From this question: The average annual rainfall amount in Yucutat, Alaska, is normally distributed with a mean of 150 inches and a standard deviation of 20 inches. The 90% confidence interval for the annual rainfall in Yucutat is closest to: A) 110 to 190 inches. B) 117 to 183 inches. C) 137 to 163 inches. D) 144 to 156 inches. The correct answer is B) 117 to 183 inches. The 90% confidence interval is µ ± 1.65 standard deviations. 150 - 1.65(20) = 117 and 150 + 1.65(20) = 183. I don’t understand - I thought the equation is 150 +/- 1.65 x (20 divided by the square root of n) Help
no in any normal distribution, 90% of the data is +/- 1.645 Standard Deviations from the mean. Sd / n^0.5 = standard error. Since they don’t give you ‘n’ you wouldn’t be able to work that out anyway
Thanks, went back to the books and I see my mistake. I think I am getting confused between samples and populations
Standard error of mean (confidence interval around mean) vs. distribution of data points
You could also think that 95% is +/- 2 standard deviations which would be 110-190. The more confident, the wider the range, the less confident, the narrower the range. 90% is a little more narrow, so B is the only answer that makes sense.