You preform a t test on the sample means.
You need to assume normality of both populations.
BUT we (you and me) need to know if you're assuming equal populatoin variances or not.
If they are equal you pool the two sample variances.
The Tampa Bay Area Chamber of Commerce wanted to know whether the mean weekly salary of nurses was larger than that of school teachers. To investigate, they collected the following information on the amounts earned last week by a sample of school teachers and nurses.
Teachers: 845, 826, 827, 875, 784, 809, 802, 820, 829, 830, 842, 832.
Nurses: 841, 890, 821, 771, 850, 859, 825, 829.
Is it reasonable to conclude that the mean weekly salary of nurses is higher? Use the .01 significance level. What is the p-value.
you can test to see if the variances are equal via an F test.
If the pop st deviations are equal then use ...
Unequal sample sizes, equal variance at http://en.wikipedia.org/wiki/T-test
N: salary of nurses
T: salary of teachers
You are testing
H0 : E(T) = E(N)
H1 : E(T) > E(N), E(T) - E(N) > 0
with 1% significance (or 99% confidence).
Some parts are in portuguese. But i think you cant get it anyway. (GL is Degrees of Freedom).
You assume variances are different, because the populations are of small dimension, and you'd have to estimate the common variance.
There's another way which is to setup an interval of confidence of 99%.
You do it for the difference of the expected values.
GL is degrees of freedom
NA is size of population A
SA is the estimator of the variance of population A (since it is of small dimension we have to use the estimator, which you calculate the same way as the variance)
alpha on this case will be 0.01
If the interval goes from positive values to positives values [+,+] you know the expected value of the 1st is bigger than the second, and vice versa. Otherwise you can't conclude anything.
For the p value, you have to go with the first method, so this second one would be just to answer the first part of the question.