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Power

TrielaM

New member
Jan 17, 2014
2
I have had some issues with an equation in one of my classes, hoping some one can point out what I am going wrong. If any one can point out some flaw in my process or calculations I would appreciate it. The problem lists:

Population µ = 100, σ = 20. Planning to have a sample of 50 participants, and expect the sample mean to be 5 points different from the population mean. Use non-directional hypothesis and α = .05.

What is the Power for this planned study?​

My calculations:
σ /(SqrRt)n = 20/(SqrRt) 50 = 20/7.071 = 2.828 Standard error
α = .05 = z = +1.96
1.96(2.828) = 5.543

z= (M-µ)/Standard Error = (105.543-105)/2.828 = .543/2.828 = 0.19

Z Table: .19 = .5753
Power = 57.53%
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
I have had some issues with an equation in one of my classes, hoping some one can point out what I am going wrong. If any one can point out some flaw in my process or calculations I would appreciate it. The problem lists:

Population µ = 100, σ = 20. Planning to have a sample of 50 participants, and expect the sample mean to be 5 points different from the population mean. Use non-directional hypothesis and α = .05.

What is the Power for this planned study?​

My calculations:
σ /(SqrRt)n = 20/(SqrRt) 50 = 20/7.071 = 2.828 Standard error
α = .05 = z = +1.96
1.96(2.828) = 5.543

z= (M-µ)/Standard Error = (105.543-105)/2.828 = .543/2.828 = 0.19

Z Table: .19 = .5753
Power = 57.53%
Welcome to MHB, TrielaM! :)

It appears you have calculated the type II error $\beta$.
The power is its complement.



In your case:
\begin{aligned}\beta &= 57.53\% \\
\text{Power} &= 42.47\% \end{aligned}
 

TrielaM

New member
Jan 17, 2014
2
Thank you very much for the reply. So in the process I used I needed to add a step for 1-B. Would this step be applicable all the time with this form of equation or no? The books I was using didn't include this but it was certainly the correct answer, instead it had you pull from a Z table in the section for the body, where as here the answer was found in the tail. At least I know where I was flubbing up though I am still not sure how to make the distinction.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Thank you very much for the reply. So in the process I used I needed to add a step for 1-B. Would this step be applicable all the time with this form of equation or no?
Yes.
I would suggest always drawing a picture though, like the one I have included in my previous post.

The books I was using didn't include this but it was certainly the correct answer, instead it had you pull from a Z table in the section for the body, where as here the answer was found in the tail. At least I know where I was flubbing up though I am still not sure how to make the distinction.
Apparently you pulled your result from a Z table that gives the area to the left, which is pretty standard and often shown in a small graph next to the table.
As a result your value of z=0.19 gives you the green area.



But you need the blue area, which is the area to the right of z=0.19.

Note that the power of a test (the blue area) is the probability that we reject the null hypothesis when it should indeed be rejected, because the alternative hypothesis is true.