---
title: "Test: performance of college A vs B"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:colleges-a-b-mean-grade-test"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/colleges-a-b-mean-grade-test
site: BCT Engineering
---

# Test: performance of college A vs B

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2074 Asoj, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/colleges-a-b-mean-grade-test
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

An examination was given to 50 students at college A and 60 students at college B. At A mean grade was 75 with standard deviation of 9. At B mean grade was 79 with a standard deviation of 7. Is there significant difference between the performance of students at A and those at B, given that α = 0.05?

## Short answer

Large-sample z-test: SE = √(9²/50+7²/60)=1.561, z = (75−79)/1.561 = −2.562. Critical values ±1.96 at 5% level. Reject H0; there is a significant difference in performance between colleges.

## Answer

Samples are large (n₁ = 50, n₂ = 60), so a two-sample z-test is used with sample SDs as estimates.

1. H0: μ_A = μ_B (no difference in performance)
   H1: μ_A ≠ μ_B (two-tailed)
2. Level of significance: α = 0.05
3. Test statistic (large samples, z-test):

```text
z = (x̄₁ − x̄₂) / √(s₁²/n₁ + s₂²/n₂)
SE = √(9²/50 + 7²/60) = √(1.6200 + 0.8167) = 1.5610
z  = (75 − 79) / 1.5610 = -2.562
```

4. Critical value: z at α = 0.05 (two-tail) = ±1.960. Reject H0 if |z| > 1.960.
5. Decision: the calculated value falls in the rejection region (p-value = 0.0104), so **reject H0**.

**Conclusion:** there is a significant difference between the performance of students at A and B; college B students performed better.

## Check yourself

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

1. Why is a z-test used?
   A. The variances are equal and known to be 9
   B. The samples are small
   C. Samples are paired
   D. Samples are large, so sample SDs estimate sigma
   Answer: D. With n1 = 50 and n2 = 60 both large, the normal test with s as estimates is used. (Key point: Choice of test)

2. How is SE computed?
   A. sqrt(9^2/50 + 7^2/60) = 1.561
   B. 9^2/50 + 7^2/60
   C. (9 + 7)/sqrt(110)
   D. sqrt(9/50 + 7/60)
   Answer: A. SE = sqrt(s1^2/n1 + s2^2/n2) = sqrt(1.62 + 0.8167) = 1.561. (Key point: Standard error)

3. What is the decision?
   A. z = -2.56; |z| > 1.96, fail to reject H0
   B. z = 4; do not reject H0
   C. z = -2.56; |z| > 1.96, reject H0
   D. z = -2.56; |z| < 1.96, reject H0
   Answer: C. z = -4/1.561 = -2.562 lies in the rejection region at 5%, so H0 is rejected. (Key point: Decision)

4. What is the conclusion?
   A. College A is better
   B. The sample is too small
   C. No difference in performance
   D. Performance differs significantly and B is better
   Answer: D. H0 is rejected; B's mean (79) is higher than A's (75). (Key point: Conclusion)

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