Audience profit data collected at the ESPN sports zone web site showed that 26% of the...

Audience profit data collected at the ESPN sports zone web site showed that 26% of the users were women. Assume that this percentage was based on a sample of 400 users What would be the 99% confidence interval estimate of the population proportion?

would you please do it step by step thanks

Solutions

Expert Solution

n = 400

= 26% = 0.26

1 - = 1 - 0.26 = 0.74

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * ( $\sqrt$ ((0.26 * 0.74) / 400)

= 0.056

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.26 - 0.056 < p < 0.26 + 0.056

0.204 < p < 0.316

The 99% confidence interval for the population proportion p is : (0.204 , 0.316)

ekkarill92 answered 1 year ago

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