1. Independent Samples: Comparing Means
Lecture 37
Sections 11.1 – 11.2, 11.4
Fri, Apr 6, 2007

2. Independent Samples
In a paired study, two observations are made on each subject, producing one sample of bivariate data.
Or we could think of it as two samples of paired data.

3. Independent Samples
Paired data are often “before” and “after” observations.
By comparing the mean before treatment to the mean after treatment, we can determine whether the treatment had an effect.

4. Independent Samples
On the other hand, with independent samples, there is no logical way to “pair” the data.
One sample might be from a population of males and the other from a population of females.
Or one might be the treatment group and the other the control group.
Furthermore, the samples could be of different.

5. Independent Samples We start with two populations.
Population 1 has mean 1 and st. dev. 1.
Population 2 has mean 2 and st. dev. 2.
We wish to compare 1 and 2.
We do so by comparing sample meansx1 andx2.

6. Independent Samples
To do this, we will usex1 –x2 as an estimator of 1 – 2.
If we want to know whether 1 = 2, we test to see whether 1 – 2 = 0 by computingx1 –x2 and comparing it to 0.

7. The Distributions ofx1 andx2
Let n1 and n2 be the sample sizes.
If the samples are large, thenx1 andx2 have (approx.) normal distributions.
However, if either sample is small, then we will need an additional assumption:
The populations are normal.
in order to use the normal distribution.

8. Further Assumption
We will also assume that the two populations have the same standard deviation.
Call it .
That is,  = 1 = 2.
If this assumption is not supported by the evidence, then it should not be made.
If this assumption is not made, then the formulas become much more complicated. See p. 658.

9. The Distribution ofx1 –x2
If the sample sizes are large enough (or the populations are normal), then according to the Central Limit Theorem,
x1 has a normal distribution with mean 1 and standard deviation 1/n1.
x2 has a normal distribution with mean 2 and standard deviation 2/n2.

10. The Distribution ofx1 –x2
It follows from theory thatx1 –x2 is
Normal, with
mean 1 – 2 and
Variance

11. The Distribution ofx1 –x2
If we assume that 1 = 2, (call it ), then the standard deviation may be simplified to
That is,

12. The Distribution ofx1 1

13. The Distribution ofx2 2 1

14. The Distribution ofx1 –x22 1
1 – 2

15. The Distribution ofx1 –x2If
then it follows that

16. Example Do Example 11.4, page 699, but
Assume that the same sizes are 100, not 10.
Then work the same example using the TI-83 and the 2-SampZTest function.

17. The t Distribution Let s1 and s2 be the sample standard deviations.
Whenever we use s1 and s2 instead of , then we will have to use the t distribution instead of the standard normal distribution, unless the sample sizes are large.

18. Estimating  Individually, s1 and s2 estimate .
However, we can get a better estimate than either one if we “pool” them together.
The pooled estimate is

19. x1 –x2 and the t Distribution
If we use sp instead of , and the sample sizes are small, then we should use t instead of Z.
The number of degrees of freedom is
df = df1 + df2 = n1 + n2 – 2.
That is

20. Hypothesis Testing
See Example 11.4, p. 699 – Comparing Two Headache Treatments.
State the hypotheses.
H0: 1 = 2
H1: 1 > 2
State the level of significance.
 = 0.05.

21. The t Statistic Compute the value of the test statistic.
The test statistic is
with df = n1 + n2 – 2.

22. Computations

23. Hypothesis Testing Calculate the p-value.
The number of degrees of freedom is
df = df1 + df2 = 18.
p-value = P(t > 1.416)
= tcdf(1.416, E99, 18) =

24. Hypothesis Testing State the decision. State the conclusion.
Accept H0.
State the conclusion.
Treatment 1 is more effective than Treatment 2.

25. The TI-83 and Means of Independent Samples
Enter the data from the first sample into L1.
Enter the data from the second sample into L2.
Press STAT > TESTS.
Choose either 2-SampZTest or 2-SampTTest.
Choose Data or Stats.
Provide the information that is called for.
2-SampTTest will ask whether to use a pooled estimate of . Answer “yes.”

26. The TI-83 and Means of Independent Samples
Select Calculate and press ENTER.
The display shows, among other things, the value of the test statistic and the p-value.

27. Paired vs. Independent Samples
The following data represent students’ calculus test scores before and after taking an algebra refresher course.
Student 1 2 3 4 5 6 7 8
Before 85 63 94 78 75 82 45 58
After 92 68 98 83 80 88 53 62

28. Paired vs. Independent Samples
Perform a test of the hypotheses
H0: 2 – 1 = 0
H1: 2 – 1 > 0
treating the samples as independent.

29. Paired vs. Independent Samples
Had we performed a test of the “same” hypotheses
H0: D = 0
H1: D > 0
treating the samples as paired, then the p-value would have been
Why so small?

30. Paired Samples Why is there a difference? 1 2 3 5 4 6 8 7 40 50 60 8090
100 70
Paired

31. Independent Samples Why is there a difference? 1 2 3 5 4 6 8 7 40 5060 80 90
100 70
Independent

32. Confidence Intervals
Confidence intervals for 1 – 2 use the same theory.
The point estimate isx1 –x2.
The standard deviation ofx1 –x2 is approximately

33. Confidence Intervals The confidence interval is or
( known, large samples)
( unknown, large samples)
( unknown, normal pops.,
small samples)

34. Confidence Intervals The choice depends on Whether  is known.
Whether the populations are normal.
Whether the sample sizes are large.

35. Example
Find a 95% confidence interval for 1 – 2 in Example 11.4, p. 699.
x1 –x2 = 3.2.
sp =
Use t =
The confidence interval is
3.2  (2.101)(2.259) = 3.2  4.75.

36. The TI-83 and Means of Independent Samples
To find a confidence interval for the difference between means on the TI-83,
Press STAT > TESTS.
Choose either 2-SampZInt or 2-SampTInt.
Choose Data or Stats.
Provide the information that is called for.
2-SampTTest will ask whether to use a pooled estimate of . Answer “yes.”