---
title: "CI: mean weight loss of grinding balls"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:grinding-balls-weight-loss-ci"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/grinding-balls-weight-loss-ci
site: BCT Engineering
---

# CI: mean weight loss of grinding balls

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2071 Shrawan, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/grinding-balls-weight-loss-ci
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

The mean weight loss of n = 16 grinding balls after a certain length of time in mill slurry is 3.42 grams with a standard deviation of 0.68 gram. Construct a 99% confidence interval for the true mean weight loss of such grinding balls under the stated conditions.

## Short answer

Using t-interval with 15 df: 3.42 ± 2.947 × 0.68/√16 = 3.42 ± 0.501, giving 99% confidence interval (2.92, 3.92) grams for the true mean weight loss.

## Answer

The sample is small (n = 16) and σ is unknown, so the t-interval x̄ ± t(α/2, n−1)·s/√n is used.

```text
n = 16, x̄ = 3.42, s = 0.68, df = 15
SE = s/√n = 0.68/√16 = 0.1700

99% CI: t(0.005, 15) = 2.947
  margin = 2.947 × 0.1700 = 0.5009
  limits = 3.42 ± 0.5009 = (2.919, 3.921)
```

**Answer: 99% confidence interval for the true mean weight loss = (2.92, 3.92) grams.**

## Check yourself

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

1. Why use t?
   A. sigma known
   B. n = 16 is small and sigma is unknown
   C. n is large
   D. Data skewed
   Answer: B. Small n with unknown sigma requires the t-interval with n-1 d.f. (Key point: Distribution)

2. What is SE?
   A. 0.68/sqrt(16) = 0.17
   B. 0.68/16 = 0.0425
   C. 0.68 x 4 = 2.72
   D. 0.68^2/16
   Answer: A. SE = s/sqrt(n) = 0.17. (Key point: Standard error)

3. Which t is used?
   A. z = 2.576
   B. t(0.005, 16) = 2.921
   C. t(0.025, 15) = 2.131
   D. t(0.005, 15) = 2.947
   Answer: D. 99% leaves 0.005 per tail with df = 15. (Key point: Critical value)

4. What is the interval?
   A. (3.25, 3.59)
   B. (3.42, 3.92)
   C. (2.80, 4.04)
   D. (2.92, 3.92)
   Answer: D. 3.42 +- 2.947 x 0.17 = 3.42 +- 0.501. (Key point: Result)

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