find the sample size

Question

Given a normal distribution with 𝜎 = 1.790. Find the required sample size for a 90% confidence level (estimating the mean) with a 1.02 margin-of-error.

Answer 1

Margin-of-error = z(σ/√n)

1.02 = 1.645(1.790/√n)

1.02 = 2.94455/√n

√n = 2.94455/1.02

n = (2.94455/1.02)²

n ≈ 8.334 ≈ 9 <-- (remember to always round up to the nearest positive integer!)

Therefore, the required sample size for a 90% confidence level with a 1.02 margin-of-error is n=9.

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