1. ## Sampling and distribution

The mass of a particular type of steel sheet produced by a factory has a normal distribution. The mean mass of a random sample of 14 sheets is 72.5kg and the standard deviation is 3.2kg. Calculate a 95% confidence interval for the mean mass of the steel sheets produced by the factory.

The mean mass of a random sample of 14 sheets is 72.5kg and the standard deviation is 3.2kg.
Is this 3.2kg a sample standard deviation?
Should i do like this?

or:
this (3.2kg)^2 is already an unbiased estimate of population variance?

2. nobody help me??

3. Originally Posted by walreinlim88
nobody help me??
To be quite frank about it: there is no clear answer to your question, imho. This is because different teachers assume different conventions: some teachers speak of the "standard deviation of a sample" already with an eye towards later estimating the "standard deviation of the population" from which the sample has been taken. That is, they define it to be $\displaystyle s=\sqrt{\frac{\sum_i(x_i-\overline{x})^2}{\color{red}n-1}}$.
If your teacher belongs to that subspecies of teachers, then the given standard deviation of the sample already is the unbiased estimator for the standard deviation of the population that you require in order to determine the boundaries of the confidence interval.
But it just might be the case that your teacher belongs to the second subspecies of teachers that assumes the standard deviation of the sample to be $\displaystyle s=\sqrt{\frac{\sum_i(x_i-\overline{x})^2}{\color{red}n}}$. If so, then you have to apply the correcting factor $\displaystyle \sqrt{\frac{n}{n-1}}$ to estimate the standard deviation of the population from which the sample has been taken, as you wrote.

To sum up: You have to rely on your notes from class - or, alternatively, your textbook - to figure out to which of the two species of teachers your particular teacher happens to belong.