 5-3 Inference on the Means of Two Populations, Variances Unknown

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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

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

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

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

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.

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

(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.

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

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.

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

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

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

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

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.

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.

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.

Hw for ,5-21

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

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

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

5-4 The Paired t-Test n=9

5-4 The Paired t-Test p.220

95% confidence interval for example 5-8:

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

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

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

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

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.

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.

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

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

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

Therefore, we have

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

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

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

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.

Confidence Interval

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

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

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

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

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

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

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

5-6 Inference on Two Population
Proportions

5-6 Inference on Two Population Proportions

5-6 Inference on Two Population
Proportions

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

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

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

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

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

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)) =

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