Lecture 9 Inference about the ratio of two variances (Chapter 13.5)
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1 Lecture 9 Inference about the ratio of two variances (Chapter 13.5) Inference about the difference between population proportions (Chapter 13.6)
2 13.5 Inference about the ratio of two variances In this section we draw inference about the ratio of two population variances.This question is interesting because:Variances can be used to evaluate the consistency of production processes.It may help us decide which of the equal- or unequal-variances t-test of the difference between means to use.Because variance can measure risk, it allows us to compare risk, for example, between two portfolios.
3 Parameter and Statistic Parameter to be tested is s12/s22Statistic used isSampling distribution of s12/s22The statistic [s12/s12] / [s22/s22] follows the F distribution with n1 = n1 – 1, and n2 = n2 – 1.
4 F DistributionTables 6(a)-(c) only give the right tail quantiles (critical values) of the F distributionTo find quantiles for the left tail of the F distribution, use the fact
5 Parameter and Statistic Our null hypothesis is alwaysH0: s12 / s22 = 1Under this null hypothesis the F statistic becomesF =S12/s12S22/s22
6 Testing the ratio of two population variances Example 13.6 (revisiting Example 13.1)(see Xm13-01)In order to perform a test regarding average consumption of calories at people’s lunch in relation to the inclusion of high-fiber cereal in their breakfast, the variance ratio of two samples has to be tested first.The hypotheses are:H0:H1:
11 Inference about the difference between two population proportions In this section we deal with two populations whose data are nominal.For nominal data we compare the population proportions of the occurrence of a certain event.ExamplesComparing the effectiveness of new drug versus older oneComparing market share before and after advertising campaignComparing defective rates between two machines
12 Parameter and Statistic When the data are nominal, we can only count the occurrences of a certain event in the two populations, and calculate proportions.The parameter is therefore p1 – p2.StatisticAn unbiased estimator of p1 – p2 is (the difference between the sample proportions).
13 Sampling Distribution of Two random samples are drawn from two populations.The number of successes in each sample is recorded.The sample proportions are computed.Sample 1Sample size n1Number of successes x1Sample proportionSample 2Sample size n2Number of successes x2Sample proportionxn1ˆ=p
14 Sampling distribution of The statistic is approximately normally distributed if n1p1, n1(1 - p1), n2p2, n2(1 - p2) are all greater than or equal to 5.The mean of is p1 - p2.The variance of is (p1(1-p1) /n1)+ (p2(1-p2)/n2)
15 The z-statistic Because and are unknown the standard error must be estimated using the sample proportions.The method depends on the null hypothesis
16 Testing the p1 – p2 There are two cases to consider: Then Then Case 1: H0: p1-p2 =0Calculate the pooled proportionCase 2:H0: p1-p2 =D (D is not equal to 0)Do not pool the dataThenThen
17 Testing p1 – p2Example 13.8The marketing manager needs to decide which of two new packaging designs to adopt, to help improve sales of his company’s soap.A study is performed in two supermarkets:Brightly-colored packaging is distributed in supermarket 1.Simple packaging is distributed in supermarket 2.First design is more expensive, therefore,to be financially viable it has to outsell the second design.
18 Testing p1 – p2 Summary of the experiment results Supermarket purchasers of Johnson Brothers soap out of a total of 904Supermarket purchasers of Johnson Brothers soap out of a total of 1,038Use 5% significance level and perform a test to find which type of packaging to use.
20 Testing p1 – p2 Example 13.9 (Revisit Example 13.8) Management needs to decide which of two new packaging designs to adopt, to help improve sales of a certain soap.A study is performed in two supermarkets:For the brightly-colored design to be financially viable it has to outsell the simple design by at least 3%.
22 Estimating p1 – p2 Estimating the cost of life saved Two drugs are used to treat heart attack victims:Streptokinase (available since 1959, costs $460)t-PA (genetically engineered, costs $2900).The maker of t-PA claims that its drug outperforms Streptokinase.An experiment was conducted in 15 countries.20,500 patients were given t-PA20,500 patients were given StreptokinaseThe number of deaths by heart attacks was recorded.
23 Estimating p1 – p2 Experiment results A total of 1497 patients treated with Streptokinase died.A total of 1292 patients treated with t-PA died.Estimate the cost per life saved by using t-PA instead of Streptokinase.
25 Practice Problems13.88, 13.92, ,13.104,Next Class: Chapters
26 15.1 Introduction to ANOVAAnalysis of variance compares two or more populations of interval data.Specifically, we are interested in determining whether differences exist between the population means.The procedure works by analyzing the sample variance.
27 One Way Analysis of Variance The analysis of variance is a procedure that tests to determine whether differences exits between two or more population means.To do this, the technique analyzes the sample variances
28 One Way Analysis of Variance Example continuedAn experiment was conducted as follows:In three cities an advertisement campaign was launched .In each city only one of the three characteristics (convenience, quality, and price) was emphasized.The weekly sales were recorded for twenty weeks following the beginning of the campaigns.
29 One Way Analysis of Variance Weekly salesSee fileXm15 -01Weekly salesWeekly sales
30 One Way Analysis of Variance SolutionThe data are intervalThe problem objective is to compare sales in three cities.We hypothesize that the three population means are equal