With n = 16, x̄ = 52, s = 6, the 95% CI for μ has a margin of error of 3.20. Which value below represents the margin of error?

• A. 1.60
• B. 2.00
• C. 4.00
• D. 3.20

Answer: (D)

The margin of error is the amount added to and subtracted from the sample mean to form the confidence interval. For a 95% interval using a t distribution, it equals t*(df) times the standard error, where standard error is s/√n.

Compute the standard error: s/√n = 6/√16 = 6/4 = 1.5.

With df = n − 1 = 15, the t critical value for a 95% interval is about 2.131.

Multiply: 2.131 × 1.5 ≈ 3.1965, which rounds to 3.20.

Thus, the margin of error is 3.20. The confidence interval would be 52 ± 3.20, i.e., (48.80, 55.20). The other numbers are not the half-width of the interval.