The professor of a large probability class randomly selected 8 students and asked the amount of time (in hours) spent for her course per week. The data are given below.
7 9 5 12 14 6 10 3
Estimate the standard error of the estimated time spent in a week for this course by the students who are taking this course (round off to first decimal place).

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BlackTom AI · Accepted Answer

We start by identifying the data and the quantity to estimate. The eight observed weekly times are: 7, 9, 5, 12, 14, 6, 10, 3 hours.

First, compute the sample mean: sum = 66, n = 8, so mean = 66/8 = 8.25 hours.

Next, determine the sample standard deviation s. Calculate each deviation from the mean and square it:
- (7 − 8.25)² = 1.5625
- (9 − 8.25)² = 0.5625
- (5 − 8.25)² = 10.5625
- (12 − 8.25)² = 14.0625
- (14 − 8.25)² = 33.0625
- (6 − 8.25)² = 5.0625
- (10 − 8.25)² = 3.0625
- (3 − 8.25)² = 27.5625
Sum of squared deviations = 95.5.
With n − 1 = 7, the sample variance is s² = 95.5 / 7 ≈ 13.642857, so s ≈ √13.642857 ≈ 3.693.

The standard error of the mean (the estimate of the population mean) is SE = s / √n. Here, SE ≈ 3.693 / √8. Since √8 ≈ 2.828, SE ≈ 3.693 / 2.828 ≈ 1.305.

Rounding to one decimal place as requested, the estimated standard error is about 1.3 hours.

To summarize, the small sample standard deviation combined with the sample size leads to a modest standard error around 1.3 hours, indicating the average weekly time estimate would vary by about 1.3 hours from sample to sample.

AMS 310 Section 7.1 Quiz

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