A random sample of 20 recent weddings in a country yielded a mean wedding cost of $\$ 26,369.38$. Assume that recent wedding costs, in this country are normally distributed with a standard deviation of $\$ 8500$. Complete parts (a) through (c) below.
a. Determine a $95 \%$ confidence interval for the mean cost, $\mu$, of all recent weddings in this country.

The $95 \%$ confidence interval is from $\$ \square$ to $\$ \square$.
(Round to the nearest cent as needed.)
example
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Step 1 ：Given values are: sample mean (\(x_{bar}\)) = \$26369.38, z-score (z) for 95% confidence level = 1.96, standard deviation (\(\sigma\)) = \$8500, and sample size (n) = 20.

Step 2 ：Calculate the margin of error using the formula: \(margin\_of\_error = z \times \frac{\sigma}{\sqrt{n}}\).

Step 3 ：Substitute the given values into the formula: \(margin\_of\_error = 1.96 \times \frac{8500}{\sqrt{20}}\), which gives \(margin\_of\_error \approx \$3725.29\).

Step 4 ：Calculate the confidence interval using the formula: \(lower\_limit = x_{bar} - margin\_of\_error\) and \(upper\_limit = x_{bar} + margin\_of\_error\).

Step 5 ：Substitute the values into the formula: \(lower\_limit = 26369.38 - 3725.29 \approx \$22644.09\) and \(upper\_limit = 26369.38 + 3725.29 \approx \$30094.67\).

Step 6 ：Final Answer: The 95% confidence interval for the mean cost, \(\mu\), of all recent weddings in this country is from \(\boxed{\$22644.09}\) to \(\boxed{\$30094.67}\).