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A sample of $10$ elements was taken from a normal population, provided a variance $25$. Calculate the sample size that evaluates the population mean with error of $2.5$ units at the $95$% confidence level.

There exists a formula to calculate the size of the sample, but in the formula it needs the population standard deviation, and I can't see a way to aproximate the standard deviation given only the variance of the sample.

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    $\begingroup$ The variance is that of the population. $\endgroup$ – Karn Watcharasupat Jul 29 '18 at 16:43
  • $\begingroup$ @KarnWatcharasupat the answer is 20, but when I plug in the values the answer is different. (1,96*5/2,5)^2 = 15.3664 $\endgroup$ – Sat Jul 29 '18 at 16:47
  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ – José Carlos Santos Jul 29 '18 at 16:48
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    $\begingroup$ If we take $25$ to be the known population variance, I get that a sample of $16$ suffices. If it's the sample variance, it's a more complicated problem. You can find the necessary sample size from Student's t-distribution with a sample variance of $25,$ but one cannot be sure the sample variance will remain $25.$ And in that case I'm getting $18$ as the needed sample size. How to take into account the uncertainty in variance in a problem like this is not something I've seen addressed anywhere. $\endgroup$ – Michael Hardy Jul 29 '18 at 17:05
  • $\begingroup$ Here's a guess: Although $18$ suffices if you use Student's distribution and get the SAME sample variance, someone decided to increase it to $20$ to take into account the uncertainty about what the sample variance will be by the time you get up to $18$ observations. And they could not claim certainty that $20$ would be enough. Presumably some formula would be provided for the occasion if this is an assigned exercise. $\endgroup$ – Michael Hardy Jul 29 '18 at 17:08
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Is your formula the same as this one:

$$ n = \left( \frac{z_{\alpha/2}\sigma }{E} \right)^2 $$

with $z_{\alpha/2} = 1.96$ for a $95\%$ confidence interval and $E=2.5$ is your margin of error?

If you think that the question has given you the sample variance ($s^2=25$), then we can estimate the population variance ($\sigma^2$) using:

$$ \begin{align} \sigma^2 &= \frac{n_0}{n_0-1}s^2 \end{align}$$

where $n_0=10$ in your case.

(You might have come across this estimation before. If not, try Googling it: "estimating the population variance from the sample variance".)

Using this estimation you get $n=17.07$, rounded up to $n=18$. Where did you get the answer $20$ from? It seems wrong to me...

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  • $\begingroup$ I got the answer in the pdf of exercises given by the professor, but some answer seem to be incorrect though. $\endgroup$ – Sat Jul 29 '18 at 18:28
  • $\begingroup$ I hope that your professor will be happy with this solution! $\endgroup$ – Malkin Jul 29 '18 at 18:38

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