Answer: The correct answer is 8.12%-11.88%.
We can use the formula for a confidence interval for a proportion:
p ± z*SE
where p is the sample proportion, z* is the z-score corresponding to the desired level of confidence (in this case, 1.96 for a 95% confidence interval), and SE is the standard error of the proportion.
SE = sqrt(p*(1-p)/n) = sqrt(0.1*0.9/1015) = 0.0094
Plugging in the values, we get:
0.1 ± 1.96*0.0094
= 0.1 ± 0.0184
So the 95% confidence interval is (0.0816, 0.1184), which is equivalent to 8.12%-11.88%.
Explanation: