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

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

Loading...

Similar Threads for Proving treatment statistically | Date |
---|---|

I Standard deviation of data after data treatment | Oct 15, 2017 |

I Prove if x + (1/x) = 1 then x^7 + (1/x^7) = 1. | Mar 9, 2016 |

I Proving Odd and Even | Feb 13, 2016 |

Prove A.(B+C) = (A.B)+(A.C) <Boolean Algebra> | Oct 24, 2015 |

**Physics Forums - The Fusion of Science and Community**