Question

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.

Answer 1

Answer:

The large sample n = 190.44≅190

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

$M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }$

The 85% confidence level $z_{\alpha } = 1.44$

$\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}$

$\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}$

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