The sample proportion, p^ = 0.8. The 99% confidence interval for the proportion is approximately from 0.73220 to 0.86780
How do we find the proportion sample and confidence level?
Given that 184 out of 230 people prefer Trydint, let's calculate p^;
p^ = 184/230 = 0.8
SE = √(p(1-p))/n
SE = √(0.8(1-0.8)/230
SE = √0.00069565217
SE = 0.0263
Z-score for 0.995 (99.5%) in a standard normal distribution is 2.575
ME = 2.5758 × 0.0263
ME = 0.06774354
CI = Xbar ± Z× s/√n
CL = 0.8−0.06779334
CL = 0.7322
CL = 0.8678
The 99% confidence interval for the proportion is approximately from 0.73220 to 0.86780