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NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6 6 . In an earlier study, the population proportion was estimated to be 0.32 0.32 . How large a sample would be required in order to estimate the fraction of people who black out at 6 6 or more Gs at the 85% 85 % confidence level with an error of at most 0.03 0.03 ? Round your answer up to the next integer.

Answer :

Answer:

The sample size must be atleast 502 to have a margin of error of 0.03

Step-by-step explanation:

We are given the following in the question:

Proportion = 0.32

[tex]\hat{p} = 0.32[/tex]

Level of significance = 0.15

Margin of error = 0.03

Formula for margin of error =

[tex]z_{stat}\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.15} = 1.44[/tex]

We have to find the sample size such that the margin of error is atmost 0.03.

Putting values, we get,

[tex]z_{stat}\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\leq 0.03\\\\1.44\times \sqrt{\dfrac{0.32(1-0.32)}{n}}\leq 0.03\\\\\sqrt{n}\geq 1.44\times \dfrac{\sqrt{0.32(1-0.32)}}{0.03}\\\\\sqrt{n}\geq 22.39\\n\geq 501.3121\\\Rightarrow n\geq 502[/tex]

Thus, the sample size must be atleast 502 to have a margin of error of 0.03 in approximation of proportion of people who would black out at 6 or more Gs.

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