if the confidence intervals (CI) are based on a geometric mean (so a geometric CI) , does that alter the equation to require a geometric "T-value"? if yes, how do u calculate the geometric t value? thx a ton!!
Very interesting indeed! But I wonder, why do you use a T distribution? Can't we use the inverse of a normal distribution? Thank you so much for this video.
If you have a sample that is large (>30 is generally considered "large enough") the distribution of the sample is approximately normal and you can use the z value rather than the t value. (They become indistinguishable as the sample size increases.)
Thank you for this informative and useful video. I have a question about Excel Formula which you used. When I try to write exactly the same formula, something went wrong and i haven't figure out what is the problem yet.. Excel didn't accept dollar symbol ($). Rest of the formula is right except this symbol.
Higgins JPT, Green S (editors). Cochrane Handbook for Systematic Reviews of Interventions Version 5.1.0 [updated March 2011]. The Cochrane Collaboration, 2011. Available from www.handbook.cochrane.org.
Thanks for the video! I can't seem to get this to work for me though. Tried it with your numbers to check my work and the formula despite being a copy gave me Group 1 4.547430071 Any ideas?
Why is the resuing SD actually greater than the confidence range?
Is there any paper describing this method which you can cite? Would really appreciate if you know of any.
Morning! I copied the formula but it shows the error. Do you have any idea WHY?! =SQRT(D2)*(B2-C2)/(T.DIST.2T(0.05;$D$2-1)*2)
Can you tell me where I went wrong?
90% CI
The SD should be 3.508
This is in R
sqrt(25)*(77-65)/qt(0.05, 24, lower.tail = FALSE)
if the confidence intervals (CI) are based on a geometric mean (so a geometric CI) , does that alter the equation to require a geometric "T-value"? if yes, how do u calculate the geometric t value? thx a ton!!
Who is behind all these fantastic videos? Thanks any way.
Very interesting indeed! But I wonder, why do you use a T distribution? Can't we use the inverse of a normal distribution? Thank you so much for this video.
If you have a sample that is large (>30 is generally considered "large enough") the distribution of the sample is approximately normal and you can use the z value rather than the t value. (They become indistinguishable as the sample size increases.)
Thank you for this informative and useful video. I have a question about Excel Formula which you used. When I try to write exactly the same formula, something went wrong and i haven't figure out what is the problem yet.. Excel didn't accept dollar symbol ($). Rest of the formula is right except this symbol.
My hunch is the order in which you placed the $ is incorrect. Other than that, I can't say.
Do you have the source/reference for this formula?
Higgins JPT, Green S (editors). Cochrane Handbook for Systematic Reviews of Interventions Version 5.1.0 [updated March 2011]. The Cochrane Collaboration, 2011. Available from www.handbook.cochrane.org.
@@ubeballer Thank you!!
Thanks for the video! I can't seem to get this to work for me though. Tried it with your numbers to check my work and the formula despite being a copy gave me Group 1 4.547430071 Any ideas?
thanks so much!!!!
Thanks, my friend!!!!
What does it mean "very accurately". Isn't it precise ??
Yes, it's both accurate and precise. I wrote that in the title, because there are methods that are less accurate.