Dependent t-test with extra information provided

M

martinl

Member
#1
#1
Dear all,

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?

Thank you in advance.

M
 
O

ondansetron

TS Contributor
#2
#2
martinl said:
Dear all,

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?

Thank you in advance.

M
Click to expand...
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.
 
M

martinl

Member
#3
#3
That sounds like a good idea, but his claim applies only to the after measurement, not the mean difference.
Would your idea still work?
 
O

ondansetron

TS Contributor
#4
#4
martinl said:
That sounds like a good idea, but his claim applies only to the after measurement, not the mean difference.
Would your idea still work?
Click to expand...
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

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(after) greater than or equal to 20
Ha: Mu(after) less than 20

Rejecting Ho allows you to conclude the mean after is normal at the selected alpha level.
 
M

martinl

Member
#5
#5
ondansetron said:
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

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(after) greater than or equal to 20
Ha: Mu(after) less than 20

Rejecting Ho allows you to conclude the mean after is normal at the selected alpha level.
Click to expand...
That would no longer be a paired t-test. What do we do with the paired data?
Thank you.
 
O

ondansetron

TS Contributor
#7
#7
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.
 
M

martinl

Member
#8
#8
ondansetron said:
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.
Click to expand...
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.

1520414356814.png
 
Last edited:
O

ondansetron

TS Contributor
#9
#9
martinl said:
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.

View attachment 119
Click to expand...
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?
 
M

martinl

Member
#10
#10
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).
 
Dason

Dason

Ambassador to the humans
#13
#13
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)
 
M

martinl

Member
#15
#15
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).
1520466455694.png
 
Last edited:
O

ondansetron

TS Contributor
#16
#16
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.
 
M

martinl

Member
#17
#17
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.
 
Last edited:
O

ondansetron

TS Contributor
#19
#19
martinl said:
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.
Click to expand...
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.
 
Last edited:
M

martinl

Member
#20
#20
ondansetron said:
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 trip.
Click to expand...
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?
 
You must log in or register to reply here.