A consumer affairs investigator records the repair cost for 2222 randomly select

Question

A consumer affairs investigator records the repair cost for 2222 randomly selected TVs. A sample mean of $83.23$83.23 and standard deviation of $22.67$22.67 are subsequently computed. Determine the 95%95% confidence interval for the mean repair cost for the TVs. 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.

Step 2 of 2: Construct the 95% confidence interval. Round your answer to two decimal places.

Explanation / Answer

Here we will use the t statistic as population standard deviation is unknown and n<30 i.e. n = 22

degree of freedom = 22-1 = 21

Step 1: Critical value of t at 5% two sided and df of 21 = 2.080

Step 2: 95% CI of t = sample mean +- t0.025,21 (s/n0.25)

= 83.23+-2.080(22.67/220.5)

=(73.18, 93.28)

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