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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:

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Anova determining which means differ in single factor models

ANOVA

Determining Which Means Differ in Single Factor Models


Single factor models review of assumptions
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.


Determining which means differ basic concept
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’?”


Example
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?”


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


Output
OUTPUT

p-value = .000000000108

Can conclude differences


Motivation for the fisher procedure
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


Best estimate for 2 and the appropriate degrees of fredom
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


Two types of tests
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 experiment wise (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.


Determining if i differs from j fisher s lsd approach
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”


When do we conclude two treatment means i and j differ
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:


Confidence intervals for the difference in two population means
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:


Equal vs unequal sample sizes
Equal vs. Unequal Sample Sizes Means

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:



What do we use for
What Do We Use For Meansα?

  • 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


    Ew the probability of making at least one type i error
    α MeansEW = 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.


    The bonferroni adjustment for
    The Bonferroni Adjustment for Meansα

    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 two means when there really is. Thus, some researchers are reluctant to make α too small because this can result in very high β values.


    What should for each test be
    What Should Meansα for Each Test Be?

    The required α values for theindividual t-testsfor αEW = .05 andαEW = .10 are:


    Lsd ew for multiple comparison tests
    LSD MeansEWFor 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:


    Procedure for testing differences among all means
    Procedure for Testing Differences Among All Means 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.


    Procedure continued
    Procedure (continued) Means

    • Then construct a matrix as follows


    Procedure continued1
    Procedure (continued) Means

    • Fill in mean of each treatment across the top row and down the left-most column; (in our example, XA = 121.67, XB = 90, XC= 100, XD = 140, XE = 150)


    Procedure continued2
    Procedure (continued) Means

    • For each cell below the main diagonal, compute the absolute value of difference of the means in the corresponding column and row


    Procedure continued3
    Procedure (continued) Means

    • Compare each difference with LSDEw(17.235 in our case). If the

    • difference between and > LSDEw. we can conclude that there is difference in µi and µj.


    Tests for the battery example
    Tests For the Battery Example Means

    • 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


    Battery example calculations
    Battery Example Calculations Means

    • 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


    Analysis of which means differ
    Analysis of Which Means Differ Means

    • We conclude that two population means differ if their sample means differ by more than LSDEW = 17.2355.

    • Construct a matrix of differences,

    • Compare with LSDEW



    Lsd for confidence intervals
    LSD For Confidence Intervals Means

     = .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)


    Lsd for battery example
    LSD for Battery Example Means

    For the battery example:


    The confidence intervals
    The Confidence Intervals Means

    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.


    Review
    REVIEW Means

    • 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


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