# How do you find standard deviation for a paired difference test?

My teacher gave an example for this kind of problem. The problem states that you work in Human Resources and you are trying to see if there is a difference in test scores after a training program. There are four pairs for the sample

Before After

Sam 85 94

Tamika 94 87

Brian 78 79

Mike 87 88

The answer for the standard deviation was 6.53. I already know how to find the mean and degrees of freedom. Can someone help me with the standard deviation? Please and Thank You.

• Before After Difference

Sam 85 94 9

Tamika 94 87 -7

Brian 78 79 2

Mike 87 88 1

Mean difference = (9-7+2+1)/4 = 1.25

Var =sum[(diff-mean diff)^2] / 3

= [(9-1.25)^2 + (-7-1.25)^2 + (2-1.25)^2 + (1-1.25)^2] / 3

= 128.75/3

= 42.92

SD = sqrt(Var) = sqrt(42.92) = 6.55

• Standard Deviation Of Differences

RE:

How do you find standard deviation for a paired difference test?

My teacher gave an example for this kind of problem. The problem states that you work in Human Resources and you are trying to see if there is a difference in test scores after a training program. There are four pairs for the sample

Before After

Sam 85 …

• Find The Sample Standard Deviation

• A. √[(40-45.8)²+(38-45.8)²+(57-45.8)²+(53… /5 =√58.96… std= 7.67 B. √[(38-36.8)²+(34-36.8)²+(35-36.8)²+(45… /5 =√20.56… std= 4.53 I believe these to be the correct answers. It is hard to explain how to do it really, but I suggest you consult wikipedia if you want to know how to use the formula. Edit: err, it appears the poster above me got a different answer than I did. I am only in introductory stats, but still, I believe I did my answer correctly, especially because the internet has std solvers, and my answers checked. Edit again: Ya thats right, I remember hearing about that, but of course I forgot.. (so much to remember!). To Blahb: Nice catch on that, I would expect nothing less from a prof though 🙂