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Assume you are using a significance level of $α=0.05$ to test the claim that $μ<20$ and that your sample is a random sample of 33 values. Find $β$ given that the population actually has a normal distribution with $μ$1 $=$ 15 and $σ=7$.

  • These are me workings out up to the point that I got stuck

$H$0 : $μ$ = 20

$H$1 : $μ$ < 20

Firstly, I'm not sure if this is right. To work out $β$, we find P(Do not reject $H$0|$μ$ = 15)

So $\bar X$ = $20 + \frac{7}{33^{0.5}}(-1.96) = 17.61$

And I think I'm supposed to find the probability that $X > 17.61$ given $μ = 15$ and $σ = 7$ but I feel like it's wrong. If so, can anyone please point out where I went wrong?

Thanks in advance.

Edit: Changed the null and alternative hypotheses

Bill Cogn
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    If the claim is $\mu < 20$, I might have presumed this was $H_1$ – Henry May 09 '23 at 21:23
  • @Henry Uh, well I'm think I'm also confused as to what should be the null and alternative hypothesis at this point. So if μ<20 is the alternative hypothesis, then is the null μ = 20? Aside from that, is my working out valid? – Bill Cogn May 09 '23 at 21:52
  • @Henry I'm really confused on this, do you mind explaining? – Bill Cogn May 09 '23 at 22:23

1 Answers1

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First, we determine the rejection criterion for the test. Clearly, we are intending to use the one-sample $Z$-test: $$Z \mid H_0 = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}} = \frac{\bar X - 20}{7/\sqrt{33}}. \tag{2}$$ Then we will reject $H_0$ at level $\alpha$ if $$\Pr[Z < z_\alpha] = \alpha; \tag{3}$$ that is to say, if $Z < -1.64485$ for $\alpha = 0.05$. (Note that your choice of $z_\alpha$ is incorrect because the test is one-sided, not two-sided.) This in turn implies $$\bar X < \frac{(-1.64485)(7)}{\sqrt{33}} + 20 \approx 17.9957, \tag{4}$$ where $\bar X$ is the sample mean.

So if the sample is drawn from a normal distribution with mean $\mu_1 = 15$ and standard deviation $\sigma = 7$, the Type II error is given by $$\beta = \Pr[\bar X > 17.9957 \mid \mu_1 = 15], \tag{5}$$ i.e., it is the probability of failing to meet the rejection criterion under the assumption that the alternative hypothesis is correct. In this case, the alternative hypothesis is the point hypothesis $H_1 : \mu = \mu_1 = 15$.

Hence $$\beta = \Pr\left[\frac{\bar X - \mu_1}{\sigma/\sqrt{n}} > \frac{17.9957 - 15}{7/\sqrt{33}} \mid \mu_1 = 15\right] = \Pr[Z > 2.45841] \approx 0.00697778. \tag{6}$$

heropup
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  • Ah I understand it now thanks. Follow up question, when I try to compute Pr[X¯>17.9957∣μ1=15], I don't get anywhere near the correct answer. When I use the standardized version I do, is there a reason for this or am I computing it wrong? – Bill Cogn May 10 '23 at 12:42
  • @BillCogn Because the standard deviation $\sigma$ is for a single observation, not for a sample of $33$ observations. In other words, you're confusing the standard deviation and the standard error of the mean. The sample mean $\bar X$ is a normal random variable with the same $\mu$, but its standard deviation (which is called the "standard error") is $\sigma/\sqrt{n}$. – heropup May 10 '23 at 13:50
  • I see now, thank you so much for showing me – Bill Cogn May 10 '23 at 22:14