# Thread:

1. ## Help with null/alternative hypothesis in Z-test of proportions

I have a clinical trail where all the patients sufferd from a stroke.

I need to assess the efficacy of the drug treatment (drug/placebo) for the entire sample by using the z-test in excel.

I have the values if the drug treatment was a succes or a failure for each patient.

My quesiton is what is the null hypothesis and the alternative, cause i have no idea what to write for z-test of proportions.

I did also a chi-square test and the null hypothesis was: the drug treatment and its result are independent / and the alternative hypothesis was : the drug treatment and its result are dependent,
maybe it can helps!

THANKS A LOT!!!!

2. ## Re: Help with null/alternative hypothesis in Z-test of proportions

Hey eladgro.

If you have two sets of data (placebo and drug) then you may want to use a t-test or an equivalent t-test in non-parametric form.

You can then test if there is a statistically significant difference between the means of the distributions and use that to help you come to a conclusion based on evidence.

If you need to look at individual cases then another test would be worth considering.

What exactly are you trying to figure out specifically? (In other words, what is the specific nature of your question and inquiry? What do you hope to answer in full?)

3. ## Re: Help with null/alternative hypothesis in Z-test of proportions

Typically, in a test like that the "null hypothesis" would be "there is no advantage to the placebo" and the alternative would be "there is an advantage".

4. ## Re: Help with null/alternative hypothesis in Z-test of proportions

Hi eladgro!

Suppose $\pi_1$ is the proportion of the success rate for the drug.
And suppose $\pi_2$ is the proportion of the success rate for the placebo.

Then your null hypothesis is: $\pi_1 = \pi_2$.
And your alternative hypothesis is: $\pi_1 \ne \pi_2$.
Unless you're only interested in whether the drug helps. In that case your alternative hypothesis is $\pi_1 > \pi_2$.

See here for the formulas that you are supposed to use: Statistics: Test for Comparing Two Proportions
Or here for a more in depth description: Statistical hypothesis testing - Wikipedia, the free encyclopedia
You probably have them in your notes as well.
This is indeed a z-test and not a t-test.

Basically you should find the same result as when you did the $\chi^2$ test.

5. ## Re: Help with null/alternative hypothesis in Z-test of proportions

Thank you !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!
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