0
$\begingroup$

I'm trying to work out how to calculate the necessary sample size for a two sample t-test using excel and need some help.

I don't think I am able to upload an excel attachment, so I will write out what I have.

Cell B1 = mu1, Cell B2 = mu2, Cell B3 = sigma

The calculation i'm doing is

=((1.96+NORM.INV(0.8,0,1))/((B1-B2)/B3))^2

For the entry of mu1 = 0.4, mu2 = 0.43, sigma = 0.1 I'm getting an n = 87.212

Plugging those same numbers into here:

https://www.stat.ubc.ca/~rollin/stats/ssize/n2.html

They are showing that I need almost double of that. Am I doing something incorrectly?

*edit: It's actually almost exactly double. They do state that their's is "sample size (for each sample separately)", so I don't see why I would need to double mine since mine should be for each sample separately as well I believe.

Improve this question
$\endgroup$
2
  • $\begingroup$ What do you expect the number you are calculating to be? The total sample size or the sample size in each of two equally size groups being compared? And presumably you are 100% sure that there will be 0% missing data, if you are using this formula? $\endgroup$
    – Björn
    Commented Jan 23, 2018 at 16:34
  • 1
    $\begingroup$ The denominator looks right (the standardized difference, also known as the effect size, squared). But the numerator should evaluate to a little less than 16: 2 * (1.96 + .84)^2, You seem to be missing that first 2. $\endgroup$
    – zbicyclist
    Commented Jan 24, 2018 at 5:09

1 Answer 1

Reset to default
0
$\begingroup$

A useful reference for these types of questions is Statistical Rules of Thumb, by Gerald Van Belle (2nd ed 2008). Wiley.

Using the formulas in chapter 2 (Lehr's equation), I get 64 in each group, and your website gives 63, so that's good agreement.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Ah ok, I appreciate the reference! I'll have to purchase. Would you mind letting me know if the equation in the excel formula is incorrect, or if I should be doubling for some reason? $\endgroup$
    – FortuneFaded
    Commented Jan 23, 2018 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.