Normal distribution: finding the probability of type II error and power of the test

Question

I have the following values:

– A sample of size 100 is taken from the population.

– Standard Deviation is 5.

– Mean is 125

– $H_0: \mu = 125 \ \ \ \text{against} \ \ \ H_a: \mu < 125$

– Accept $H_0$ if the sample mean $\bar{\mathbb{x}}$ is $\ge 124$.

– Reject $H_0$ if the sample mean $\bar{\mathbb{x}}$ is $< 124$.

I am trying to find (1) the probability of type II error if the true mean is 123.75 and (2) the power of the test.

I did the following:

$$ Z = \dfrac{\bar{\mathbb{x}} - \mu }{ \dfrac{\text{S.D.}}{\sqrt{n}} } = \dfrac{ 125 - 123.75 }{ \dfrac{5}{\sqrt{100}} } = 2.5 $$

I then found that $\phi(2.5) = 0.99379$ by using a table of values of the normal distribution.

The power of the test is $1 - 0.99379 = 0.00621$.

Is this correct?

Answer 1

The significance level of the test you describe is $$\alpha = P(\text{Rej} H_0 \mid H_0\, \text{True}) = P(\text{Rej} H_0 \mid \mu = 125) = P(\bar X < 124 \mid \mu = 125)\\ = P\left(\frac{\bar X - 125}{\sigma/\sqrt{100}} < \frac{124 - 125}{0.5}\right) = P(Z < -2) = \Phi(-2) = 2.275\%.$$

With this significance level (rejection rule), the power against alternative $H_a: \mu = 123.75$ is $$\pi(\mu = 123.75) = P(\text{Rej} H_0 \mid \mu = 123.75) = P(\bar X < 124 \mid \mu = 123.75)\\ = P\left(\frac{\bar X - 123.75}{\sigma/\sqrt{100}} < \frac{124 - 123.75}{0.5}\right) = P(Z < 0.5) = \Phi(0.5) = 69.14\%.$$

Here is a power curve from Minitab. Power for the particular alternative of interest in your problem is denoted with a red dot. (The commands shown were generated by menu choices. Notice that this procedure requires the significance level $\alpha$ as input.)

 MTB > Power;
 SUBC>   ZOne;
 SUBC>     Sample 100;
 SUBC>     Difference -1.25;
 SUBC>     Sigma 5;
 SUBC>     Alternative -1;
 SUBC>     Alpha 0.02275;
 SUBC>   GPCurve.

 Power and Sample Size 

 1-Sample Z Test

 Testing mean = null (versus < null)
 Calculating power for mean = null + difference
 Alpha = 0.02275  Assumed standard deviation = 5

             Sample
 Difference    Size     Power
      -1.25     100  0.691462