Math Is Fun Forum

a sample of 9465 Chinese babies has a mean birth weight of 109.9 oz, a median of 107.5 oz, and a standard deviation of 13.593oz. set 95% confidence intervals to the mean, median, and coefficient of variation. of the confidence intervals for mean and median, which is wider? will this generally be true.

I think for this first one here i think i just take the mean and impose +/- limits on it by taking the standard error multiplied by the t value i looked up in the table [sd/√(n-1)][1.980]

... to look something like this 109.9 +/-  (13.593/√9465)(1.980)

answer for confidence interval of mean: 109.9 +/- .277

next,

for this one i think i take the median then impose +/- limits on it by taking standard error of mean(.277) multiplied by 1.2533(a number my stat book gave me for standard error of median).

... to look something like this 107.5 +/- (.277)(1.253)

answer for confidence interval of median: 107.5 +/- .347

the next one ins the one that confuses me to no end

i have to find confidence interval of the coefficient of variation(CV), i started by finding the CV (13.593/109.9) = .1237

the i have this equation where standard error of V <15 = V/(2√n). is V representing CV here?? should i enter my CV as V here and solve then multiply by my t value (1.980)

please explain this very carefullt to me, as i am clueless.