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CI: sample of 16, mean 52, SD 4

Short answer

Small sample (n=16) with unknown σ, use t-interval: 99% CI = 52 ± 2.947*(4/√16) = (49.05, 54.95); 95% CI = 52 ± 2.131*1 = (49.87, 54.13).

A random sample of size 16 showed a mean of 52 with a standard deviation 4. Obtain 99% and 95% confidence limits population mean.

n = 16 is small and σ is unknown, so the t-interval x̄ ± t(α/2, n−1)·s/√n with 15 d.f. is used.

n = 16, x̄ = 52, s = 4, df = 15
SE = s/√n = 4/√16 = 1.0000

99% CI: t(0.005, 15) = 2.947
  margin = 2.947 × 1.0000 = 2.9467
  limits = 52 ± 2.9467 = (49.053, 54.947)

95% CI: t(0.025, 15) = 2.131
  margin = 2.131 × 1.0000 = 2.1314
  limits = 52 ± 2.1314 = (49.869, 54.131)

Answer: 99% limits: (49.05, 54.95); 95% limits: (49.87, 54.13).

✓ Answer checked by Claude Sonnet 5.5

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