Posted by kelly on Tuesday, March 12, 2013 at 1:49pm.
This is a one sample test. It is one sample because you aren't comparing groups, etc.
Here's the rest of the data
155 160 150 145 200 155
190 140 170 175 142 168
So I will have to find the mean and standard deviation of the sample.
Then how will I set up the hypothesis to begin running the test
Ho: Mu = 150
Ha: Mu >150
some books set up Ho: as Mu is less than or equal to because the go in the opposite direction of the Ha.
The Ho always includes the =
You are testing to see if your mean is significantly different from the known 150.
When you do the test you will compare your outcome to a given significance level. Usually it is 10% or 5% or 1%
Also written as .10, .05, .01
So I have the mean as 162.5 and standard deviation 18.86
standard error =s/square root n =18.86/12= 5.44
t=m-mean/standard error = 150-162.5/5.44= -2.29
So we fail to reject Ho
You have to compare the -2.29 either with the significance level of say .05
You will go to a t-table look up the -2.29 with the correct degrees of freedom and compare that value with the .05. If it is less then you reject, if it is more you fail to reject.
Another way would be to go to the t-table. use .05 in the body and the correct degrees of freedom. You will find a t-value and you compare it to -2.29
You will get a positive number from the t-table, but you are talking about the lower tail, so you want it to be negative.
If the -2.29 is farther out in the tail than the t-value then you reject HO.
yes I used it against the alpha level .05 of the z score +/- 1.96 so I reject because it is less than stating that those who consumed salty foods didn't have an significant impact on whether the intake of salt affects weight