So that is the entire day wasted.
I have no idea what I am supposed to be doing here, an assignment due in next week and only in the second question (which is the above). I distance learn so can't just pop and see a teacher.
I am just reading this and don't even understand half of the words and can't find anything similar anywhere.
Basically, I have two sets of information - comparing the effects of alcohol on two similar illnesses - I am given the sample number for the first, and the number of drinkers- and the same for the second illness.
I have to create distribution models for both - I created Binomial distribution for both of them.
Then I need to write the null and alternative hypotheses used to test the proportion differences/similarities.
I am sorry to be so emotional haha - but if I can't do this, I have no hope with the rest of the assignment and I don't know what to do :-(
Thanks for replying IlikeSerena,
I have managed to calm myself down a bit - temporarily!
I can't really say anymore because it is a marked assignment. But I look into ANOVA testing - it is only 2 points, but the rest is on minitab so will have to try figure that out!
Actually, I do have one :-)
I thought I was doing really well with this - took a few days - but seem to have got there.
What I missed in the question was, although n=10 they told us to 'assume normal distribution'
So back to square one!
I have used xbar-mu / s/sqrtn to get the test stat - I also need the null distribution which I think is;N(mu,sigma^2) - So I have used mu from my null hypothesis and s^2 for the pop variation - is that correct?
But now I am trying to get the rejection zone - and I keep getting +/-1.96 as it is a 5% confidence interval, but when I use minitab I get totally different answers, and have no idea where they came from.
Any clues? I am fairly sure I am making errors all the way through!
When you have n=10 and no information on sigma, you are supposed to do a t-test instead of a z-test.
Your formula is correct, but you need to look it up in a t-table with n-1=9 degrees of freedom.
This is what minitab will have done.
Oh - I thought it was two totally seperate tests - one for below 30 samples and one for above 30 samples.
So actually, my original answer should be correct - although in that I showed a T distribution - T~T(n-1) rather than an N one - N(mu, sigma^2)- which one is correct?
When n is large enough the corresponding t-distribution is equal to the normal distribution.
You can see this if you check the t-table.
Btw, n=30 is still a bit low to switch to a z-test.
In your case, the assumed null distribution would be the t(n-1) distribution.
Btw, if n > 30, you can assume that sample mean of any distribution behaves like a normal distribution (central limit theorem).
To do a t-test, the assumption is that the sample mean is normally distributed.