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Lesson 13 - 1 Comparing Three or More Means ANOVA (One-Way Analysis of Variance)

Lesson 13 - 1

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Lesson 13 - 1. Comparing Three or More Means ANOVA (One-Way Analysis of Variance). Objectives. Verify the requirements to perform a one-way ANOVA Test a claim regarding three or more means using one way ANOVA. Vocabulary. - PowerPoint PPT Presentation

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Page 1: Lesson 13 - 1

Lesson 13 - 1

Comparing Three or More Means ANOVA

(One-Way Analysis of Variance)

Page 2: Lesson 13 - 1

Objectives

• Verify the requirements to perform a one-way ANOVA

• Test a claim regarding three or more means using one way ANOVA

Page 3: Lesson 13 - 1

Vocabulary• ANOVA – Analysis of Variance: inferential method that is used

to test the equality of three or more population means

• Robust – small departures from the requirement of normality will not significantly affect the results

• Mean squares – is an average of the squared values (for example variance is a mean square)

• MST – mean square due to the treatment

• MSE – mean square due to error

• F-statistic – ration of two mean squares

Page 4: Lesson 13 - 1

One-way ANOVA Test Requirements

• There are k simple random samples from k populations

• The k samples are independent of each other; that is, the subjects in one group cannot be related in any way to subjects in a second group

• The populations are normally distributed

• The populations have the same variance; that is, each treatment group has a population variance σ2

Page 5: Lesson 13 - 1

ANOVA Requirements Verification

● ANOVA is robust, the accuracy of ANOVA is not affected if the populations are somewhat non- normal or do not quite have the same variances● Particularly if the sample sizes are roughly equal● Use normality plots

● Verifying equal population variances requirement:● Largest sample standard deviation is no more than two

times larger than the smallest

Page 6: Lesson 13 - 1

mean square due to treatment MST F = ------------------------------------- = ---------- mean square due to error MSE

ni(xi – x)2 (ni – 1)si2

MST = -------------- MSE = ------------- k – l n – k

Σ Σn = 1 n = 1

k k

ANOVA – Analysis of Variance

Computing the F-test Statistic

1. Compute the sample mean of the combined data set, x

2. Find the sample mean of each treatment (sample), xi

3. Find the sample variance of each treatment (sample), si2

4. Compute the mean square due to treatment, MST

5. Compute the mean square due to error, MSE

6. Compute the F-test statistic:

Page 7: Lesson 13 - 1

MSE and MST

● MSE - mean square due to error, measures how different the observations, within each sample, are from each other It compares only observations within the same sample Larger values correspond to more spread sample means This mean square is approximately the same as the

population variance

● MST - mean square due to treatment, measures how different the samples are from each other It compares the different sample means Larger values correspond to more spread sample means Under the null hypothesis, this mean square is

approximately the same as the population variance

Page 8: Lesson 13 - 1

Source of Variation

Sum of Squares

Degrees of

FreedomMean

SquaresF-test

StatisticF Critical

Value

Treatment Σ ni(xi – x)2 k - 1 MST MST/MSE F α, k-1, n-k

Error Σ (ni – 1)si2 n - k MSE

Total SST + SSE n - 1

ANOVA – Analysis of Variance Table

Page 9: Lesson 13 - 1

Excel ANOVA Output

• Classical Approach:– Test statistic > Critical value … reject the null hypothesis

• P-value Approach:– P-value < α (0.05) … reject the null hypothesis

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TI Instructions

• Enter each population’s or treatments raw data into a list

• Press STAT, highlight TESTS and select F: ANOVA(

• Enter list names for each sample or treatment after “ANOVA(“ separate by commas

• Close parenthesis and hit ENTER• Example: ANOVA(L1,L2,L3)

Page 11: Lesson 13 - 1

Summary and Homework

• Summary– ANOVA is a method that tests whether three, or more,

means are equal– One-Way ANOVA is applicable when there is only one

factor that differentiates the groups

– Not rejecting H0 means that there is not sufficient evidence to say that the group means are unequal

– Rejecting H0 means that there is sufficient evidence to say that group means are unequal

• Homework– pg 685-691; 1-4, 6, 7, 11, 13, 14, 19

Page 12: Lesson 13 - 1

Problem 19 TI-83 Calculator Output

• One-way ANOVA– F=5.81095– p=.013532– Factor

• df=2• SS=1.1675• MS=0.58375

– Error• df=15• SS=1.50686• MS=.100457• Sxp=0.31695