A 90% confidence intervals…

Question

A 90% confidence intervals for the proportion of first-year statistics students with brown hair is given to be (0.25, 0.41).

(d) What is the standard error of the estimate? (Round to 3 decimal places, if needed)

(e) What was the sample size? (Round to the appropriate integer)

Answer 1

margin of error = width of confidence interval/2 = 0.41 - 0.25 / 2 = 0.08

margin of error = critical z * SE = 1.645 * SE

so 0.08 = 1.645 * SE

0.08/1.645 = SE

SE = 0.049

SE = sigma/sqrt(sample size)

0.049 = sigma/sqrt(sample size)

(0.049//sigma)^2 = sample size

were you given sigma?