1. Comparing Two Populations or Treatments
Chapter 11 Lesson 11.2a
Comparing Two Populations or Treatments
11.2a: Inferences Concerning the Difference Between 2 Population or Treatment Means Using Paired Samples

2. Paired Data
Data are paired when the observations are collected in pairs or the observations in one group are naturally related to observations in the other group.
Paired data arise in a number of ways:
Before and After
Husbands and wives
Twins

3. Paired Data (cont.)
If you know the data are paired, you can (and must!) take advantage of it.
There is no test to determine whether the data are paired. You just have to think about how the data is collected.
Once we know the data are paired, we can examine the paired differences.
Because it is the differences we care about, we treat them as if they were the data and ignore the original two sets of data.

4. Paired Data (cont.)
Now that we have only one set of data to consider, we can return to the simple one-sample t-test.
Mechanically, a paired t-test is just a one-sample t-test for the mean of the differences.
The sample size is the number of pairs.

5. Where md is the mean of the differences in the paired observations
Summary of the Paired t test for Comparing Two Population or Treatment Means
Null Hypothesis: H0: md = hypothesized value
Test Statistic:
Where n is the number of sample differences and xd and sd are the mean and standard deviation of the sample differences. This test is based on df = n – 1.
Alternative Hypothesis: P-value:
Ha: md > hypothesized value Area to the right of calculated t
Ha: md < hypothesized value Area to the left of calculated t
Ha: md ≠ hypothesized value 2(area to the right of t) if +t
or 2(area to the left of t) if -t
The hypothesized value is usually 0 – meaning that there is no difference.
Where md is the mean of the differences in the paired observations

6. Summary of the Paired t test for Comparing Two Population or Treatment Means Continued . . .
Assumptions:
The n sample differences can be viewed as a random sample.
The number of sample differences is large (n>30) so the CLT applies….or the population differences are approximately normal.
The samples are paired.

7. Is this an example of paired samples?
An engineering association wants to see if there is a difference in the mean annual salary for electrical engineers and chemical engineers. A random sample of electrical engineers is surveyed about their annual income. Another random sample of chemical engineers is surveyed about their annual income.
No, there is no pairing of individuals, you have two independent samples

8. Is this an example of paired samples?
A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to volunteers, company researchers weigh each person. After a month of using the drug, each person’s weight is measured again.
Yes, you have two observations on each individual, resulting in paired data.

9. First, find the differences pre-test minus post-test.
Can playing chess improve your memory? In a study, students who had not previously played chess participated in a program in which they took chess lessons and played chess daily for 9 months. Each student took a memory test before starting the chess program and again at the end of the 9-month period.
If we had subtracted Post-test minus Pre-test, then the alternative hypothesis would be the mean difference is greater than 0.
Student 1 2 3 4 5 6 7 8 9 10 11 12
Pre-test
510
610
640
675
600
550
625
450
720
575
Post-test
850
790
775
700
690
540
680
Difference
-340
-180
-210
-100
-225
-90
-240
-55 35 -5
H0: md = 0
Ha: md < 0
(d = Pretest – Posttest)
First, find the differences
pre-test minus post-test.
State the hypotheses.

10. Playing Chess Continued . . .
Student 1 2 3 4 5 6 7 8 9 10 11 12
Pre-test
510
610
640
675
600
550
625
450
720
575
Post-test
850
790
775
700
690
540
680
Difference
-340
-180
-210
-100
-225
-90
-240
-55 35 -5
H0: md = 0
Ha: md < 0
Assumptions: 1) The samples are paired.
2) Although the sample of students is not a random sample, the
Where md is the mean memory score difference between students with no chess training and students who have completed chess training
Verify assumptions
investigator believed that it was reasonable to view the 12 sample differences as representative of all such differences.
3) A boxplot of the differences is approximately symmetrical with no outliers so the assumption of normality is plausible.

11. Playing Chess Continued . . .
Student 1 2 3 4 5 6 7 8 9 10 11 12
Pre-test
510
610
640
675
600
550
625
450
720
575
Post-test
850
790
775
700
690
540
680
Difference
-340
-180
-210
-100
-225
-90
-240
-55 35 -5
H0: md = 0
Ha: md < 0
Test Statistic:
Where md is the mean memory score difference between students with no chess training and students who have completed chess training
State the conclusion in context.
Compute the test statistic and P-value.
P-value ≈ 0 df = 11 a = .05
Since the P-value < a, we reject H0. There is convincing evidence to suggest that the mean memory score after chess training is higher than the mean memory score before training.

12. Homework
Pg.564: #11.42, 43