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enter image description here

I'm reading the introductory statistics and get stuck at chi square test for variances. Like the image shown below, the alternative hypothesis is that

$σ^2 < 7.2^2$

. But if we plug this inequality into the formula for the calculation of chi square value $(n-1)*S^2/σ^2$ (S is the sample standard deviation), the result is supposed to be $χ^2 > (n-1)*S^2/7.2^2 = 5.67$, which suggests that this is a upper sided test. However, the truth is that this is a lower sided test. Is there anything wrong? Thanks!

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  • $\begingroup$ Where do you get the inequality in your last expression? 5.67, the test statistic, is the same for any test. The sidedness affects the p value or inference. $\endgroup$
    – AdamO
    Commented Dec 6, 2017 at 21:37
  • $\begingroup$ @AdamO Just as the image shown, the formula is $χ^2 = (n-1)S^2/sigma^2$. If I plug $sigma^2 < 7.2^2$ into it, I can get the inequality. $\endgroup$
    – Yujian
    Commented Dec 6, 2017 at 21:58
  • $\begingroup$ The shaded area of the chisquare curve represents the critical area for the test. It is left tailed as you say. The test statistic is in the critical region. Are you having trouble finding the 13.85 critical value? $\endgroup$
    – AdamO
    Commented Dec 6, 2017 at 22:02
  • $\begingroup$ 5.67 is the correct statistic. It suggests the test is statistically significant. $\endgroup$
    – AdamO
    Commented Dec 6, 2017 at 22:02
  • $\begingroup$ @AdamO No, I have no trouble finding the correct statistic, but transforming it to p-value. The image shows that for an alternative hypothesis $\sigma^2 < 7.2^2$, the p-value should be obtained by a lower tail area. However, I'm thinking that as the $\sigma^2$ is at the denominator, this question should be solved with a upper tail test. $\endgroup$
    – Yujian
    Commented Dec 6, 2017 at 22:09

1 Answer 1

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I know it's late but I think your problem is that you wonder why for a given alpha like 0.05, your test statistic is on the right side and 1-alpha on the left side. because in chi square distribution area you have under the curve is being computed from right not left(unlike t-test for example)

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