An automotive repair shop has determined that the average service time on an automobile is 20 hours with a standard deviation of 320 minutes. A random sample of 160 services is selected. Assuming the population consists of 2,000 services. Determine the standard error of the mean. Hint: Use the finite population correction factor.

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BlackTom AI · Accepted Answer

We start by identifying the given quantities and units. The population standard deviation is 320 minutes, the sample size is n = 160, the population size is N = 2000, and the sample mean is based on these 160 observations. To account for the finite population, we use the finite population correction factor (FPC): sqrt((N - n) / (N - 1)).

Step 1: Compute the ordinary standard error (ignoring FPC) in the same units as the data. SE_ordinary = sigma / sqrt(n) = 320 / sqrt(160).
- sqrt(160) ≈ 12.6491, so SE_ordinary ≈ 320 / 12.6491 ≈ 25.30 minutes.

Step 2: Apply the finite population correction. FPC = sqrt((N - n) / (N - 1)) = sqrt((2000 - 160) / (2000 - 1)) = sqrt(1840 / 1999).
- 1840 / 1999 ≈ 0.92046, and sqrt(0.92046) ≈ 0.9594.

Step 3: Multiply the ordinary SE by the FPC to obtain the standard error of the mean with finite population correction. SE_FPC = SE_ordinary × FPC ≈ 25.30 × 0.9594 ≈ 24.28 minutes.

Step 4: If desired, convert to hours for interpretation. 24.28 minutes ≈ 0.4047 hours. Since the problem’s data are in minutes for the spread, the standard error of the mean is about 24.3 minutes when reported in minutes.

Summary note: The finite population correction reduces the standard error from the unconstrained 25.3 minutes to about 24.3 minutes due to sampling from a relatively small population (N = 2000 with n = 160). The key formula used is SE_FPC = (sigma / sqrt(n)) × sqrt((N - n) / (N - 1), and all quantities were kept in minutes to maintain consistency.

22803 Quantitative Skills in Accounting and Finance - Autumn 2025 Quiz 3

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