Question

A supervisor records the repair cost for 22 randomly selected VCRs. A sample mean of $75.50 and standard deviation of $18.07 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the VCRs. Assume the population is approximately normal.

Answer 1

Answer:

 The 99% confidence interval for the mean repair cost for the VCRs

(65.801, 85.199)

Step-by-step explanation:

Explanation:-

Mean of the sample(x⁻) = $75.50

Given the standard deviation of the sample (S) =  $18.07

Given the size of the sample 'n' = 22

Degrees of freedom = n-1 =22-1 =21

critical value t₍₀.₀₁, ₂₁₎ = 2.5176

The 99% confidence interval for the mean repair cost for the VCRs

$(x^{-} - t_{0.01} \frac{S.D}{\sqrt{n} } , x^{-} +t_{0.01} \frac{S.D}{\sqrt{n} })$

$(75.50 - 2.5176 \frac{18.07}{\sqrt{22} } , 75.50 +2.5176 \frac{18.07}{\sqrt{22} })$

( 75.50 - 9.699 , 75.50 + 9.699 )

(65.801, 85.199)

Final answer:-

The 99% confidence interval for the mean repair cost for the VCRs

(65.801, 85.199)