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CI: weekly study hours of students

Short answer

Using z-interval with SE=9.2/√50=1.301: 99% CI = 41.5±2.576×1.301 = (38.15, 44.85) hours; 95% CI = 41.5±1.96×1.301 = (38.95, 44.05) hours.

Suppose that in the previous year all graduates at a certain university reported the number of hours spent on their studies during a certain week; the average was 40 hours and the standard deviation was 10 hours. Suppose we want to investigate the problem whether students now are studying more than they used to. This year a random sample of n = 50 students is selected. Each student in the sample was interviewed about the number of hours spent on his/her study. This experiment produced the following statistics: x̄ = 41.5 hours, s = 9.2 hours. i) Estimate μ, the mean number of hours spent on study, using a 99% confidence interval. ii) Construct a 95% confidence interval for mean weekly time spent on study of all students in the university this year.

Here n = 50 (large) and σ is unknown, so s = 9.2 is used as its estimate and the normal (z) interval x̄ ± z·s/√n is applied.

x̄ = 41.5, s = 9.2, n = 50
SE = s/√n = 9.2/√50 = 1.3011

i) 99% confidence interval

z = 2.576

margin = 2.576 × 1.3011 = 3.352
μ = 41.5 ± 3.352 = (38.15, 44.85)

ii) 95% confidence interval

z = 1.960

margin = 1.960 × 1.3011 = 2.550
μ = 41.5 ± 2.550 = (38.95, 44.05)

Answer: 99% CI = (38.15, 44.85) hours; 95% CI = (38.95, 44.05) hours.

Remark: with the t-distribution (d.f. = 49, t = 2.680 and 2.010) the intervals are slightly wider: (38.01, 44.99) and (38.89, 44.11). Both intervals contain last year's mean of 40 hours, so there is no evidence that students now study more.

✓ Answer checked by Claude Sonnet 5.5

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