Suppose we observe a random sample of five measurements: 10, 13, 15, 15, 17, from a normal distribution with unknown mean $\mu_1$ and unknown variance $\sigma_1^2$. A second random sample from another normal population with unknown mean $\mu_2$ and unknown variance $\sigma_2^2$ yields the measurements: 13, 7, 9, 15, 11.

If I test

a) $\sigma_1 > 1.0$ with a chi-squared test: $\chi^2 = \frac{(n-1)S^2}{\sigma_0^2}$

b) $\frac{\sigma_2^2}{\sigma_1^2}$ = 1 with an F-test

what desirable statistical property is absent in these tests?

I think that the answer is power, because the sample size is low. However, I'm not absolutely sure.

  • $\begingroup$ Please tell us how you test, because the answers depend on that. $\endgroup$ – whuber Apr 24 at 2:30
  • $\begingroup$ @whuber: I've updated my question to respond to your comment. $\endgroup$ – MSE Apr 24 at 2:39
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    $\begingroup$ Thank you. Because t-tests usually compare means, could you explain how you conceive of using a t-test to test a standard deviation and why the hypothesis appears to be unrelated to the second sample altogether? $\endgroup$ – whuber Apr 24 at 2:42
  • $\begingroup$ Sorry - I meant to use the chi-squared test for part a) $\endgroup$ – MSE Apr 24 at 2:58
  • $\begingroup$ Strange question(s): If you mean to test $H_0: \sigma_1 = 1$ vs $H_a: \sigma_1 > 1,$ then you can reject at 5% level, so it doesn't seem reasonable to complain about power. // If you mean to test $H_0: \sigma_1 = \sigma_2$ vs $H_a: \sigma_1 \ne \sigma_2$, then you can't reject and maybe you could complain about power. // Personally, I would complain about doing statistical analysis without any context of purpose, but that's probably not the answer they're looking for. Certainly, there's not enough data to test for normality, so you have only someone's allegation for that. $\endgroup$ – BruceET Apr 24 at 7:47

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