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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. In an earlier study, the population proportion was estimated to be 0.46. 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 0.03?

Respuesta :

JeanaShupp

Answer: 1499

Step-by-step explanation:

As per given , we have

Prior Population proportion : [tex]p=0.46[/tex]

Significance level : [tex]\alpha=1-0.98=0.02[/tex]

Critical value for 98% confidence interval (refer to z-value table) = [tex]z_{\alpha/2}=2.33[/tex]

Margin of error : [tex]E=0.03[/tex]

Formula for sample size :

[tex]n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2\\\\ n=0.46(1-0.46)(\dfrac{2.33}{0.03})^2\\\\ n=1498.3764\approx1499[/tex]

Therefore , the required minimum sample size = 1499

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