ethanjohnsons
asked on
Critical value and test statistic
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 α =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
(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 α =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
ASKER
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 α =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.
"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 α =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.
ASKER CERTIFIED SOLUTION
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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
sum(n=0..15:5000!(5/1000)^
is 5000 the number of samples? how can the test sample have more than the gerneral population?