Dependent t-test with extra information provided
Thread starter martinl
martinl
I did a dependent t-test for a medical research, it was comparing before and after measurements.
So the dependent t-test should be the right test to use.
The after measurements have a wide range.
And then the client provided me with another piece of information: "it is very normal for the after
measurements to go up to 20. <20 is considered normal, 20-30 prolonged and >30 abnormal".
I am wondering whether this impacts how I should run the paired t-test? Do I need to do some
something extra to include this information to my test?
martinl said:
Why not set the null hypothesis to
Ho: Mu(diff after minus before) less than or equal to 30
Ha: Mu (diff after minus before) greater than 30
If you reject Ho you conclude the mean measurement after is abnormal at the selected alpha level.
You could also set it to
Ho: Mu(diff after minus before) greater than or equal to 20
Ha: Mu (diff after minus before) less than 20
Rejecting Ho allows you to conclude the mean after is normal at the selected alpha level.
The choice would depend on what is more important to conclude, abnormal (use #1) or normal (use #2). Keep in mind that Nonsignificant p-values do not allow for you to conclude Ho.
That sounds like a good idea, but his claim applies only to the after measurement, not the mean difference.
Would your idea still work?
Sorry, I thought you were describing the difference of "up to 20". You also mentioned "comparing before and after" which also sent me down that road.
Sure just test mean of the after measurement instead of mean difference.
Ho: Mu(after) less than or equal to 30
Ha: Mu(after) greater than 30
Ho: Mu(after) greater than or equal to 20
Ha: Mu(after) less than 20
ondansetron said:
That would no longer be a paired t-test. What do we do with the paired data?
Right. You just said you don't want to look at mean differences (can be for dependent samples t test), only an after measurement. Maybe we're not on the same page.
The only thing you need to do with the paired test is figure about normal and abnormal differences and use that to set the null and alternative in a similar manner we discussed earlier.
Oh, sorry I was wrong. His claim applied to the mean. So you were right.
So I did a one tailed test, Ho is "mean difference <20".
The SPSS test I used was one sample t test, I set the test value = 20. The resulting p value = 0.000 (see images below).
I am wondering: how should I do the rest ? Thanks in advance.
Ho wouldn't be < 20 in this case. It would need to be greater than or equal to 20 if you wanted a rejection of Ho to tell you the difference is normal (<20). It looks like this set your null hypothesis to Ho mu(diff)=20 with a two-tail test. Depending whether you subtracted before from after or after from before, you will have a different conclusion from the 2-tailed p-value and CI.
Why not just use the paired t test and specify Ho value of 20 and conduct a one-tailed test?
How did you create the difference variable and are POSITIVE differences what the research cares about or an absolute difference?
Can I calculate like this :
The p value from SPSS is <0.001, so half of that will be <0.0005, that's the probability of mean >20.
So the probability of mean <=20 is (1 - 0.0005).
You're really misinterpreting what a p-value represents here
Likes: ondansetron
I don't think so, I am doing a one tailed test now, and 0.000 is the probability of getting >20.
Can you tell me what you think the definition of p-value is? (It's a serious question - I think you have it wrong and I want to help you correct your misconception)
I agree with Dason, based on what you've written in both posts.
This image illustrates my calculation.
The p value I got from the test is 0.000, and the test is a two tailed test with test value=20.
That p value is represented by the far right orange-coloured area.
It is a p value from a two tailed test, but I am doing a one tailed test (Ho: mu < 20), so my p value should be
0.000 divided by 2, which is still approximately 0.000.
What I am after is the probability of getting a result of <20 -- which is represented by the grey area, so it should be
(1 - 0.000) or approximately 1. So I shouldn't reject the Ho (mu < 20).
It is a two tailed p value that SPSS spit out, but a p value doesn't represent the probability of a result < 20 in your example.
A p value is the probability of receiving a summary measure at least as extreme as the observed one, assuming the null hypothesis is true.
Your null hypothesis, as it's constructed, is not Ho mu < 20. This isn't an equivalence/noninferiority-type study and you're not using appropriate methodology to conduct a test that way.
The pvalue you're using is calculated using a null of mu=0 and your drawing demonstrates this. The probability you calculated is that of obtaining a value less than 20 assuming the true difference is zero.
I recalculated it, it is still not significant under Ho: mu <=20. The t-value is negative, but the Critical Value is 1.729 (df=19).
So cannot reject Ho.
It was 1 tailed.
I'm not sure why you conducted the test that way. Additionally, it seems you need review in these areas.
I would recommend this before giving instructions on this topic.
Edited due to autocorrect.
I would recommend this before giving instructions on this trip.
I was trying to perform a one tailed t-test, with Ho: mu <=20 (given that mu can go up to 20 and still be considered normal).
Isn't this correct?