From a random sample $X_1;...;X_8\sim \mathcal N(3,σ^2)$, we would like to test whether the variance $σ^2$ can be considered to be at least $5$, against the hypothesis that it is less than 5. We have recorded $\sum_{i=1}^8 (x_i-3)^2 = 37.5$.

(i) Perform the relevant test at $1\%$ signicance level and conclude.

(ii) In terms of the cumulative distribution function of a known distribution, express the power of the test when (a) $σ^2 = 4$, (b) $σ^2 = 3$ and (c) $σ^2 = 2$.

(iii) Explain briefly what would change if we did not know the true mean in the population, but we had recorded the sample mean $x = 3$.

I got the answer for part i,the observed t-statistic is $7.5$ so do not reject the null hypothesis. But I have problem with part ii. I try to use the power method but it seems not working as in the statistic T there is only one variance $σ^2$, can anyone helps me out?

  • $\begingroup$ I assume in part (ii), the question is still referring to the null hypothesis given in the problem description. $\endgroup$
    – A. Webb
    Mar 7 '17 at 17:59

In order to understand the power computation, it will help to write out the original test in (1). I think it is bad for to call the test statistic $T$ because this has nothing to do with t tests or Student's t distribution; so I'll use $Q$.

Testing the null hypothesis at level $\alpha.$

You want to test $H_0: \sigma^2 \ge 5$ against the alternative $H_a: \sigma^2 < 5.$ Your estimate of the variance is $$V = \frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2 = \frac{1}{8}\sum_{i=1}^8 (X_i - 3)^2 = \frac{37.5}{8} = 4.6875.$$

Thus your estimate $V < 5,$ but the question is whether it is enough smaller than $5$ to reject $H_0$ at the 1% level of significance.

The test statistic is

$$Q = \frac{nV}{\sigma_0^2} =\sum_{i=1}^8 \frac{(X_i - 3)^2}{\sigma_0^2} = \sum_{i=1}^8 \left(\frac{X_i - 3}{\sigma_0} \right)^2 = \sum_{i=1}^8 Z_i^2,$$ where $Z_i \stackrel{iid}{\sim} \mathsf{Norm}(0,1),$ because $\mu = 3$ is known. Thus $Q \sim \mathsf{Chisq}(df=8).$

We will reject $H_0$ for sufficiently small values of $Q.$ Specifically, the critical value $q^* = 1.646$ cuts 1% of the area from the lower tail of $\mathsf{Chisq}(8).$ So we reject $H_0$ at the $\alpha =1\%$ level of significance against the left-sided alternative if $Q < 1.646.$ (You can get $q^*$ from a suitable printed table of chi-squared distributions or by using software; the result from R statistical software is shown below.)

qchisq(.01, 8)
## 1.646497

In summary, we can write $\alpha = P\{Q = nV/\sigma_0^2 \le q^* = 1.65\; |\; \sigma_0^2 = 5\} = 0.01.$

Power against alternative $\sigma_a.$ In finding the critical value $q^*$ we have used the value of $\sigma = \sigma_0 = 5,$ specified by the $=$-sign in $H_0.$ One can find the power of the tests for any of the values $\sigma_a$ of $\sigma$ in the alternative hypothesis. Specifically, let's choose $\sigma_a = 2.$ The power is the probability of rejecting $H_0$ assuming that $\sigma = \sigma_a = 2.$

$\pi(\sigma_a=2) = P\{Q = nV/\sigma_0^2 \le q^* = 1.65\; |\; \sigma_0^2 = 2\} = ??.$

The test procedure and significance level $\alpha = .01$ remain the same. What is different is that the data $X_i$ are now a random sample from $\mathsf{N}(\mu = 3, \sigma^2 = 2).$ You no longer have $V = 4.6875.$ You are imagining new data, for which you no longer have $Q = nV/\sigma_0^2 \sim \mathsf{Chisq}(8).$ Your task in solving this problem is to decide how to take this change into account.

[I get $\pi(2) \approx 0.154.$ From simulation and from the R code below:]

 pchisq(qchisq(.01,8)*2.5, 8)
 ## 0.1535139

When $\mu$ is unknown. If $\mu$ is unknown, you would estimate it by $\bar X.$ Then the estimate of $\sigma^2$ would be $S^2 = \frac{1}{n-1}\sum_{i=1}^8 (X_i - \bar X)^2$ and $Q = (n-1)S^2/\sigma^2 \sim \mathsf{Chisq}(n - 1).$

  • $\begingroup$ My problem is how to calculate the power, without using R. Could you show me how to calculate it? $\endgroup$
    – Berry
    Mar 10 '17 at 9:23
  • $\begingroup$ I have given you a page of notation and framework--and a numerical answer. It is still not a trivial problem. But I think you will learn something important--as a graduate student--by comparing what I have written with your text, and putting the pieces together for yourself on the 2nd part of the problem. $\endgroup$
    – BruceET
    Mar 10 '17 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.