In: Statistics and Probability
Suppose the High-density lipoprotein (HDL) cholesterol levels of 20 subjects were measured before and after the treatment. We want to find out if, in general, the treatment will lead to improvements of health, i.e., increasing the HDL cholesterol level. The mean and standard deviation of the difference are 2.05, 2.837, respectively. We assume the differences follow normal. Use ? =0.05.
Null hypothesis
Alternative hypothesis
Sample mean for difference=2.05
Sample standard deviation for difference=2.837
Sample size=20
Degree of freedom =n-1= 19
Test statistic for paired t test is
=3.232
Test statistic value is =3.232
P value is 0.0022......................by using Excel command TDIST(3.232 , 19 , 1)
P value is 0.0022< 0.05
Therefore, we reject H0.
We have sufficient evidence at to say that, the treatment will lead to improvements of health