2. 5-3 Inference on the Means of Two Populations, Variances Unknown
5-3.1 Hypothesis Testing on the Difference in
Means
Use the data in both samples to estimate
Each is an estimator of
The pooled sample variance is the sum of all those terms divided by the total degrees of freedom

3. 5-3 Inference on the Means of Two Populations, Variances Unknown
5-3.1 Hypothesis Testing on the Difference in
Means
Replace in the test statistics by
The degrees of freedom of T is the degrees of freedom of

4. 5-3 Inference on the Means of Two Populations, Variances Unknown
5-3.1 Hypothesis Testing on the Difference in
Means
p.226

5. 5-3 Inference on the Means of Two Populations, Variances Unknown
5-3.1 Hypothesis Testing on the Difference in
Means
p.227

6. Do not reject
Since both samples are from normal distributions with equal variance. Now data shows that the means are equal, so two samples are from the same distribution. See p.210 for plots.

7. 5-3 Inference on the Means of Two Populations, Variances Unknown
Estimate them separately
(1) Equal sample sizes ( )

8. (2) Unequal sample sizes This is the Behrens-Fisher problem, the test (T) is not exactly t-distributed
p.229
If the sample sizes are moderate, can be used.
If the sample sizes are large, use z-value.

9. 5-3 Inference on the Means of Two Populations, Variances Unknown
Compare to the plots on Fig.5-2

10. 5-3 Inference on the Means of Two Populations, Variances Unknown
5-3.1 Hypothesis Testing on the Difference in
Means
Round down to 13.
p.231
Since the CI does not contain 0, the difference in mean is significantly different from0.

11. Type II Error Probability and sample size can be
obtained from Table V for the case of
And
with

12. 5-3 Inference on the Means of Two Populations, Variances Unknown
5-3.3 Confidence Interval on the Difference
in Means
p.232

13. 5-3 Inference on the Means of Two Populations, Variances Unknown
5-3.3 Confidence Interval on the Difference
in Means

14. 5-3 Inference on the Means of Two Populations, Variances Unknown
5-3.3 Confidence Interval on the Difference
in Means
p.234

15. Example #5-19 (p235)
1) The parameter of interest is the difference in mean battery life, Assuming 2)
3)  = 0.01
4) The test statistic is 5)
6) Since 4.67 > 2.508, reject the null hypothesis and conclude that the mean battery life of Type A significantly exceeds that of Type B at  = 0.01.

16. b) Calculate the P-value for this test.
P-value = P(t > 4.67) <
c) Construct a 99% CI.
99% lower-side confidence bound:
Since zero is not contained in this interval, we conclude that the
null hypothesis can be rejected and the alternative accepted.
The life of Type A is significantly longer than that of Type B.

17. d) Suppose that if the mean of A batteries exceeds that of B by
2 months, it is important to detect this difference with
probability at least Is the choice of
of this problem adequate?
Form Table V(d), we obtain that
Should be adequate.

18. Hw for ,5-21

19. 5-4 The Paired t-Test
A special case of the two-sample t-tests of Section 5-3 occurs when the observations on the two populations of interest are collected in pairs.
Each pair of observations, say (X1j , X2j ), is taken under homogeneous conditions, but these conditions may change from one pair to another.
The test procedure consists of analyzing the differences between two means.
We will transform the two-sample problem to a one-sample problem by defining

20. Let be n pairs of independent observations from a bivariate normal distribution.
where is the correlation coefficient between and
Define Then the distribution of is normal with
mean and
variance
i.e.,
is reduced if

21. 5-4 The Paired t-Test
Original null hypothesis:

22. 5-4 The Paired t-Test
n=9

23. 5-4 The Paired t-Test
p.220

24. 95% confidence interval for example 5-8:

25. 5-4 The Paired t-Test
Paired Versus Unpaired Comparisons

26. 5-4 The Paired t-Test
Confidence Interval for D

27. 5-4 The Paired t-Test
p.241
Example 5-9
The 90% CI

28. Hw for 5-4
5-39, 5-40

29. 5-5 Inference on the Ratio of Variances of Two Normal Populations
5-5.1 The F Distribution
Let and
be two independent samples with both and unknown.

30. 5-5 Inference on the Ratio of Variances of Two Normal Populations
5-5.1 The F Distribution
We wish to test the hypotheses:
The development of a test procedure for these hypotheses
requires a new probability distribution, the F distribution.

31. 5-5 Inference on the Ratio of Variances of Two Normal Populations
5-5.1 The F Distribution

32. Example: 5-44.p.249
p.464
p.465
p.466

33. 5-5 Inference on the Ratio of Variances of Two Normal Populations
5-5.1 The F Distribution
Why?

34. Therefore, we have

35. 5-5 Inference on the Ratio of Variances of Two Normal Populations
The Test Procedure
Recall: If

36. 5-5 Inference on the Ratio of Variances of Two Normal Populations
The Test Procedure

37. 5-5 Inference on the Ratio of Variances of Two Normal Populations
The Test Procedure

38. 5-21
Example #5-49(p.249)
vs.
Reject if
Since < 3.34 < 4.03 do not reject the null hypothesis and conclude the etch rate variances do not differ at the 0.05 level of significance.

39. Confidence Interval

40. 5-5 Inference on the Ratio of Variances
of Two Normal Populations

41. 5-5 Inference on the Ratio of Variances
of Two Normal Populations

42. 5-5 Inference on the Ratio of Variances
of Two Normal Populations

43. 5-6 Inference on Two Population Proportions
5-6.1 Hypothesis Testing on the Equality of
Two Binomial Proportions
Let and be two independent Binomial random variables.
The sampling distribution of is approximately

44. 5-6 Inference on Two Population Proportions
5-6.1 Hypothesis Testing on the Equality of
Two Binomial Proportions

45. 5-6 Inference on Two Population Proportions
5-6.1 Hypothesis Testing on the Equality of
Two Binomial Proportions
Under the null hypothesis,
we pull the information in both samples together to estimate the
variance of , ,by first estimating by
The sampling distribution becomes approximately

46. 5-6 Inference on Two Population Proportions
5-6.1 Hypothesis Testing on the Equality of Two
Binomial Proportions

47. 5-6 Inference on Two Population
Proportions

48. 5-6 Inference on Two Population Proportions

49. 5-6 Inference on Two Population
Proportions

50. 5-6 Inference on Two Population Proportions
5-6.2 Type II Error and Choice of Sample Size
Where

51. 5-6 Inference on Two Population Proportions
5-6.2 Type II Error and Choice of Sample Size

52. 5-6 Inference on Two Population Proportions
5-6.2 Type II Error and Choice of Sample Size

53. 5-6 Inference on Two Population Proportions
5-6.3 Confidence Interval on the Difference in
Binomial Proportions

54. 5-6 Inference on Two Population Proportions
5-6.3 Confidence Interval on the Difference in
Binomial Proportions
Discuss Example 5-13

55. Example #5-60(p.255) Two types of injection-molding machine.
Is it reasonable to conclude that the two machines produces the same fraction of defective parts? Using
Since 1.96 < 1.49 < 1.96 do not reject the null hypothesis and conclude that the evidence indicates that there is not a significant difference in the fraction of defective parts produced by the two machines at the 0.05 level of significance.
P-value = 2(1P(z < 1.49)) =