I have been given a problem for homework with which I need a little help. In part A, I suspect that the solution has to do something with the degrees of freedom where we use n-1 for sample and n for population when calculating the standard deviation, but I do not like the idea of estimating for population. Is it not an extrapolation? Furthermore since we are not given the sum of x^2, how would you compute the standard deviation? In part B, since we have found the standard deviation of population in part A, would you use t-distribution or normal distribution? I am asking especially because the standard deviation of the population will already be an estimation. I appreciate your kind help! The problem is following:

Michael regularly buys 150g packets of tea. He has noticed recently that he gets more cups of tea than usual out of one packet, and suspects that the packets contain more than 150g on average. He weighs eight packets and finds that their mean mass is 153g and the standard deviation of their masses is 4.2g.

A) Find the unbiased estimate of the standard deviation of the mass based on Michael´s sample.

B) Assuming that the masses are normally distributed, test Michael´s suspicion at 5% level of significance.


1 Answer 1


For part A), basically the whole point of dividing by $n-1$ instead of $n$ is to give us an unbiased estimate of the population variance. (Variance, not standard deviation! In fact one can show using Jensen's inequality that it is a biased estimate for the population standard deviation. See here for further reading on this if you are interested. I am guessing for your question though they just want you to use the sample standard deviation.) Of course, it is only an estimate -- we are not saying the population standard deviation is equal to this estimate.

For part B), yes, use $t$-distribution. (It will be a "one-tailed one sample $t$ test".)

  • $\begingroup$ How would you compute the standard deviation in part A since we are not given the sum of x^2? $\endgroup$ Mar 8, 2019 at 15:43
  • 1
    $\begingroup$ Excellent answer and link (+1). // No need to compute the usual (slightly biased) sample SD, that is given as 4.2. // I fear the person who wrote the question may be thinking of unbaisedness of $S^2$ (with denominator $n-1)$ for $\sigma^2$ and forgetting that $S$ is slightly biased for $\sigma.$ $\endgroup$
    – BruceET
    Mar 10, 2019 at 19:48

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