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
[tex]M.E = Z_{0.05} \frac{S.D}{\sqrt{n} }[/tex]
Level of significance = 0.05
Z₀.₀₅ = 1.96
Step(ii):-
The Margin of error is
[tex]M.E = Z_{0.05} \frac{S.D}{\sqrt{n} }[/tex]
[tex]0.9 = 1.96 \frac{4}{\sqrt{n} }[/tex]
Cross multiplication , we get
[tex]\sqrt{n} = \frac{4 X 1.96}{0.9 }[/tex]
√n = 8.711
Squaring on both sides ,we get
n = 75.88≅ 76
Conclusion:-
The sample size 'n' = 76
