Calculating Power for a Study: Population µ = 100, σ = 20

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SUMMARY

The discussion revolves around calculating the statistical power for a study with a population mean (µ) of 100 and a standard deviation (σ) of 20, using a sample size of 50 participants. The initial calculations incorrectly yielded a power of 57.53%, which was identified as the type II error (β). The correct power, calculated as 1 - β, is 42.47%. Participants are advised to always consider the area in the tail of the Z table when determining power, as this represents the probability of correctly rejecting the null hypothesis.

PREREQUISITES
  • Understanding of statistical power and type I/type II errors
  • Familiarity with Z tables and their applications in hypothesis testing
  • Knowledge of standard error calculations
  • Ability to perform basic statistical calculations involving means and standard deviations
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  • Learn how to calculate statistical power using different sample sizes and effect sizes
  • Explore the implications of type I and type II errors in hypothesis testing
  • Study the use of graphical representations in understanding statistical concepts
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Students in statistics courses, researchers conducting hypothesis testing, and anyone involved in designing studies requiring power analysis.

TrielaM
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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 anyone 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%
 
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TrielaM said:
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 anyone 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.

Power.png


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

Power.png


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.
 

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