Question

a researcher records the repair cost for 4 randomly selected refrigerators. a sample mean of $85.03 and standard deviation of $22.34 are subsequently computed. determine the 80 % 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.

Explanation:

Step1: Identify distribution and parameters

We have a small sample (\(n = 4\)), population is normal, and we know the sample standard deviation, so we use the t - distribution. The degrees of freedom \(df=n - 1=4 - 1 = 3\). The confidence level is \(80\%\), so the significance level \(\alpha=1 - 0.80 = 0.20\). Since it's a two - tailed test for the confidence interval, we use \(\alpha/2=0.10\).

Step2: Find critical value

We need to find \(t_{\alpha/2,df}\), where \(\alpha/2 = 0.10\) and \(df = 3\). Using a t - table or a calculator with t - distribution functions, for \(df = 3\) and the upper - tail probability of \(0.10\), the critical value \(t_{0.10,3}\) is approximately \(1.638\).

Answer:

\(1.638\)