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

Answer:

The sample size 'n' = 76

Step-by-step 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

$M.E = Z_{0.05} \frac{S.D}{\sqrt{n} }$

Level of significance = 0.05

Z₀.₀₅ = 1.96

Step(ii):-

The Margin of error is

                    $M.E = Z_{0.05} \frac{S.D}{\sqrt{n} }$

                   $0.9 = 1.96 \frac{4}{\sqrt{n} }$

Cross multiplication , we get

              $\sqrt{n} = \frac{4 X 1.96}{0.9 }$

             √n  = 8.711

Squaring on both sides ,we get

             n =  75.88≅ 76

Conclusion:-

The sample size 'n' = 76