Lets say you got 10 nails and are given the length of every single one (average = 59,98mm... standard deviation =0,26)

How do you calculate theminimumaverage nail length? (The nails are normally distributed and you're allowed to choose your level of significance freely.)

This question doesn't make any sense.

Do you mean to ask "what is the lower bound of the confidence interval of the mean nail length?"

And we choose the alpha (significance) for the confidence interval?

Sample size too small - n should be greater than 30

But minimum size doesn't make sense.

Should be some lower limit like a quartile or standard deviation below mean

so minimum average is 59.98 - 0.26. then you calculate de CI95 or CI99 with alpha = 0.01 or 0.05 ?

idk

okay, I believe you may need to do that "degrees of freedom" or something like that. i believe "mininmum average" refers to how much possible error may be going on in the small sample of 26 nails, then you would subtract the estimated potential error from the average of the 26 nails.

I'm not familiar though with exactly what to do, but I understand the question. It may not be a "degrees of freedom" question, but you would need to find "percent error" if that is a statistic term

edit: yes the "degree of freedom" maybe totally irrelevant. and it may be more so related to "alpha" or the confidence interval being .01 or .05.

I'm just using intuition right now because I did statistics like a year ago

Ikr the question confuses me aswell, thats why i posted it here. But it literally says so.

Yeah i think they're talking about lower bounds here.

Forgive me, on phone so I can't use mathematical notation.

The equation of the lower bound is as follows:

Sample Mean -( Z-Score(1-alpha/2)×(Standard Deviation / root(Sample Size)))


So with what you have given us, and an alpha of .05, the answer follows(so lower bound of 95% Confidence Interval):

59.98 mm - 1.96(.26/root(10))

= 59.81885 (Rounded to 5 decimals, using calculator on phone)