Hi,(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Proving that treatment is statistically effective - paired t, P value, CI?

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