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 0.4. 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 98% confidence level with an error of at most 0.04

Answer 1

Answer:

The sample size is = 814(approximately)

Step-by-step explanation:

Given -

Population proportion ${\widehat{(p)}$ = 0.4

Confidence interval = 98$\%$

$\alpha =$ 1 -  Confidence interval = 1 - .98

=  .02

$\frac{\alpha}{2}$  = .001

$Margin\; of\; error$ = .04

Let sample size = n

$Margin\; of\; error = z_{\frac{\alpha }{2}}\sqrt{\frac{{\widehat{(p)}}{(1 - \widehat{p})}}{n}}$

.04 = $z_{.01}\sqrt{\frac{(.4){(1 - (0.4)}}{n}}$

For 98$\%$ level of confidence, z = 2.33

.04 = $2.33\times\sqrt{\frac{(.4){(.6)}}{n}}$

(Squaring both side)

$(.04)^2$ = $(2.33)^2\times\frac{(.4)(.6)}{n}$

n = $\frac{(2.33)^2}{(.04)^2}\times{(.4)(.6)}$

n = 814.33

Approximately n =814