Question

A student records the repair cost for 13 randomly selected TVs. A sample mean of $72.19 and standard deviation of $15.88 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

1) Find the critical value that should be used in constructing the confidence interval. Round you answer to three decimal places.

Answer 1

To find the 98% confidence interval for mean TV repair costs, you calculate the margin of error using the critical t-value for 12 degrees of freedom from the t-distribution and the sample's standard deviation and size.

Step-by-step explanation:

Calculating the 98% Confidence Interval for Mean Repair Cost

To calculate the 98% confidence interval for the mean repair cost of TVs, we need to take into account the given sample mean, sample standard deviation, and the sample size. Since the population standard deviation is unknown and the sample size is less than 30, we use the t-distribution to find the critical t-value.

To find the critical value for a 98% confidence interval with 12 degrees of freedom (n-1 where n=13), we look up the t-distribution table or use statistical software to find t* value. For a two-tailed test at 98% confidence, we have a 1% significance level on each tail, so we look for the t-value corresponding to a 99% confidence level. Let's assume the t-value is approximately 2.627.

Next, we use the formula for the confidence interval:

1. Calculate the margin of error (ME): ME = t* * (s/√n), where s is the sample standard deviation, and n is the sample size.
2. Substitute the values: ME = 2.627 * (15.88/√13).
3. The margin of error is calculated, and we then add and subtract it from the sample mean to obtain the confidence interval.

The confidence interval provides a range in which we are reasonably sure the true population mean of TV repair costs lies.

The critical value and margin of error are key concepts in constructing a confidence interval.