Question

please help!!!! an oatmeal manufacturer wants to test the claim that the standard deviation of the amount of oatmeal in a box is 18 grams. Assume that the amount of oatmeal in a box is normally distributed. 24 boxes of oatmeal are selected at random. These 24 boxes have a standard deviation (of the amount of oatmeal in a box) of 14 grams. At a level of significance of 0.05, is there enough evidence to reject the manufacturer's claim?

Answer 1

The answer is pretty obviously 'no, there is nowhere near enough evidence to reject this claim', but we can solve this with math as well.

Answer 2

Let us assume that the population Standard Deviation is 18 grams, giving us a variance of 18^2 or 324 grams squared. If we select 24 random boxes as our sample, we can calculate a Sample Standard Deviation:
\[s_{24}=\sqrt{\frac{1}{24-1}\sum_{i=1}^{24}(x_i-\bar{x})^2}\]
Surely if we are only picking 24 boxes to sample from, we could by chance select boxes with similar amounts of oatmeal (and thus calculate a low sample standard deviation, or, by chance, select boxes with dissimilar amounts of oatmeant (and thus calculate a high sample standard deviation.) The question becomes: \[\text{What is the distribution of }s_{24}?\]

Answer 3

I think the distribution of the sample standard deviation is proportional to a Chi Distribution with 24-1 degrees of freedom.

Answer 4

\[\text{Yes, }s_{24}\text{ is distributed by: }\sqrt{\frac{18^2}{24-1}}\huge\chi_{\large 24-1}^{\small2}\]

Answer 5

I meant:
\[s_{24}=\sqrt{\frac{18^2}{24-1}\huge\chi_{\large 24-1}^{\small2}}\]
That means we can see how unlikely a sample standard deviation of 14 would be with 24 observations using Excel like this:
=CHISQ.DIST((24-1)*(14^2)/(18^2),24-1,1)
That tells us the percent of the time we we observe that result or smaller. If this result is less than 2.5% or greater than 97.5% we can reject the manufacturer's claim with a 0.05 level of significance.

Answer 6

okay that makes sense thanks so much!

Answer 7

Really? I was worried. Cheers!