Question

scores on a certain test are normally distributed with a variance of 22. a researcher wishes to estimate the mean score achieved by all adults on the test. find the sample size needed to assure with 95 percent confidence that the sample mean will not differ from the population mean by more than 2 units.

Answer 1

The sample size must be 30 for the sample mean to vary from the population mean by no more than 2 units.

Variance = 22

σ² = 22

σ = √22

Make a 95% confidence interval with a sample mean that departs from the population mean with no more than 2 units.

The error margin is determined as follows:

Z critical × (σ ÷ √n)

Substitute the values,

Z critical at $\alpha_{0.05}$ = 2.33

2 = 2.33 × (√22 ÷ √n)

√n = (2.33 × √22) ÷ 2

√n = 5.464

n = 5.464²

n = 29.85 ≈ 30

Therefore, The sample size must be 30 in order for the sample mean to deviate from the population mean by no more than two units.

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