The standard deviation of the household incomes of all households in St. Lucie county is known to be $\$ 955$.

Using a $96 \%$ confidence level, determine the margin of error, $E$, and a confidence interval for the average household income in St. Lucie county. Report the confidence interval using interval notation. Round solutions to two decimal places, if necessary.

The margin of error is given by $E=$

A $96 \%$ confidence interval is given by
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Step 1 ：First, we are given the following values: the sample size \(n = 193\), the sample mean \(\overline{x} = \$45267\), the standard deviation \(\sigma = \$955\), and the confidence level \(0.96\).

Step 2 ：We calculate the Z-score for a 96% confidence level. The Z-score is approximately \(2.0537489106318225\).

Step 3 ：Next, we calculate the margin of error using the formula \(E = Z \times \frac{\sigma}{\sqrt{n}}\). The margin of error is approximately \$141.18.

Step 4 ：Finally, we calculate the confidence interval using the formula \((\overline{x} - E, \overline{x} + E)\). The confidence interval is approximately (\$45125.82, \$45408.18).

Step 5 ：This means that we are 96% confident that the true average household income in St. Lucie county is between \$45125.82 and \$45408.18.

Step 6 ：Final Answer: The margin of error is \(\boxed{\$141.18}\). A 96% confidence interval for the average household income in St. Lucie county is \(\boxed{(\$45125.82, \$45408.18)}\).