In: Accounting
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
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 = Z0.005 = 2.576
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 2.576 * (((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)