If you are asked to do a t-test on this data, see the following:
t = 5.15/(11.45/√322)
You can probably get away with doing a z-test for this problem because the sample size is so large. However, the formulas are the same for a t or z test.
If you use a z-table for a one-tailed z-test, check the table at .05 level of significance. If the test statistic exceeds the critical or cutoff value from the table, then you conclude there is a difference. If the test statistic does not exceed the critical or cutoff value from the table, then you cannot conclude a difference.
If you use a t-test, determine the degrees of freedom, then check the t-table for a one-tailed test at the .05 level.
I'll let you take it from there.
T-Test
The article “Does Smoking Cessation Lead to Weight Gain?” described an experiment in which 322
subjects, selected at random from those show successfully participated in a program to quit smoking
were weighed at the beginning of the program and again one year later . The mean change in weight
was 5.15 lb and the standard deviation was 11.45 lb. Is there sufficient evidence at =.05 to conclude
that the true mean change is positive?
1 answer