Question

g What sample size should the managers use to ensure their 5% level 2-sided test has power of at least 0.9 to detect a true mean of 33 ounces (assuming σ = 4)?

Answer 1

The sample size 'n' = 76

Explanation:

Step(i):-

Given mean of the Population = 33 ounces

Given standard deviation of the Population = 4 ounces

Given the margin of error ( M.E) = 0.9

The Margin of error is determined by

`M.E = Z_(0.05) (S.D)/(√(n) )`

Level of significance = 0.05

Z₀.₀₅ = 1.96

Step(ii):-

The Margin of error is

`M.E = Z_(0.05) (S.D)/(√(n) )`

`0.9 = 1.96 (4)/(√(n) )`

Cross multiplication , we get

`√(n) = (4 X 1.96)/(0.9 )`

√n = 8.711

Squaring on both sides ,we get

n = 75.88≅ 76

Conclusion:-

The sample size 'n' = 76