Hi, i have a question and i like to understand how to calculate. An analyst determined that the sample mean and variance for a normal distribution are 42 and 9 respectively. The 99% confidence interval for this random variable is closest to: a) 15.0 to 69.0 b) 18.8 to 65.2 c) 34.3 to 49.7 d) 39.0 to 45.0 The correct ans is c. The 99% confidence interval for a normally distributed random variable is equal to the sample mean +/- (2.58 * sample standard deviation) = 42 +/- (2.58*90.5) = 42 +/- (2.58 * 3) = 42 +/- 7.74 which is 34.3 to 49.7 Actually how does is derive the value 7.74? How is it calculate? thanks:)
CI = mean +/- Std error = mean +/- (z-value)*std Deviation/sqroot(n) where n = sample size Since , sample size is not given, we’ll consider std error = (z-value)*std.Dev Std dev = sqroot(9) = 3 z-value for 99% confidence interval = 2.58 therefore, CI = 42 +/- (2.58 *3) = 42 +/- 7.74
smuggycfa is confusing confidence interval for a variable with confidence variable for the mean. All you need to know that if a variable has normal distribution with mu, sigma, then its 99% confidence interval is mu - 2.58*sigma, mu + 2.58* sigma (2.58 is from a table, quantile of 99.5%: since there will be .5% in the left tail and 0.5% in the right tail -> 1 - 0.005% = 99.5%). sigma = sqrt(variance) = sqrt(9)=3. If you need to esimate a confidence interval for a MEAN of a variable (not variable itself), then keep in mind that standard deviation of the mean is sigma/sqrt(n) just as smuggycfa said.
Oh thanks for the explanation…