Calculating power of a one sample two-sided test, what is d in this example?

Question

My book presents a formula and an exercise how to calculate the power of a one sample two-sided test, but I'm not clear on one variable used in that formula.

First, the exercise wants us to find the power of an appropriate test, given the following information:

• We have one sample of $20$ observations from a normally distributed population with standard deviation 5.
• We want to test at 5% significance level $H_0:\mu=0$ and $H_1:\mu\not=0$
• Assume the true value of the population mean is $2$.

The formula in the book for such a case, where "$d$ is not small" is as follows:

$$ 1 - \Phi\bigg(q_{1-(\alpha/2)} - \frac{d}{\sigma / \sqrt{n}}\bigg) $$

What is clear to me:

• $n=20$, the sample size
• $\sigma=5$, the standard deviation of the population
• 5% significance level implies $\alpha=0.05$

But what is $d$? In the book they say "the true value of $\mu$ is $\mu_0 \pm d$", so since we assume $\mu_0=0$ and we are given $\mu=2$, what I would conclude is $d = 2$?

Maybe it's an obvious question but I can't really grok what they want to say with "the true value of $\mu$ is $\mu_0 \pm d$".

Answer 1

Here $d=\mu_1-\mu_0$ or in the example $d=2-0=2$, where $\mu_1$ is the value you are measuring the power for. The argument goes:

• If your sample observations are $x_1,\ldots x_{20}$ with average $\bar x$ then

• your test will be to reject $H_0$ if $\Phi\left(\frac{\bar x-\mu_0}{\sigma/\sqrt{n}}\right)< \frac{\alpha}{2}$ or if $\Phi\left(\frac{\bar x-\mu_0}{\sigma/\sqrt{n}}\right)>1- \frac{\alpha}{2}$

• which you can rewrite as rejecting if $\bar x < \mu_0 +\frac{\sigma}{\sqrt{n}}q_{\alpha/2}$ or $\bar x > \mu_0 +\frac{\sigma}{\sqrt{n}}q_{1-\alpha/2}$.

• The power, i.e. the probability of rejection if in fact $X_i\sim \mathcal N(\mu_1,\sigma^2)$ is $$\Phi\left(\frac{\mu_0 +\frac{\sigma}{\sqrt{n}}q_{\alpha/2}-\mu_1}{\sigma/\sqrt{n}}\right) + 1 - \Phi\left(\frac{\mu_0 +\frac{\sigma}{\sqrt{n}}q_{1-\alpha/2}-\mu_1}{\sigma/\sqrt{n}}\right) \\=\Phi\left(q_{\alpha/2}-\frac{\mu_1-\mu_0}{\sigma/\sqrt{n}}\right) + 1-\Phi\left(q_{1-\alpha/2}-\frac{\mu_1-\mu_0}{\sigma/\sqrt{n}}\right) \\=\Phi\left(q_{\alpha/2}-\frac{\mu_1-\mu_0}{\sigma/\sqrt{n}}\right) + \Phi\left(\frac{\mu_1-\mu_0}{\sigma/\sqrt{n}}-q_{1-\alpha/2}\right)$$

• If $\mu_0 < \mu_1$, the first term in those sums is relatively small (it would involve rejecting the null hypothesis for the observations being too small when you are interested in the expectation being higher than in the null hypothesis) and is ignored in the quotation