How do I calculate $P(S^2 > 1.8307(\mathrm{PopVariance}))$ if $n =11$?
I think I should use the Chi square formula:
$$X^2 = \frac{(n-1)s^2}{\mathrm{PopVariance}}$$
But I can't really understand the textbook application of this formula.
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1The usual pronunciation is "chi squared" if I understand your meaning. See here for an introduction to posting mathematical notation. – hardmath Mar 16 '19 at 16:02
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While we're mentioning English pronunciation, the ch in 'chi' is pronounced like k as in 'Christmas', 'choir', and 'arachnid' (all of Greek origin). – BruceET Mar 16 '19 at 21:24
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What is PopVariance ? I don't understand this term ? Can you explain if S2 is equal to SUM(x-xi)2 divided by n or divided by (n-1) ? – schlebe Oct 27 '19 at 13:45
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If $Q = \frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(n-1)$ and $n = 11,$ then you have $$P(S^2 > 1.8397\sigma^2) = P\left(Q=\frac{10S^2}{\sigma^2} > 18.307\right) = 0.05.$$
My computation below uses R statistical software, but you should be able to find this result in a printed chi-squared table (row $\nu = 10,$ column headed .05 for the tail probability).
1 - pchisq(18.307, 10) # in R, 'pchisq' is a CDF
## 0.05000059
BruceET
- 51,500