1 / 30

ANOVA Determining Which Means Differ in Single Factor Models

ANOVA Determining Which Means Differ in Single Factor Models. Single Factor Models Review of Assumptions. Recall that the problem solved by ANOVA is to determine if at least one of the true mean values of several different treatments differs from the others. For ANOVA we assumed:

harlan-duncan
Download Presentation

ANOVA Determining Which Means Differ in Single Factor Models

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ANOVA Determining Which Means Differ in Single Factor Models

  2. Single Factor ModelsReview of Assumptions • Recall that the problem solved by ANOVA is to determine if at least one of the true mean values of several different treatments differs from the others. • For ANOVA we assumed: • The distribution of the population for each treatment is normal. • The standard deviations of each population, although unknown, are equal. • Sampling is random and independent.

  3. Determining Which Means DifferBasic Concept • Suppose the result of performing a single factor ANOVA test is a low p-value, which indicates that at least one population mean does, in fact, differ from the others. • The natural question is, “Which differ?” • The answer is that we conclude that two population means differ if their two sample means differ by “a lot”. • The statistical question is, “What is ‘a lot’?”

  4. Example • The length of battery life for notebook computers is of concern to computer manufacturers. • Toshiba is considering 5 different battery models (A, B, C, D, E) that have different costs. • The question is, “Is there enough evidence to show that average battery life differs among battery types?”

  5. Data A B C D E 130 90 100 140 160 115 80 95 150 150 130 95 110 150 155 125 98 100 125 145 120 92 105 145 165 11085 90130125 x =121.67 90 100 140 150 Grand Mean x =120

  6. OUTPUT p-value = .000000000108 Can conclude differences

  7. Motivation for The Fisher Procedure What do we use for these two values? Reject H0 (Accept HA) if: • Fisher’s Procedure is a natural extension of the comparison of two population means when the unknown variances are assumed to be equal • Recall this is an assumption in single factor ANOVA • Testing for the difference of two population means (with equal but unknown σ’s) has the form: H0: μ1 – μ2 = 0 HA: μ1 – μ2 ≠ 0

  8. Best Estimate for σ2 and the Appropriate Degrees of Fredom • Recall that when there were only 2 populations, the best estimate for σ2 is sp2 and the degrees of freedom is (n1-1) + (n2-1) or n1 + n2 - 2. • For ANOVA, using all the information from the k populations the best estimate for σ2 is MSE and the degrees of freedom is DFE. Two populations With Equal VariancesANOVA Best estimate for σ2sp2MSE Degrees of Freedom n1 + n2 – 2DFE

  9. Two Types of Tests • There are two types of tests that can be applied: • A test or a confidence interval for the difference in two particular means • e.g. µE and µB • A set of tests which determine differences among all means. • This is called a set of experimentwise (EW) tests. • The approach is the same. • We will illustrate an approach called the Fisher LSD approach. • Only the value used for α will be different.

  10. Determining if μi Differs From μjFisher’s LSD Approach Reject H0 (Accept HA) if: That is, we conclude there is a differences between μi and μj if LSD H0: μi – μj = 0 HA: μi – μj ≠ 0 LSD stands for “Least Significant Difference”

  11. When Do We Conclude Two Treatment Means (µi and µj) Differ? • We conclude that two means differ, if their sample means,xi andxj, differ by “a lot”. • “A lot” is LSD given by:

  12. Confidence Intervals for the Difference in Two Population Means Confidence Interval for μi – μj Confidence Interval for μi – μj • A confidence interval for μi – μj is found by:

  13. Equal vs. Unequal Sample Sizes Reject H0 (Accept HA) if: LSD • If the sample sizes drawn from the various populations differ, then the denominator of the t-statistic will be different for each pairwise comparison. • But if the sample sizes are equal (n1 = n2 = n3 = ….) , we can designate the equal sample size by N • Then the t-test becomes:

  14. LSD For Equal Sample Sizes

  15. What Do We Use For α? • Recall that α is • In Hypothesis Tests: the probability of concluding that there is a difference when there is not. • In Confidence Intervals: the probability the interval will not contain the true difference in mean values • If doing a single comparison test or constructing a confidence interval, • For an experimentwise comparison of all means, • We will actually be conducting 10 t-tests: • μE - μD, (2) μE - μC, (3) μE - μB, (4) μE - μA, (5) μD - μC, (6) μD - μB, (7) μD - μA, (8) μC – μB, (9) μC - μA, (10) μB - μA select α as usual Use αEW

  16. αEW = The probability ofMaking at least one Type I Error • Suppose each test has a probability ofconcluding that there is a difference when there is not (making a Type I error) =α. • Thus for each test, the probability of not making a Type I error is 1-α. • So the probability of not making any Type I errors on any of the 10 tests is: (1- α)10 • For α = .05, this is (.95)10 = .5987 • The probability of making at least one Type I error in this experiment, is denoted by αEW. • Here,αEW = 1 - .5987 = .4013 -- That is, the probability we make at least one mistake is now over 40%! • To have a lower αEW, α for each test must be significantly reduced.

  17. The Bonferroni Adjustment for α α for each Test For an experimentwise value, αEW, for each test use α = αEW/c c = number of tests For k treatments, c = k(k-1)/2 • To make αEW reasonable, say .05, α for each test must be reduced. • The Bonferroni Adjustment is as follows: NOTE:decreasingα, increasesβ, the probability of not concluding that there is a difference between to means when there really is. Thus, some researchers are reluctant to make α too small because this can result in very high β values.

  18. What Should α for Each Test Be? The required α values for theindividual t-testsfor αEW = .05 andαEW = .10 are:

  19. LSDEWFor Multiple Comparison Tests α for each test When ni≠ nj When ni = nj = N When doing the series of multiple comparison tests to determine which means differ, the test would be to conclude that µi differs from µj if : Where LSDEW is given by:

  20. Procedure for Testing Differences Among All Means • We begin by calculating LSDEW which we have shown will not change from test to test if the sample sizes are the same from each sample. That is the situation in the battery example that we illustrate here. • A different LSD would have to be calculated for each comparison if the sample sizes are different. • Then we order the x’s and begin doing the tests, comparing the x’s in descending order. (In our example, xE = 150,  xD = 140,  xA = 121.67,  xc = 100,  xB = 90.)

  21. Procedure (continued) • Subtract the second largest from the first largest sample mean. If the difference is not less LSDEW, then subtract the third largest from the largest sample mean and so forth until we find a difference larger than LSDEw. We just determined that the μi associated with the largest sample mean differs from another μj. • Once we conclude that some μi differs from another μj, we can conclude that μi differs from all other μ’s whose corresponding x’s < xj. • Repeat this process subtracting from the second largest sample mean, then the third, etc.

  22. Tests For the Battery Example • For the battery example, • Which average battery lives can we conclude differ? • Give a 95% confidence interval for the difference in average battery lives between: • C batteries and B batteries • E batteries and B batteries Use LSDEW Multiple Comparisons Use LSD Individual Comparisons

  23. Battery Example Calculations • Experimental error of EW = .05 • For k = 5 populations, α = αEW /10 = .05/10 = .005 • From the Excel output:  xE = 150,  xD = 140,  xA = 121.67, xc = 100,  xB = 90 MSE = 94.05333, DFE = 25,N = 6 from each population • Use TINV(.005,25) to generate t.0025,25 = 3.078203

  24. Analysis of Which Means Differ • We conclude that two population means differ if their sample means differ by more than LSDEW = 17.2355. • Order the sample means and start with the largest: • E and D • E and A • D and A • A and C • C and B

  25. LSD For Confidence Intervals  = .05 LSD(not LSDEW) Confidence Interval for Confidence intervals for the difference between two mean values, i and j, are of the form: (Point Estimate) ± t/2,DFE(Standard Error)

  26. LSD for Battery Example For the battery example:

  27. The Confidence Intervals Confidence Interval for μC – μB Confidence Interval for μE – μB • 95% confidence interval for the difference in mean battery lives between batteries of type C and batteries of type B. • 95% confidence interval for the difference in mean battery lives between batteries of type E and batteries of type B.

  28. =TINV(.005,C23)*SQRT(D23*(2/6)) =TINV(.05,C23)*SQRT(D23*(2/6)) Copy and pasteAverage and Groups Then do a Z to A ordering Compare differences To cell H3 D15-D14-K3 D15-D14+K3 D17-D14-K3 D17-D14+K3

  29. REVIEW • The Fisher LSD Test • What to use for: • Best Estimate of σ2 = MSE • Degrees of Freedom = DFE • Calculation of LSD • Bonferroni Modification • Modify α so that αEW is reasonable • α = αEW/c, where the # of tests, c = k(k-1)/2 • Calculation of LSDEW • Excel Calculations

More Related