# Finding type II error for testing mean in normal distribution

Let's consider variable $$X \sim N(\mu, 4)$$. I want to check hypothesis $$H_0:\mu=-1$$ versus $$H_1: \mu = 1$$.

Critical region is given as $$R = \{\overline{X}_n > c\}$$

I want to find such $$c$$ that this test has size of $$\alpha$$ and I want to find probability of committing II type error.

My solution

Finding proper c

We are intereted in finidng such a C that $$P_{\mu = -1}(\overline X_n > c) = 1 - \alpha$$.

Assuming null hypothesis we know that $$\frac{\overline X + 1}{\frac{2}{\sqrt n}} \sim N(0, 1)$$. So:

$$P_{\mu = -1} (\overline X_n > c) = P_{\mu = -1}(\frac{\overline X_n + 1}{\frac{2}{\sqrt{n}}} > \frac{c + 1}{\frac{2}{\sqrt n}}) = 1 - P_{\mu = -1}(\frac{\overline X_n + 1}{\frac{2}{\sqrt{n}}} \le \frac{c + 1}{\frac{2}{\sqrt n}})$$ $$=1 - F_{N(0, 1)}(\frac{c + 1}{\frac{2}{\sqrt n}}) = 1 - \alpha$$

So the $$c$$ that we are searching for is formulated as:

$$c = \frac{2 q_{N(0, 1)}(1 - \alpha)}{\sqrt n} - 1$$

where $$F_{N(0, 1)}$$ and $$q_{N(0, 1)}$$ are respectively CDF and quantile function of standard normal distribution.

Finding probability of committing II type error

If we define our decision function as:

$$\phi(x_1, ..., x_n)= \begin{cases} 1, & \textrm{when} \; (x_1, ..., x_n) \in R \\ 0, & \textrm{when} \; (x_1, ..., x_n) \in R^c \end{cases}$$

II type error is probability of not rejecting null hypothesis when its false.

In other words we are searching for such a probablity that:

$$P_{\mu = 1}(\overline X < c) = P_{\mu = 1}(\frac{\overline X - 1}{\frac{2}{\sqrt n}} < \frac{c - 1}{\frac{2}{\sqrt n}}) = F_{N(0, 1)}(\frac{c - 1}{\frac{2}{\sqrt n}})$$

Do my calculations make sense to you?

• The critical region $\mathcal{R}$ is a subset of the sample space that leads to a rejection of the $H_0$. So, I think you should be finding $C$ so that $$\mathbb{P}\left(\text{Reject }H_0\big|H_0\text{ true}\right)=\mathbb{P}\left(\overline{X}>C\big|\mu=-1\right)$$ equals $\alpha$, not $1-\alpha$.
– user801306
Commented Jan 16, 2022 at 17:08

Your answer for $$c$$ is correct, but your intermediate work contains errors.
As already stated in the comments, the type $$I$$ error probability is $$\Pr[\text{reject } H_0 \mid H_0] = \Pr[\bar X_n > c \mid \mu = -1] = \alpha, \tag{1}$$ whereas you wrote $$P_{\mu = -1}(\bar X_n > c) = 1 - \alpha$$. Then the equation you obtained $$1 - F_{N(0,1)}\left(\frac{c+1}{2/\sqrt{n}}\right) = 1 - \alpha \tag{2}$$ should actually be $$1 - F_{N(0,1)}\left(\frac{c+1}{2/\sqrt{n}}\right) = \alpha \tag{3}$$ in order for the result $$c = \frac{2 q_{N(0,1)}(1 - \alpha)}{\sqrt{n}} - 1 \tag{4}$$ to be correct. Your equation $$(2)$$ would give the incorrect result $$c = \frac{2 q_{N(0,1)}(\alpha)}{\sqrt{n}} - 1, \tag{5}$$ which we can see is wrong because for example, with $$\alpha = 0.05$$ and $$n = 100$$, Equation $$(5)$$ yields $$c = -1.32897 < -1$$ hence the rejection region for the test contains the null mean, which is absurd.
For the Type II error, your result is correct, but simplifies a little further: $$\beta = F_{N(0,1)} \left( \frac{c-1}{2/\sqrt{n}} \right) = F_{N(0,1)} \left( q_{N(0,1)}(1-\alpha) - \sqrt{n} \right).$$