NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier study, the population proportion was estimated to be 0.36. How large a sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04? Round your answer up to the next integer.

The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

Step-by-step explanation:

Given population proportion was estimated to be 0.3

p = 0.3

Given maximum of error E = 0.04

we know that maximum error

[tex]M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }[/tex]

The 85% confidence level [tex]z_{\alpha } = 1.44[/tex]

[tex]\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}[/tex]

[tex]\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}[/tex]

now calculation , we get

√n=13.80

now squaring on both sides n = 190.44

large sample n = 190.44≅190

Conclusion:-

Hence The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44