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I have these values:

A sample of size 100 is taken from the population.

Standard Deviation is 5

Mean 125

H0: $ \mu=$ 125 against Ha:$ \mu<125 $

Accept H0 if the sample mean $\bar{\mathbb{x}} $ is ≥ 124

Reject H0 if the sample mean $\bar{\mathbb{x}} $ is < 124

1) Find the probability of Type II error if the true mean is 123.75 the power of the test.

I did:

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

$ \phi(2.5) =0.99379$ by using table book of normal distribution

The power of the test is 1-0.99379=0.00621

Am I correct?

thanks

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  • $\begingroup$ Can you show details of your answer? $\endgroup$ – DeepSea Feb 12 '16 at 16:06
  • $\begingroup$ I edited and I showed what I did $\endgroup$ – user290335 Feb 12 '16 at 16:18
  • $\begingroup$ Right track. See Answer for corrected result. $\endgroup$ – BruceET Feb 13 '16 at 1:52
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The significance level of the test you describe is $$\alpha = P(\text{Rej } H_O | H_0\, True) = P(\text{Rej } H_0 | \mu = 125) = P(\bar X < 124|\mu = 125)\\ = P\left(\frac{\bar X - 125}{\sigma/\sqrt{100}} < \frac{124 - 125}{.5}\right) = P(Z < -2) = \Phi(-2) = 2.275\%.$$

With this significance level (rejection rule), the power against alternative $H_a: \mu = 123.75$ is $$\pi(\mu = 123.75) = P(\text{Rej } H_0 | \mu = 123.75) = P(\bar X < 124|\mu = 123.75)\\ = P\left(\frac{\bar X - 123.75}{\sigma/\sqrt{100}} < \frac{124 - 123.75}{.5}\right) = P(Z < 0.5) = \Phi(0.5) = 69.14\%.$$

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

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  • $\begingroup$ Thank you for your help . It was really helpful $\endgroup$ – user290335 Feb 13 '16 at 18:12
  • $\begingroup$ Mon plasir. Power in an important idea, but first contact with the concept is often confusing. $\endgroup$ – BruceET Feb 13 '16 at 19:09

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