0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Can you show details of your answer? $\endgroup$
    – DeepSea
    Feb 12, 2016 at 16:06
  • $\begingroup$ I edited and I showed what I did $\endgroup$
    – user290335
    Feb 12, 2016 at 16:18
  • $\begingroup$ Right track. See Answer for corrected result. $\endgroup$
    – BruceET
    Feb 13, 2016 at 1:52

1 Answer 1

2
$\begingroup$

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

enter image description here

$\endgroup$
4
  • $\begingroup$ Thank you for your help . It was really helpful $\endgroup$
    – user290335
    Feb 13, 2016 at 18:12
  • $\begingroup$ Mon plasir. Power in an important idea, but first contact with the concept is often confusing. $\endgroup$
    – BruceET
    Feb 13, 2016 at 19:09
  • $\begingroup$ Hi Bruce, would you please explain the reasoning for how you got that $$P(\bar X < 124 \mid \mu = 125) = P\left(\frac{\bar X - 125}{\sigma/\sqrt{100}} < \frac{124 - 125}{0.5}\right)$$? $\endgroup$ May 26, 2021 at 19:58
  • $\begingroup$ Inside $(\;)$ standardize $\bar X$ to get $Z = \frac {\bar x - 125} {\sigma/\sqrt{100}},$ and $\frac{124-125}{\sigma/\sqrt{100}},$ where $\sigma/\sqrt{100} = 5/10 = 0.5).$ $\endgroup$
    – BruceET
    May 26, 2021 at 20:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .