Download presentation

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(1P(z < 1.49)) =

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google