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CI: mean weight loss of grinding balls

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.

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.

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

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.

✓ Answer checked by Claude Sonnet 5.5

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