Math / Data & Statistics

QuestionA manager records the repair cost for 12 randomly selected refrigerators. A sample mean of $\$ 95.47$ and standard deviation of $\$ 18.62$ are subsequently computed. Determine the $90 \%$ confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal.
Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Tables Keypad
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Studdy Solution
For a $90\%$ confidence interval, the level of significance ( $\alpha$ ) is $0.10$ . Since the confidence interval is two-tailed, divide the significance level by 2:
$\alpha/2 = 0.10/2 = 0.05$
Using a $t$ -distribution table or calculator, find the critical value $t^$ for $df = 11$ and $\alpha/2 = 0.05$ .
The critical value $t^$ is approximately:
$\boxed{1.796}$

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