If a sample size of 16 yields an average of 12 and standard deviation of 3, estimate the 95% confidence interval for the population (assume a normal distribution).

Question

Options

• A10.40 < m < 13.60
• B10.45 < m <13.55
• C10.53 < m <13.47
• D10.77 < m <13.23

• A
  74% (25)
• B
  15% (5)
• C
  9% (3)
• D
  3% (1)

Explanation

Option A is correct because with n = 16, you must use the t-distribution (df = n − 1 = 15), giving a critical value of t* ≈ 2.131. The standard error is s/√n = 3/√16 = 0.75, so the margin of error is 2.131 × 0.75 ≈ 1.60, yielding 12 ± 1.60 → (10.40, 13.60).

Why the distractors are wrong:

• C (10.53–13.47) is the most tempting trap - it uses z* = 1.96 (the standard normal 95% critical value) instead of the t-critical value; this underestimates uncertainty for small samples.
• D (10.77–13.23) uses z* = 1.645, which is the critical value for a 90% confidence interval, not 95%.
• B (10.45–13.55) reflects a slightly off critical value - likely from misreading a t-table or using the wrong degrees of freedom.

Memory tip: When sample size is small (n < 30), t is taller/wider than z - so the t-based confidence interval is always wider than the z-based one. If your answer looks like the narrowest interval, you probably used z when you should have used t.

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