In a survey of 8000 ?women, 5431 say they change their nail polish once a week....

In a survey of 8000 ?women, 5431 say they change their nail polish once a week. Construct a 90 % confidence interval for the population proportion of women who change their nail polish once a week. A 90?% confidence interval for the population proportion is ?( nothing?, nothing?). ?(Round to three decimal places as? needed.)

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Expert Solution

Solution :

Given that,

n = 8000

x = 5431

= x / n = 5431 / 8000 = 0.679

1 - = 1 - 0.679 = 0.321

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_{/ 2} * $\sqrt$ (( * (1 - )) / n)

= 1.645 * ( $\sqrt$ ((0.679 * 0.321) / 8000)

= 0.009

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.679 - 0.009 < p < 0.679 + 0.009

0.670 < p < 0.687

(0.670 , 0.687)

orchestra answered 1 year ago

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