annoying s3 question :(

when conducting a sample, the sample variance is basically the population variance multiplied by n/n-1 right. Well in this qiestion it doesnt appear to be so. And i dont know why

on each of 100 days a conservationist took a sample of 1 litre of water from a particular place along a river and measured the amount, xmg, on cholrine in the sample. The results she obtained are shown in the table

x/1/2/3/4/5/6/7/8/9
d/4/8/20/22/16/13/10/6/1

d=no of days

calculate the mean amount of cholrine present per litre of water and estimate to three decimal places the standard error.

standard error = sqr(variance/n)

when i put variance of the population over n i get the right answer

however when i use varaiance of the sample (i.e n/n-1) the answer is wrong.

Why is this :confused:

Reply 1

Jonny W

Standard error = sqrt(

\sigma^2

/n), where

\sigma^2

is the unknown variance of the distribution of chlorine readings.

The sample variance (3.4233) and the population variance of the sample (3.3891) both approximate

\sigma^2

. It's debatable which estimator is better. The former is correct on average (unbiased). Assuming the distribution of chlorine readings is normal, the latter has a lower mean-squared-error.

I think you could use either estimate and get full marks. Since n is large, they only differ slightly.

Reply 2

BLUREMI

Jonny W

Standard error = sqrt(

\sigma^2

/n), where

\sigma^2

is the unknown variance of the distribution of chlorine readings.

The sample variance (3.4233) and the population variance of the sample (3.3891) both approximate