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In order to prevent damages to the wind turbines, the windspeed is monitored every 5 seconds, and if it is determined that the variance of the windspeed is high the turbine switches off to a safe mode.

Let $X_1, . . . , X_{12}$ be independent random variables representing the measurements made in a one minute period (in km/h). We can assume these are normally distributed with mean $\mu$ and variance $\sigma^2$, both unknown. At every minute the following hypothesis test is performed: $\left\{ \begin{array}{ll} H_0: \sigma = 75 km/h\\ H_A: \sigma \geq 75 km/h \end{array} \right. $

The turbine is programmed to stop if $S ≥ 83.7km/h$ (this is the critical region of the test), where S is the sample standard deviation

Suppose that $\sigma = 110km/h$. Is it true that the power of the test in this situation is larger than $0.75$?

The power of a test is the probability that the null hypothesis is rejected when it is false. So what is the probability that $S \geq 83.7$, given that $\sigma \geq 75$, and in this case $\sigma=110$. $$Power=P(S \geq 83.7 | \sigma=110)$$

I don't know how to solve this. I thought about standardizing, but $\mu$ is unknown so I don't think that's going to work.

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    $\begingroup$ It is not necessary to know $\mu$ in order to do this test on $\sigma^2.$ The computation of the sample variance $S^2$ uses the estimate $\bar X$ of $\mu.$ Then you have $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(n-1).$ Try using that relationship to find $P(S\ge 83.7|\sigma=110),$ and edit your result into your Question. Then maybe one of us can check your work. $\endgroup$
    – BruceET
    Commented Dec 14, 2018 at 7:19
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    $\begingroup$ It looks as if you're not testing at the 5% level of significance. What level? $\endgroup$
    – BruceET
    Commented Dec 14, 2018 at 8:46

1 Answer 1

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Comment continued:

My guess is that you are expected to obtain an analytic solution using the chi-squared distribution according to the relationship in my earlier comment. However, the following simulation in R provides a pretty good hint that the answer to the main question is that the power is indeed greater than $0.75.$

set.seed(1213);  m = 10^5; n = 12;  sg = 110;  mu = 100
s = replicate(m, sd(rnorm(n, mu, sg)))
mean(s > 83.7)
[1] 0.84682

Note: In the simulation it is crucial to have $\sigma = 110$ (code sg = 110). It is also necessary to have some value of $\mu$ in order to do the simulation. But notice that the result doesn't depend on what you pick. (For normal data, $\bar X$ and $S^2$ are independent random variables.)

  set.seed(1213);  m = 10^5; n = 12;  sg = 110;  mu = 200   # value of mu irrelevant
  s = replicate(m, sd(rnorm(n, mu, sg)))
  mean(s > 83.7)
  [1] 0.84682

If you knew the value of $\mu$ then the natural estimate of $\sigma^2$ would be $V = \frac 1 n \sum_{i=1}^n (X_i - \mu)^2$ and you would have $\frac{nV}{\sigma^2} \sim \mathsf{Chisq}(n).$ This is easy to show. But some beginning students are surprised that the corresponding quantity involving $S^2$ has $n-1$ degrees of freedom, which is a little more difficult to prove.

By making a histogram of the simulated values (first simulation) of $Q = 11S^2/\sigma^2$ one can make the graph below illustrating that $\mathsf{Chisq}(11)$ is the correct distribution and $\mathsf{Chisq}(12)$ is not.

enter image description here

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