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A supervisor records the repair cost for 11 randomly selected refrigerators. A sample mean of $82.43 and standard deviation of $13.96 are subsequently computed. Determine the 99% 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.

Respuesta :

Answer:

Critical value z = 2.574

Confidence interval = ( $71.596 to $93.264)

Step-by-step explanation:

Mean x = $82.43

Standard deviation r = $13.96

Number of sample n = 11

Confidence interval of 99%

Critical value = z = t(a/2)

a = 1 - 0.99 = 0.01

a/2 = 0.01/2 = 0.005

z= t(0.005) = 2.57

z = 2.574

Confidence interval = x+/-(z×r/√n)

= 82.43 +/- (2.574 × 13.96/√11)

= 82.43 +/-10.834

Confidence interval = ( 71.596 to 93.264)

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