Asked by Peter
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?
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?
Answers
Answered by
MathGuru
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 = 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.
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