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 2 of 2: construct the 80 % confidence interval. round your answer to two decimal places.

Explanation:

Step1: Identify the formula for confidence interval

For a small sample (n < 30) with a normal population, we use the t - distribution. The formula for the confidence interval for the population mean \(\mu\) is \(\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, \(n\) is the sample size, and \(t_{\alpha/2}\) is the t - critical value.

We know that \(\bar{x} = 85.03\), \(s=22.34\), \(n = 4\), and the confidence level is \(80\%\). The significance level \(\alpha=1 - 0.8=0.2\), so \(\alpha/2=0.1\). The degrees of freedom \(df=n - 1=4 - 1 = 3\).

Looking up the t - critical value in the t - distribution table, for \(df = 3\) and \(\alpha/2=0.1\), \(t_{0.1,3}=1.638\).

Step2: Calculate the margin of error (E)

The margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\). Substitute the values: \(t_{\alpha/2}=1.638\), \(s = 22.34\), \(n = 4\).

\(\sqrt{n}=\sqrt{4} = 2\), so \(\frac{s}{\sqrt{n}}=\frac{22.34}{2}=11.17\)

Then \(E=1.638\times11.17\approx18.20\) (rounded to two decimal places)

Step3: Calculate the confidence interval

The lower limit of the confidence interval is \(\bar{x}-E=85.03 - 18.20 = 66.83\)

The upper limit of the confidence interval is \(\bar{x}+E=85.03+18.20 = 103.23\)

Answer:

The \(80\%\) confidence interval is \((66.83, 103.23)\)