For some reason, I've been stuck on this question, even though other questions of this type weren't that much of a problem to me before (with for instance the binomial distribution). The situation is as follows:

Suppose $X_1...X_{25}$ is a sample from an $N(\mu,2^2)$-distribution. We would like to test for $H_0: \mu\leq 0$; $H_1:\mu>0$, with significance $a_o=0.05$.

Calculate the power function of this test at the point $\mu=0.5$.

My approach:

Earlier in this problem I had to calculate the critical values and found the following: $K_T = \{T\in \mathbb{R}:T\geq 1.64\}$} and $K=\{\mu\in\mathbb{R}:\mu\geq0.656\}$} where $T=\sqrt{n}\frac{\mu-\mu_0}{\sigma}$

For $H_0$ we use $\mu_0=0$ and find (with a table or calculator) these values for $c$ in $P(T\geq c)\leq0.05$.

Now we have to use $\mu_0=0.5$. Given this information, we have to calculate $P_{0.5}(X\in K)$ which is the power function. I thought of two things, but they both give me wrong answers.

First I tried to just calculate $P(T\in K_T$) given $\mu=0.5$, but I stumbled upon a problem; I didn't seem to have used the fact that $\mu=0$ in my calculation for the critical values for $K_T$. I just chose $T=\sqrt{n}\frac{\mu-\mu_0}{\sigma} =\frac{5}{2}\mu$ and said that $T\sim N(0,1)$ and proceeded to use the table. Only for $K$ I used that $T=\frac{5}{2}\mu\geq1.64 \iff \mu\geq0.656$. (so here we used $\mu_0=0\implies T=\frac{5}{2}\mu$)

So, changing to $\mu_o=0.5$ didn't really give me anything in this calculation, if I would just put it in there I would get that $P(T\in K_T)=0.05=a_0$ again.

Then I tried to calculate $$P(X\in K)=P(N(0.5,2^2)\geq1.64)=$$ $$P(4N(0,1)+0.5\geq1.64)=P(N(0,1)\geq 0.285)=0.39$$

This comes closer, but the correction model says it's $0.34$. I did find, that $P(N(0,1)\geq a)=0.34$ for $a=0.41$ which is $1.64/4$, the original critical value divided by $4$.

I really feel like I'm missing a (relatively simple) key element of the way to approach this problem. Any corrections, tips, hints or explanations are highly appreciated!


2 Answers 2


Under $H_o$, find the critical value (C) for $\bar X$, which is $$\Pr(\bar X > C) =\Pr(\frac {\bar X}{2/5}>\frac C{2/5}) = 0.05$$ We have $\frac C{2/5} = 1.64$, so $C= 1.64\times \frac 25=0.656$. You did it correctly.

Then think about if $H_a: \mu = 0.5$ is true, $\bar X \sim N\left(0.5, \frac {2^2}{25}\right)$.

The Probability of $\bar X>0.656$, which is power. is

$$\Pr(\bar X >0.656) = \Pr(\frac{\bar X -0.5}{2/5}>\frac {0.656-0.5}{2/5})$$ It equals $Z = (0.656 - 0.5) \times 5/2 = 0.39$

From standard normal distribution, $\Pr(Z>0.39) = 0.3483$


Comment: The power computation in the Answer by @a_statistician is correct (+1) for $n=25, \alpha= .05, \sigma = 2,$ and $\Delta = \mu_a - \mu_0 = 0.5$ (one-sided alternative).

Because the power for these parameters seems lower than is usually considered desirable, I show power functions from Minitab software that illustrate what is required for larger power values. I use both $n = 25$ (lower curve) and $n = 150$ (upper).

Power and Sample Size 

1-Sample Z Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.05  Assumed standard deviation = 2

Difference    Size     Power
       0.5      25  0.346475
       0.5     150  0.921760

enter image description here

From the upper curve we see that the power increases from about 0.35 to about 0.92 if we increase $n$ from 25 to 150. Alternatively, from the lower curve we see that with $n = 25$ the power against the alternative $\mu_a = 1.2$ is about 0.9.

Notes: (a) The slight difference between power 0.3465 and 0.3483 seems to be because of rounding necessary to use a printed standard normal table.

(b) To be fussy, the vertical axis might be labeled "P(Reject)" instead of "Power": Except for one point, the points on the curve are power, that is, $P(\text{Reject } H_0 | \mu_a).$ The exception is that the left-most point shows the significance level, $\alpha = P(\text{Reject } H_0 | \mu_0 = 0).$

(c) Many statistical software programs include procedures for making power curves. R has a library devoted to power curves.

(d) Some textbooks give a formula for the power of such a (one-sided) z-test: $$ \pi(\mu_a) = P\left(Z > z_{\alpha} - \frac{|\mu_0-\mu_a|}{\sigma/\sqrt{n}}\right).$$ The Answer by @a_statistician can be used as a template for a proof of this formula. With a slight change in notation, I copied it from Ott & Longnecker, 5e, Chapter 5. It is the basis for making power curves such as the one above. In R, the specific computation in your Question is as follows:

1 - pnorm(qnorm(.95) - .5/(2/5))
[1] 0.3464755

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