Thread: wilcoxon rank sum test or t-test?

1. wilcoxon rank sum test or t-test?

Hello,

I have a question for my analysis and reporting.

For my analysis, simply said I measured the time needed to achieve sufficient pain relief in hospitalized patients.

I have 2 groups with "time needed to achieve sufficient pain relief in hours"
Group 1: 4, 8, 16, 20, 20, 20, 24, 24, 24, 24, 24, 24, 44, 52, 56, 72, 72, 76, 92, 92, 96, 120, 120, 144
Group 2: 4, 8, 24, 24, 24, 28, 48, 48, 48, 120, 120

I compared the 2 groups with a unpaired t-test and reported the results as follows:

For 11 patients in group 2 a mean of 45.1 hours (SD 39.9) was measured to achieve sufficient pain relief versus 52.8 hours (SD 40.4) for 24 patients in group 1. (p=0.6)

Here I got the written remark normal distribution? --> Median

So ok I had to check for normality. With some reading, I think I figured out how to do this. And came to conclusion data in both groups are not normally distributed.
So t-test no good. There is no similar data in the literature to see if this type of data is normally distributed.
So i should do the nonparametric alternative test. I looked for it: the Mann Withney U test. I did the analysis in SPSS for 2 independent samples and p=0.6 for the t-test became p=0.77 with the Mann Withney U test.
So no influence on the interpretation of the results. Should I report with mean or median?? Is my analysis ok?

Please respond, also when everything was done correctly.

2. Re: wilcoxon rank sum test or t-test?

The assumption of normality of the sample is for a one sample t test. You have a two sample and the expectation is for normality of the populations. This means your sample does not have to be normally distributed (this works on the central limit theorem).

A nice write up of assumptions of t-tests can be found here.

3. The Following User Says Thank You to trinker For This Useful Post:

noob (08-04-2011)

4. Re: wilcoxon rank sum test or t-test?

So because I have 2 samples I can use the t-test. The assumption for normality is for the populations not necessarily for the samples you have.

From your attachment I can make

1) Is it fair and honest to describe the 2 distributions using means and SDs?

I think so? Why did I get the written remark normal distribution? --> median
Like you said normallity is not the issue, also it doesn't matter if I report with median or mean

2) the 2 variances may not differ 4 times.

The standard deviations in both groups are the same. So ok

--> my initial reporting using the t-test was ok. I didn't have to change anything.

"For 11 patients in group 2 a mean of 45.1 hours (SD 39.9) was measured to achieve sufficient pain relief versus 52.8 hours (SD 40.4) for 24 patients in group 1. (p=0.6)"

5. Re: wilcoxon rank sum test or t-test?

Originally Posted by noob
Here I got the written remark normal distribution? --> Median
Where did this remark come from? a prof? A computer program?

6. Re: wilcoxon rank sum test or t-test?

From a prof in medicine (not statistician)

7. Re: wilcoxon rank sum test or t-test?

Originally Posted by noob
I think so? Why did I get the written remark normal distribution? --> median
Like you said normallity is not the issue, also it doesn't matter if I report with median or mean
With such a small sample getting a normal distribution is highly unlikely. The point of the t-test was to generalize to the population (beer in its original use) based on a very small sample size when of the population was unknown.
Is it fair and honest to describe the two distributions using means and SDs?
The difference between the mean and medians of the groups...
Code:
group: group 1
group         time
group 1:24   Min.   :  4.00
group 2: 0   1st Qu.: 23.00
Median : 34.00
Mean   : 52.83
3rd Qu.: 80.00
Max.   :144.00
------------------------------------------------------------
group: group 2
group         time
group 1: 0   Min.   :  4.00
group 2:11   1st Qu.: 24.00
Median : 28.00
Mean   : 45.09
3rd Qu.: 48.00
Max.   :120.00 
...is quite large, and a box plot revels highly skewed data and a possible outlier. The use of the nonparametric is a judgment call. After running your data and plotting out the groups (I used a density plot, histogram, and boxplot) I would say it is not fair to describe this data with mean.

Running those plots of the data really helps with making judgments about tests. In conclusion I'd say go with the non parametric Mann-Whitney, but at the end of the day the outcome is the same, non significance.

8. The Following User Says Thank You to trinker For This Useful Post:

noob (08-04-2011)

9. Re: wilcoxon rank sum test or t-test?

Thanks a lot, Trinker!
Now it's clear to me.
Mann-Whitney and median instead of mean (+- SD)c it is.

10. Re: wilcoxon rank sum test or t-test?

Well it's not that clear cut: The choice is up to you. Both sample density plots are skewed and to the left. A log transformation (see the article) would bring both distributions to more normal. If you run the Welch's t-test (does not assume = variance) with a log transformation you actually get a p-value of .5532 and for Student's t-test p-value = .5321. Closer to, but still not, significant. I don't want you to walk away with a black and white view of this decision process. This is a research/statistician call. You need to be able to justify your call. I really encourage you to reread the link (click here) and use that to guide your decision making. A transformation may be appropriate, a non-parametric, or rely on the robustness of the t-test. In this case the skew is pretty apparent so I would lean more towards the transformation or the non-parametric.

11. The Following User Says Thank You to trinker For This Useful Post:

noob (08-07-2011)

12. Re: wilcoxon rank sum test or t-test?

I have decided to do the log transformation followed by student's t-test (p=0.53). The variances are equal so Welch's t-test is not necessary.
Thanks again.

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