- #1
MBCT
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Hi,
I'm really struggling with a statistical problem:
5 people had a medical treatment, they were tested before and after using a questionnaire which recorded some numerical data. I want to analyse the before and after results to see whether there was any change, whether it is statistically significant and then write this up correctly.
I don't have SPSS so I have been using the Data Analysis functions of Microsoft Excel as described here:http://www.stattutorials.com/EXCEL/EXCEL_TTEST2.html
So (before score 1-5, after score 1-5)
Person 1 - 2.75, 3.65
Person 2 - 1.75, 3.25
Person 3 - 3.75, 3.50
Person 4 - 1.75, 2.50
Person 5 - 2.75, 4.00
So first I did a t-Test Paired Two Sample for Means which gave me the following:
Mean 2.55, 3.384
Variance 0.7, 0.31853
Observations 5
Pearson Correlation 0.600384
Hypothesized Mean Difference 0
df 4
t Stat -2.77529
P(T<=t) one-tail 0.02503
t Critical one-tail 2.131847
P(T<=t) two-tail 0.050059
t Critical two-tail 2.776445
So, the important information seems to be that on a one tail P I have got 0.02 which is less than 0.05 and therefore I could reject the null hypothesis and say that the change is not due to chance and the treatment is statistically significant.
But, I only have a sample of 5 and therefore can't really say that can I? What sample size would I need to reject the null hypothesis and how can I calculate this?
Also, as tempting as it is to use one-tail, actually the treatment might be ineffective or make things worse and so maybe I should use two-tail, in which case is 0.050059 enough above 0.05 to make rejecting the null hypothesis impossible even if the sample size was much higher?
I then worked out the difference between the two sets of data by doing "after minus before" treatment giving me:
Person 1 - 0.92
Person 2 - 1.50
Person 3 - -0.25
Person 4 - 0.75
Person 5 - 1.25
Using Excel's Descriptive Statistics gave me this information:
Mean 0.834
Standard Error 0.30051
Median 0.92
Mode #NA
Standard Deviation 0.67196
Sample Variance 0.45153
Kurtosis 1.856022
Skewness -1.24463
Range 1.75
Minimum -0.25
Maximum 1.5
Sum 4.17
Count 5
Confidence Level (95%) 0.834348
Confidence Interval -0.00035, 1.668348
Now, as far as I can tell all I really need to know is that the Mean change is +0.834, the Standard Deviation is 0.67 and the Confidence Level is 0.834.
But, I don't really understand what this tells me about the treatment, was it effective? Was its effectiveness only held back by the small sample size?
As an added complication, I also have some data where I would expect the scores to get worse, where the treatment is the same but what I'm looking at is something like tumour size where I expect the treatment to reduce it. When I am looking at the difference between the before and after scores, do I do "after minus before" the same as above? If so then I would be happier with minus differences rather than plus... or is it just a case of how you then describe the data?
I hope that this makes any sense at all, and someone can help me in a jargon free way.
Thanks.
MBCT
I'm really struggling with a statistical problem:
5 people had a medical treatment, they were tested before and after using a questionnaire which recorded some numerical data. I want to analyse the before and after results to see whether there was any change, whether it is statistically significant and then write this up correctly.
I don't have SPSS so I have been using the Data Analysis functions of Microsoft Excel as described here:http://www.stattutorials.com/EXCEL/EXCEL_TTEST2.html
So (before score 1-5, after score 1-5)
Person 1 - 2.75, 3.65
Person 2 - 1.75, 3.25
Person 3 - 3.75, 3.50
Person 4 - 1.75, 2.50
Person 5 - 2.75, 4.00
So first I did a t-Test Paired Two Sample for Means which gave me the following:
Mean 2.55, 3.384
Variance 0.7, 0.31853
Observations 5
Pearson Correlation 0.600384
Hypothesized Mean Difference 0
df 4
t Stat -2.77529
P(T<=t) one-tail 0.02503
t Critical one-tail 2.131847
P(T<=t) two-tail 0.050059
t Critical two-tail 2.776445
So, the important information seems to be that on a one tail P I have got 0.02 which is less than 0.05 and therefore I could reject the null hypothesis and say that the change is not due to chance and the treatment is statistically significant.
But, I only have a sample of 5 and therefore can't really say that can I? What sample size would I need to reject the null hypothesis and how can I calculate this?
Also, as tempting as it is to use one-tail, actually the treatment might be ineffective or make things worse and so maybe I should use two-tail, in which case is 0.050059 enough above 0.05 to make rejecting the null hypothesis impossible even if the sample size was much higher?
I then worked out the difference between the two sets of data by doing "after minus before" treatment giving me:
Person 1 - 0.92
Person 2 - 1.50
Person 3 - -0.25
Person 4 - 0.75
Person 5 - 1.25
Using Excel's Descriptive Statistics gave me this information:
Mean 0.834
Standard Error 0.30051
Median 0.92
Mode #NA
Standard Deviation 0.67196
Sample Variance 0.45153
Kurtosis 1.856022
Skewness -1.24463
Range 1.75
Minimum -0.25
Maximum 1.5
Sum 4.17
Count 5
Confidence Level (95%) 0.834348
Confidence Interval -0.00035, 1.668348
Now, as far as I can tell all I really need to know is that the Mean change is +0.834, the Standard Deviation is 0.67 and the Confidence Level is 0.834.
But, I don't really understand what this tells me about the treatment, was it effective? Was its effectiveness only held back by the small sample size?
As an added complication, I also have some data where I would expect the scores to get worse, where the treatment is the same but what I'm looking at is something like tumour size where I expect the treatment to reduce it. When I am looking at the difference between the before and after scores, do I do "after minus before" the same as above? If so then I would be happier with minus differences rather than plus... or is it just a case of how you then describe the data?
I hope that this makes any sense at all, and someone can help me in a jargon free way.
Thanks.
MBCT