# Help with CHI-SQUARE test and HYPOTHESIS

• Dec 31st 2012, 03:38 AM
Help with CHI-SQUARE test and HYPOTHESIS
I have a clinical trail where all the patients are suffering from a stroke.
Half of the patients got drug and the other half got placebo.
I need to assess the efficacy of the drug treatment for the entire sample by using the chi-square test in excel.
I have the values if the treatment(drug/placebo) was a succes or a failure for each patient (attached file).
My quesiton is what chi-square do i need to do in order to determine which drug is better? goodness-of-fit or independence?
I think independence but im not sure...
And also if anyone can help me with the null and the alternative hypothesis, cause i have no idea what to write :\
THANKS A LOT!!!!
• Dec 31st 2012, 04:56 AM
ILikeSerena
Re: Help with CHI-SQUARE test and HYPOTHESIS
Quote:

I have a clinical trail where all the patients are suffering from a stroke.
Half of the patients got drug and the other half got placebo.
I need to assess the efficacy of the drug treatment for the entire sample by using the chi-square test in excel.
I have the values if the treatment(drug/placebo) was a succes or a failure for each patient (attached file).
My quesiton is what chi-square do i need to do in order to determine which drug is better? goodness-of-fit or independence?
I think independence but im not sure...
And also if anyone can help me with the null and the alternative hypothesis, cause i have no idea what to write :\
THANKS A LOT!!!!

A goodness-of-fit is when you want to know if a set of frequencies matches a known distribution.
Since you do not have a known distribution, you're not testing for goodness-of-fit.

You do want to know if treatment and result are dependent or not.
So indeed, you have a test for independence.

H0: treatment and result are independent
Ha: treatment and result are dependent

You have a p-value of 0,000000062
Typically you would test against an alpha of 0,05.
The statistical conclusion is that treatment with the drug has a significant effect on the success failure rate ( $\chi^2(1)$, p = 0,000000062).