Finding unbiased point estimate of population variance

Question

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?

Answer 1

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.

Answer 2

The unbiased sample variance is $\hat{\sigma}^2 \times \frac{n}{n-1}$. So, it should be 2.97^2*100/99.