Sample size for mean test score at 90%

Question

A researcher wants to estimate a mean test score to within a margin of error of $E = 2$ points, with 90% confidence. The population standard deviation is $\sigma = 12$ points.

What is the minimum sample size?

Accepted Answer

Setup. Use n = (z^{}\sigma/E)^{2}. For 90% confidence, z^{} = 1.645. Compute. n = \left(\frac{1.645 	imes 12}{2}ight)^{2} = \left(\frac{19.74}{2}ight)^{2} = (9.87)^{2} = 97.42. Round up: n = 98. Interpret. At 90% confidence the critical value is only 1.645, so the required sample is smaller than it would be at 95% or 99%. Takeaway. The three levers of required sample size are the critical value, the spread, and the target margin. Relaxing any of them lowers the count.

Sample size for mean test score at 90%

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