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

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The discussion focuses on calculating the power of a study with a population mean (µ) of 100 and standard deviation (σ) of 20, using a sample of 50 participants. The initial calculations incorrectly determined the power to be 57.53%, which was identified as the type II error (β). The correct power is 42.47%, calculated as 1 - β. A key point raised is the importance of distinguishing between areas in Z tables, specifically the need to reference the area to the right of the calculated z-value for power. Understanding this distinction is crucial for accurate statistical analysis in hypothesis testing.
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.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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