1. Chapter 9 Comparing Means

2. Comparing Two Means
Most of the time we want to compare two different populations
Males vs. Females
Students who use SAT study program vs. students who do not use SAT study programs
What do we do?

3. Comparing Two Means μ1 = mean of first population
μ2 = mean of second population
Interested in the quantity μ1 - μ2
These are unknown parameters
Take samples from each group and estimate the quantity with

4. Testing Difference in Means
σ1 and σ2 are parameters (use when n ≥ 30)
Replace with s1 and s2 (use when n < 30)
Changes distribution (goes from z to t)

5. Testing Difference in Means
The standardized sample difference between the means of two independent groups (aka the test statistic) is

6. Hypothesis Test for μ1 - μ2
Step 1:
Hypotheses
HA: μ1 - μ2 > 0 (m1 > m2)
HA: μ1 - μ2 < 0 (m1 < m2)
HA: μ1 - μ2 ≠ 0 (m1 ≠ m2)
Step 2:
Alpha Level (assume .05)

7. Hypothesis Test for μ1 - μ2
Step 3:
Model is Normal
Step 4
Test statistic:

8. Hypothesis Test for μ1 - μ2
Step 5:
Decision:
Reject HO when p-value is smaller than α
Do not reject HO p-value is larger than α
Step 6:
Conclusion
Always stated in terms of Ha in the problem.

9. HT Example Difference in Means
In June 2002, the Journal of Applied Psychology reported on a study that examined whether the content of TV shows influenced the ability of viewers to recall brand names of items featured in commercials. Researchers randomly assigned 108 volunteers to watch a program with violent content and 108 to watch a program with neutral content. Both programs featured the same 9 commercials. After the shows ended, subjects were asked to recall the brands in the commercials. Is there evidence that viewer memory for ads is less than zero? Use α = 0.05.

10. HT Example Difference in Means
Step 1:
μ1 = mean number of brands recalled for violent program group.
μ2 = mean number of brands recalled for neutral program group.
HO: μ1 - μ2 ≥ 0
HA: μ1 - μ2 < 0

11. HT Example Difference in Means
Step 2:
Alpha Level: α=0.05
Step 3:
Normal

12. HT Example Difference in Means
Step 4:
Test statistic:

13. HT Example Difference in Means
Step 4 (cont.):
P-value
= P(z < )
= way less than 0.001

14. HT Example Difference in Means
Step 5:
Decision: Since p-value < α = 0.05, we will reject the null hypothesis.
Step 6:
Conclusion: The mean number of brands recalled by the neutral program group is different than the mean number of brands recalled by the violent program group. Type of program affects the recall of brands from commercials.

15. Independent vs. Dependent Samples
We would like to determine if students taking an ACT prep course will score better than students not taking the course. A random sample of 25 students was chosen who took the course and a random sample of another 25 students was chosen who did not take the course. At the end of the prep course, both groups were given the ACT.

16. Independent vs. Dependent Samples
Group #1 – score on ACT from students taking prep course
Group #2 – score on ACT from students NOT taking prep course
Observations taken from two independent samples

17. Independent vs. Dependent Samples
We would like to determine if students can improve their ACT score by taking a prep course. A random sample of 25 students was chosen. They first took the ACT test. Then they spent 6 weeks taking the prep course. At the end of the 6 weeks, they took the ACT test again.

18. Independent vs. Dependent Samples
Group #1 – score on ACT before prep course
Group #2 – score on ACT after prep course
Two observations taken for each subject

19. Important Difference If the values come from: two independent samples
use inference for (μ1 – μ2) = difference in means for two groups
dependent samples (e.g. values collected twice from same subject)
use inference for μd = mean difference between two values
The second situation is called Matched Pairs

20. Inference for μd d = pairwise difference Confidence Interval
Hypothesis Test (degrees of freedom = n-1)

21. Pairwise Differences ACT Prep Course: Does the course improve scores?
Before 18 21 25 13
After 23 26 20
Difference 5 -2 7

22. Matched Pairs Hypothesis Test Example
Measure effectiveness of exercise program in lowering blood cholesterol levels
SRS of men from a population
Measure cholesterol before program starts
Go through exercise program for 12 weeks
Measure cholesterol after program ends
Is the exercise program effective?

23. Matched Pairs Hypothesis Test Example - Data
Subject 1 2 3 4 5 6 7 8 9 10
Pre
210
236
220
250
209
240
222
Post
204
218
200
208
234
246

24. Matched Pairs Hypothesis Test Example
Two observations from each subject
Post value depends on pre value
Observations are dependent
Do not have independent samples
Cannot use inference for μ1 - μ2
Violates independent samples assumption

25. Matched Pairs Hypothesis Test Example - Data
Look at pre-program levels minus post-program levels
Subject 1 2 3 4 5 6 7 8 9 10
Pre
210
236
220
250
209
240
222
Post
204
218
200
208
234
246
Diff. 26 50 -6 -4 14
-10

26. Matched Pairs
Because it is the differences we are interested in, we will treat them as the data and ignore the original two groups.
A matched pairs t-test is just a one-sample t-test (from Chapter 23) for the mean of the pairwise differences.

27. Matched Pairs Hypothesis Test Example
Let μd = mean difference in cholesterol levels in population
Sample mean difference is
Sample standard deviation of differences is sd = 17.99
n = number of differences = 10

28. Matched Pairs Hypothesis Test Example
If exercise program has no effect, mean difference will be 0
If exercise program is effective, mean difference will be positive
Step 1:
HO: μd = 0
HA: μd > 0

29. Matched Pairs Hypothesis Test Example
Step 2:
Alpha Level (default=.05)
Step 3:
Model: t9

30. Matched Pairs Hypothesis Test Example
Step 4:
Test statistic:

31. Matched Pairs Hypothesis Test Example
Step 4 (cont.):
P-value
= P(t9 > 1.42)
= between 0.05 and 0.10

32. Matched Pairs Hypothesis Test Example
Step 5:
Decision: Since p-value > 0.05 = α, we will fail to reject the null hypothesis.
Step 6:
Conclusion: There is no evidence that the exercise program reduced the mean cholesterol level of men in this population.

33. Matched Pairs Confidence Interval Example
For a random sample of 12 European cities, the average high temperatures in January and July are given on the following slide.
Find a 90% confidence interval for the mean temperature difference between summer and winter in Europe. Assume the Nearly Normal assumption is satisfied.

34. Matched Pairs Confidence Interval Example
Jan. 34 36 42 35 54 40 47 44 43 21 37
July 75 72 76 74 90 88 69 87 73 65 84
Diff. 41 39 29 22 55

35. Matched Pairs Confidence Interval Example

36. Matched Pairs Confidence Interval Example

37. Matched Pairs Confidence Interval Example
I am 90% sure that the true mean difference is between and

38. Difference in Proportions
Hypotheses
HA: p1 - p2 > 0
HA: p1 - p2 < 0
HA: p1 - p2 ≠ 0
Test Statistic
On the calculator use 2PropZTest

39. HT Difference in Proportions
A sample of 200 teenagers shows that 50 believe that war is inevitable, and a sample of 300 people over age 60 shows that 93 believe war is inevitable. Is the proportion of teenagers who believe war is inevitable different from the proportion of people over 60 who do ? Use α = 0.01.

40. HT - Difference in Proportions
Step 1
HO: p1 - p2 = 0
HA: p1 - p2 ≠ 0
Step 2
α = 0.01
Step 3
Normal

41. HT - Difference in Proportions
Step 4
Step 5
Do Not Reject
Step 6
There is not enough evidence to suggest that
p1 - p2 ≠ 0.

42. Difference in Variance/Std. Dev.
Hypotheses
HA: σ12 – σ22 > 0
HA: σ12 – σ22 < 0
HA: σ12 – σ22 ≠ 0
Use σ if testing standard deviation
Test Statistic
The larger variance is always in the numerator
d.f.N = (n1 – 1) d.f.D = (n2 -1)

43. HT - Difference Between Std. Dev.
A researcher claims that the standard deviation of the ages of cats is smaller than the standard deviation of the ages of dogs who are owned by families in a large city. A randomly selected sample of 29 cats has a standard deviation of 2.7 years, and a random sample of 16 dogs has a standard deviation 3.5 years. Using α = 0.05 test the researchers claim.

44. HT - Difference Between Std. Dev.
Dogs have the higher s.d. so they are s1 and the test is that cats’ s.d. is smaller
Step 1
HO: σ12 – σ22 ≤ 0
HA: σ12 – σ22 > 0
Step 2
α = 0.05

45. HT - Difference Between Std. Dev.
Step 3
F(15, 28)
Step 4 - use 2-sampFtest on calculator

46. HT - Difference Between Std. Dev.
Step 5
0.114 > 0.05
Do Not Reject Ho
Step 6
There is not enough evidence to suggest
σ12 – σ22 > 0.