- 167 Views
- Uploaded on
- Presentation posted in: General

ANOVA Determining Which Means Differ in Single Factor Models

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

ANOVA

Determining Which Means Differ in Single Factor Models

- 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.

- 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’?”

- 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?”

A B C D E

130 90 100140 160

115 80 95150 150

130 95 110150 155

125 98 100125 145

120 92 105145 165

11085 90130125

x =121.67 90 100140 150

Grand Mean

x =120

p-value = .000000000108

Can conclude differences

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

- 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

- 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.

- A test or a confidence interval for the difference in two particular means
- The approach is the same.
- We will illustrate an approach called the Fisher LSD approach.
- Only the value used for α will be different.

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”

- We conclude that two means differ, if their sample means,xi andxj, differ by “a lot”.
- “A lot” is LSD given by:

Confidence Interval for μi – μj

Confidence Interval for μi – μj

- A confidence interval for μi – μj is found by:

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:

- 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

- 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

- 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.

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.

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

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:

- 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 construct a matrix as follows

- 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)

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

- 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.

- 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

- 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

- 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

= .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)

For the battery example:

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.

- The Fisher LSD Test
- What to use for:
- Best Estimate of σ2 = MSE
- Degrees of Freedom = DFE

- Calculation of LSD

- What to use for:
- 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