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# Critical value and test statistic

Posted on 2006-11-09
Medium Priority
4,436 Views
The general population has the mean (5/1000); and the test sample has
(15/5000).
However, we don't know the sd.  In this case, how to calculate t value?

If I use the critical-value method with &#945; =0.05, then I come up with
the critical value, -1.645158

My question is how to calculate t value without knowing the sd at all?

The formula is t = (x bar - mu 0) / (sd/sqrt(n)).
so, it could be:

(5/1000 - 15/5000)
--------------------------
sd /sqrt(5000)

thx

ej
0
Question by:ethanjohnsons
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LVL 84

Expert Comment

ID: 17909872
you can't without some way of estimating the sd
is 5000 the number of samples?   how can the test sample have more than the gerneral population?
0

Author Comment

ID: 17911007
The context is:
"Suppose the incident rate of MI (myocardial infraction) per year was 5
per 1000 men in 1990.  To look at the changes in incidence over time,
5000 men were followed for 1 year starting in 2000.  15 new cases of MI
were found.", and we know &#945; =0.05 for the critical-value method.
and I am trying to test if the incident rate of MI has changed from
1990 to 2000.
0

LVL 84

Accepted Solution

ozo earned 2000 total points
ID: 17911052
Oh, ok.
If each person has a 5/1000 chance of having a MI incident in a year,
the expeted number of incidenents in a sample of 5000 men would have a binomial distribution
which might be approximated with a Gaussian distribution with mean 5000*5/1000
and sd sqrt(5000*(5/1000)*(1-5/1000))
0

LVL 84

Expert Comment

ID: 17911121
more precisely the probaility that there will be <= 15 incidents will be
sum(n=0..15:5000!(5/1000)^n(1-5/1000)^(5000-n)/(n!(5000-n)!) = 0.022
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