Question

A consumer affairs investigator records the repair cost for 23 randomly selected refrigerators. a sample mean of $88.05 and standard deviation of $10.52 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.

Answer 1

To construct an 80% confidence interval for the mean repair cost, use the formula: Confidence Interval = sample mean ± (critical value) * (standard deviation / sqrt(sample size)).

Step-by-step explanation:

To construct an 80% confidence interval for the mean repair cost, we can use the formula:

Confidence Interval = sample mean ± (critical value) * (standard deviation / sqrt(sample size))

In this case, the sample mean is $88.05, the standard deviation is $10.52, the sample size is 23, and we want an 80% confidence level. We can find the critical value using a t-distribution table or a t-distribution function in statistical software. For an 80% confidence level and 22 degrees of freedom (sample size minus 1), the critical value is approximately 1.7207.

Plugging in the values, we get:

Confidence Interval = $88.05 ± 1.7207 * ($10.52 / sqrt(23))

Simplifying the expression, we get:

Confidence Interval = $88.05 ± $2.4865

Rounding to two decimal places, the 80% confidence interval for the mean repair cost is $85.56 to $90.54.