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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. Step 2 of 2 : Suppose a sample of 502 people is drawn. Of these people, 140 passed out. Using the data, construct the 85% confidence interval for the population proportion of people who black out at G forces greater than 6. Round your answers to three decimal places.

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ogorwyne

The confidence interval is given as 0.250 and 0.308

How to solve for the confidence interval?

n = 502

x = 140

x/ n = proportion

= 140/502

= 0.2789

To find the proportion using 85%

1-0.85 = 0.15

0.15/2 = 0.075

Find the z value to be 1.44

[tex]p +- z*\sqrt{p'(1-p')/n\\}\\ = 0.279 +-1.44\sqrt{0.279(1-0.279)/502}[/tex]

= 0.279 +- 0.029

= 0.250, 0.308

Hence the confidence interval is 0.250, 0.308.

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