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"A random sample, of size 9, from a normally distributed population, with variance 9, yielded as sample mean of 7 and a sample variance of 7.

I. What is the probability of a random sample of size 9 yielding a sample variance of 4 or less, given that the population variance is 9"

Should I be using the chi-squared distribution to solve this? However, from what I remember, chi-squared distribution is used when we have the sample variance rather than the population variance to derive the confidence interval. How should I approach this question? I really don't know where to start.

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  • $\begingroup$ Doesn’t this sound like a one-sided hypothesis test of one variance? $\endgroup$
    – Dave
    Mar 8 at 12:03
  • $\begingroup$ Now that I think about it, it kind of does. $\endgroup$ Mar 8 at 12:14

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This is asking for a one-sample hypothesis test of the variance, assuming a Gaussian distribution with $\sigma^2=9$ and a sample size of nine.

$$ H_0:\sigma^2=4\\ H_a: \sigma^2 <4 $$

For a Gaussian distribution, this is addressed by computing a test statistic and comparing to a $\chi^2$ distribution. I like the JBStatistics video on testing one variance.

Note the importance of the Gaussian assumption. While a t-test of the mean has solid robustness to deviations from the Gaussian ideal (not perfect, but solid), a $\chi^2$ test of the variance does not. A possibly worthwhile exercise is to demonstrate this in a simulation.

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    $\begingroup$ Thank you for the guidance. $\endgroup$ Mar 8 at 12:37

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