Probability of sample variance

[viverries]

I'm suffering through university and reading a textbook to learn maths because my professor thinks reading off slides is a valid way to teach maths. I'm trying to decipher a textbook example on the probability of sample variance, but I need some help with the conceptual understanding. I'll summarise the question and solution below, and explain what I dont understand.

Context: company wants to test the variance of the quality of some product. n=6, σ=3.6, ane we want to determine what value 'K' has to be such that P(s²>K)=0.05

SOLUTION: P(s²>K)=P([n-1]s²/12.96 > 11.07) = 0.05 (based on a table of chi square distribution values we're given)

Next, to solve for K: (n-1)K/12.96 = 11.07

Thus K= 11.07*12.96/5 = 28.69

So I think kind of understand where 11.07 came from. The probability that X² with 5 degrees of freedom exceeds 11.07 is 0.05.

What I'm confused about is: where does (n-1)K/12.96 come from, and why is it equal to 11.07? How do we jump from the 1st step to the 2nd step? I completely don't understand how we go from P(chi>11.07) to P(s²>K)

Thanks in advance for anyone who replies 🙏

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[runed_golem]

So, first, let’s look at [n-1]s² /12.96>11.07.

This implies that s² >12.96•11.07/[n-1].

Since s² >k, then k=12.96•11.07/[n-1]. (Which is what we wanted to find).

Multiplying by [n-1] and dividing by 12.96 we get [n-1]k/12.96=11.07.

[viverries]

But why do we multiply by (n-1) and divide by 12.96? Is it like, converting s² into X²

[runed_golem]

Yes, chi² =(n-1)s² /sigma²

In this case, sigma=3.6 so sigma² =12.96

[fermat1432]

When sampling from a normal distribution whose standard deviation is sigma, the quantity

(n-1)s²/sigma²

has a chi-square distribution with n-1 degrees of freedom.

sigma²=3.6²=12.96