Question

Use the Neyman-Pearson Lemma to obtain the best critical region for testing H0: mu=0
against H1: mu<>0
in the case of a normal population N(mu , sigma^2), where sigma^2 is known.
Hence find the power of the test.

Answer 1

Answer on Question #81194 – Math – Statistics and Probability

Question

Use the Neyman-Pearson Lemma to obtain the best critical region for testing

H0: mu=0 against H1: mu<>0 in the case of a normal population N(mu, sigma^2), where sigma^2 is known. Hence find the power of the test.

Solution

Consider the relation of likelihood

$$\frac {L _ {1} (\overline {{x}})}{L _ {0} (\overline {{x}})} = \frac {\prod_ {i = 1} ^ {n} \frac {1}{\left(\sqrt {2 \pi} \sigma^ {2}\right)} \exp \left(- \frac {(x _ {i} - \mu) ^ {2}}{2 \sigma^ {2}}\right)}{\prod_ {i = 1} ^ {n} \frac {1}{\left(\sqrt {2 \pi} \sigma^ {2}\right)} \exp \left(- \frac {x _ {i} ^ {2}}{2 \sigma^ {2}}\right)} = \prod_ {i = 1} ^ {n} \exp \left(- \frac {(x _ {i} - \mu) ^ {2}}{2 \sigma^ {2}} + \frac {x _ {i} ^ {2}}{2 \sigma^ {2}}\right) = \exp \left(\frac {2 n \mu \overline {{x}} - n \mu^ {2}}{2 \sigma^ {2}}\right)$$

Then the critical region by Neyman-Pearson lemma will be of the form

$$\frac {2 n \mu \bar {x} - n \mu^ {2}}{2 \sigma^ {2}} > C _ {0}$$

If $\mu > 0$ it yields $\overline{x} > C_1$, if $\mu < 0$ it yields $\overline{x} < C_2$.

Totally the critical region has form $\overline{x} \in (-\infty; C_2) \cup (C_1; +\infty)$. We take $C_2 = -C_1$ because in this case the length of the interval when $H_0$ is accepted is maximum. So the critical area has a form $|\overline{x}| > C$. We find C from the equation

$$P _ {0} (| \overline {{x}} | > C) = \alpha .$$

Under the hypothesis $H_0$ the distribution of $\overline{x}$ is $N\left(0, \frac{\sigma^2}{n}\right)$. Then

$$P _ {0} (| \overline {{x}} | > C) = P _ {0} \left(\left| \frac {\overline {{x}} \sqrt {n}}{\sigma} \right| > \frac {C \sqrt {n}}{\sigma}\right) = P _ {0} \left(| z | > \frac {C \sqrt {n}}{\sigma}\right) = 2 F \left(- \frac {C \sqrt {n}}{\sigma}\right).$$

We have an equation

$$2 F \left(- \frac {C \sqrt {n}}{\sigma}\right) = \alpha ,$$

from which

$$C = - \frac {\sigma F ^ {- 1} (\alpha / 2)}{\sqrt {n}}$$

If $|\overline{x}| > C$ the hypothesis $H_0$ is declined, otherwise it is accepted.

The power of the test is

$$\begin{array}{l} P _ {1} (| \overline {{x}} | > C) = 1 - P _ {1} (- C < \overline {{x}} < C) = 1 - P _ {1} \left(\frac {- C - \mu}{\sigma \sqrt {n}} < \frac {\overline {{x}} - \mu}{\sigma \sqrt {n}} < \frac {C - \mu}{\sigma \sqrt {n}}\right) = \\ = 1 - \left(F \left(\frac {C - \mu}{\sigma \sqrt {n}}\right) - F \left(\frac {- C - \mu}{\sigma \sqrt {n}}\right)\right) \\ \end{array}$$

This value is a minimum for $\mu = 0$, hence the power of the test is $1 - \alpha$.

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