---
title: "CI: weekly study hours of students"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:study-hours-mean-ci-99-95"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/study-hours-mean-ci-99-95
site: BCT Engineering
---

# CI: weekly study hours of students

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2068 Jestha (old course), 6 marks
- Video: [6.3 CI for mean with unknown population SD | Basic Statistics | Confidence Intervals | UvA](https://www.youtube.com/watch?v=F-ERPYLvghI) — Research Methods and Statistics (FMG, UvA)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/study-hours-mean-ci-99-95
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

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.

## 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.

## Answer

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.

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

#### i) 99% confidence interval

z = 2.576

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

#### ii) 95% confidence interval

z = 1.960

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is the z-interval x̄ ± z·s/sqrt(n) used here?
   A. The population is binomial
   B. The population sigma is known
   C. n is small
   D. n = 50 is large, so s can replace unknown sigma
   Answer: D. For a large sample, s estimates sigma and the normal approximation is valid. (Key point: Choice of interval)

2. What is SE = s/sqrt(n) for s = 9.2, n = 50?
   A. 1.301
   B. 0.184
   C. 9.2
   D. 0.65
   Answer: A. 9.2/sqrt(50) = 9.2/7.071 = 1.301. (Key point: Standard error)

3. Which interval is wider and why?
   A. 95%, because it has more data
   B. 99%, because it uses the larger z (2.576 vs 1.96)
   C. They are equal width
   D. 95%, because it uses the larger z
   Answer: B. Higher confidence needs a larger z, so the 99% interval (38.15, 44.85) is wider than the 95% one (38.95, 44.05). (Key point: Confidence level and width)

4. Both intervals contain 40 hours. What does this suggest?
   A. Students study significantly more now
   B. Students study significantly less now
   C. No evidence that students study more than before
   D. The sample is invalid
   Answer: C. Last year's mean lies inside the intervals, so the data give no evidence of a change. (Key point: Interpretation)

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