Hello,
I aim to compare the Length of Stay (LOS) in hours b/n two independent groups, with sample sizes of 300 and 310. The LOS data does not follow a normal distribution and shows right-skewed dis in both groups. I expected a p-value less than 0.05 to reject the null hypothesis; however, the result from the Wilcoxon test indicates a p-value of 0.3, which is greater than 0.05.
Median/mean for one group is 76/99 For another is 96/112
could you plz help me what can I do or what should I report?
I think there may be lots of tie, that is why P-value is confusing
Thanks
as I expected to have a decrease in LOS in the second group, I think I need to look at the 0.15 not 0.3 , but still 0.15 is >0.05
Z-value is negative, which means the second group has a lower los, I was confused why p-value is more than 0.05
Statistic Z: -1.0066
Pr |Z| Pvalue= 0.3
In Excel you could use the function =RAND()*0.05, in R you may use runif(1, max = 0.05) to generate a p-value between 0 and 0.05 with guaranteed success. This works for any data, makes no assumptions whatsowever about the distribution and test a hypothesis that is similarly undefined. The wonderful extra feature is that it will work even in the absence of any data, saving lots of time and money, and facilitating research.
I think what Jochen is trying to say is that your question is already ill formulated. Statistics is not about finding a method, so that your p-value meets your expectations (i.e. < .05), but to find a model, which describes your data in a best possible way, independent of your hypotheses and independent if you get a statistically significant result. If you try it the other way around, it is simply p-hacking and fraud!
Looking at your data, I would guess that neither a t-test is appropriate, nor does the Wilcoxon test answers your question. Contact a local statistician to help you with your data.
Thank you for your answer, I am not looking for a method to change the p-value; I am just looking for the justification/explanation.
The research aims to compare the (LOS) pre-post intervention b/n two independent groups. It was anticipated that the second group would decrease LOS.
For example, both the mean and median LOS are lower in the post-intvn group compared to the pre-intvn group.
- Pre-intvn group: Mean LOS = 82, Median LOS = 72
- Post-intvn group: Mean LOS = 71, Median LOS = 55
However, when I conduct the Wilcoxon test, the p-value does not indicate statistical significance. My question is: Why is there a noticeable decrease in LOS in the second group, yet the p-value does not reflect this significant change?
If you want to compare the medians of two independent groups, you could use quantile regression an indicator variable for Group as the explanatory variable. It looks like you use SAS, so this might help:
https://support.sas.com/documentation/onlinedoc/stat/131/qreg.pdf
> "Why is there a noticeable decrease in LOS in the second group, yet the p-value does not reflect this significant change?"
The Wilcoxon-test is not testing this hypothesis (it's neither about the equality fo means nor of the equality of medians). It's testing the hypothesis that the probability that a random member of one group will have a higher rank than a random member of the other group.
For your question about the p-values: even IF you would have chosen an appropriate test for the data AND there ist a true difference between the two populations, statistical significance is directly dependent on sample size (if everything else remains equal). You could try it out yourself simply with a onesample t-test, either by simulating data (e.g. take a sample of size n from normally distributed population, keep the mean and standard deviation constant and use different values for n) or even simpler, just calculate the test statistic for the one sample t-test and incorporate values for the mean and standard deviation. You will see that the ratio of (mean-mu)/SD will be multiplied by sqrt(n). Hence, even for a constant ratio, the test statistic will increase with increasing sample size, and in turn the p-value decreases.
This behavior is generally true and may be the explanation. Your sample size is probably not large enough to detect the size of the difference your are interested in. I suggest to do a power analysis (a priori) to get the needed sample size.
But all this has nothing to do with the right choice of an appropriate test, of course, and it seems that this is another problem you have. Therefore, as mentioned already above, it may be a good idea to consult a local statistician.
P.S.: How can you calculate changes from t1 to t2 with unequal group sizes. Shouldnt they be dependent and equally large?
P.P.S.: your sample size "appears" large in absolute values, but if you look at the very skewed distributions, the variances are very large and a t-test would probably not be able to detect a statistically significant difference (although it would be about the intraindividual differrences, which we cannot see here), but the shapes are quite equal, why the Wilcoxon may not be significant.
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