Q.The contents of each of a random sample of 100 cans of a soft drink are measured. The results have a mean of 331.28 ml and a standard deviation of 2.97 ml. Show that an unbiased estimate of the population variance is 8.91 ml.
I'm only able to get 2.97^2 which is 8.8209 but that is not what the question wants. How do I obtain 8.91? From my knowledge, an unbiased estimate of population variance is the same as sample variance, so 8.8209 should have been the correct answer, but it isn't. Why?
$\endgroup$2 Answers
$\begingroup$Hint:
The unbiased estimator for the variance of the population is
$$s_u^2=\frac1{n-1}\cdot\sum_{i=1}^n ( x_i-\overline x)^2$$
While the variance of the sample is
$$s^2=\frac1n\cdot\sum_{i=1}^n ( x_i-\overline x)^2=\frac{n-1}{n}\cdot s_u^2$$
I think you can go on.
Remark:
I´ve found out, that you can paste 2.97^2*100/99 into the google search box without making any formatting. After pressing enter immediately the result is shown. See here.
$\endgroup$ $\begingroup$The unbiased sample variance is $\hat{\sigma}^2 \times \frac{n}{n-1}$. So, it should be 2.97^2*100/99.
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