A certain forum reported that in a survey of 2004 American adults, 28% said they believed...

A certain forum reported that in a survey of 2004 American adults, 28% said they believed in astrology.

(a) Calculate a confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology. (Round your answers to three decimal places.) ,

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Expert Solution

Solution :

Given that,

n = 2004

Point estimate = sample proportion = = 28%=0.28

1 - = 1- 0.28 =0.72

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* ( $\sqrt$ ((0.28*0.72) /2004 )

E = 0.026

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.28- 0.026 < p < 0.28+0.026

0.254< p < 0.306

orchestra answered 1 year ago

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