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

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?

• Can you show details of your answer? Feb 12, 2016 at 16:06
• I edited and I showed what I did Feb 12, 2016 at 16:18
• Right track. See Answer for corrected result. Feb 13, 2016 at 1:52

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


• Thank you for your help . It was really helpful Feb 13, 2016 at 18:12
• Mon plasir. Power in an important idea, but first contact with the concept is often confusing. Feb 13, 2016 at 19:09
• Hi Bruce, would you please explain the reasoning for how you got that $$P(\bar X < 124 \mid \mu = 125) = P\left(\frac{\bar X - 125}{\sigma/\sqrt{100}} < \frac{124 - 125}{0.5}\right)$$? May 26, 2021 at 19:58
• Inside $(\;)$ standardize $\bar X$ to get $Z = \frac {\bar x - 125} {\sigma/\sqrt{100}},$ and $\frac{124-125}{\sigma/\sqrt{100}},$ where $\sigma/\sqrt{100} = 5/10 = 0.5).$ May 26, 2021 at 20:06