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

Respuesta :

Answer:

The sample size is = 814(approximately)

Step-by-step explanation:

Given -

Population proportion [tex]{\widehat{(p)}[/tex] = 0.4

Confidence interval = 98[tex]\%[/tex]

[tex]\alpha =[/tex] 1 -  Confidence interval = 1 - .98

=  .02

[tex]\frac{\alpha}{2}[/tex]  = .001

[tex]Margin\; of\; error[/tex] = .04

Let sample size = n

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

.04 = [tex]z_{.01}\sqrt{\frac{(.4){(1 - (0.4)}}{n}}[/tex]

For 98[tex]\%[/tex] level of confidence, z = 2.33

.04 = [tex]2.33\times\sqrt{\frac{(.4){(.6)}}{n}}[/tex]

(Squaring both side)

[tex](.04)^2[/tex] = [tex](2.33)^2\times\frac{(.4)(.6)}{n}[/tex]

n = [tex]\frac{(2.33)^2}{(.04)^2}\times{(.4)(.6)}[/tex]

n = 814.33

Approximately n =814

The required large sample will be "814". To understand the calculation, check below.

Confidence level and Margin of error

According to the question,

Confidence interval (CI) = 98% or,

                                        = 0.98

Margin of error = 0.04

α - 1 - CI

  = 1 - 0.98

  = 0.02

[tex]\frac{\alpha}{2}[/tex] = [tex]\frac{0.02}{2}[/tex]

  = 0.01

Let, the sample size be "n".

We know the formula,

→ Margin or error = [tex]z_{\frac{\alpha}{2} }[/tex] [tex]\sqrt{\frac{\hat p (1 - \hat p)}{n} }[/tex]

By substituting the values,

                    0.04 = [tex]z_{0.01}\sqrt{\frac{0.04 (1-0.4)}{n} }[/tex]

                    0.04 = 2.33 × [tex]\sqrt{\frac{0.4\times 0.6}{n} }[/tex]

By squaring both the sides,

                          n = [tex]\frac{(2.33)^2}{(0.04)^2}[/tex] × 0.4 × 0.6

                             = 814.33 or,

                             = 814

Thus the approach above is correct.

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