This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 13 (13.113.2) Analysis of Variance (ANOVA)
An Introduction to Analysis of Variance s Analysis of Variance: Testing for the Equality of k Population Means
s Slide 1 Example: Reed Manufacturing
J. R. Reed would like to know if the mean number of hours worked per week is the same for the department managers at her three manufacturing plants (Buffalo, Pittsburgh, and Detroit). A simple random sample of 5 managers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide. Slide 2 Example: Reed Manufacturing
Plant 1 Plant 2 Plant 3 Buffalo Pittsburgh 48 54 57 54 62 73 63 66 64 74 68 26.5 Observation 1 2 3 4 5 Detroit 51 63 61 54 56 57 24.5 Slide 3 Sample Mean 55 Sample Variance 26.0 Example: Reed Manufacturing
s GOAL: Answer the question: Is the mean number of hours worked per week same for all three plants? • Test the Hypotheses H0: µ 1 = µ 2 = µ Ha: Not all the means are equal
3 where: µ 1 = mean number of hours worked per 1 week by the managers at Plant 1 µ 2 = mean number of hours worked per 2 week by the managers at Plant 2 µ 3 = mean number of hours worked per 3 week by the managers at Plant 3 Slide 4 An Introduction to Analysis of Variance
s Reed Manufacturing example has two variables: 1. Plant location (independent variable or factor) 2. Number of hours worked by each manager (dependent or response variable) A treatment is a level of a factor. s Q. What are the treatments in Reed Manufacturing example? Slide 5 Assumptions for Analysis of Variance
s For each population, the response variable is normally distributed. s The variance of the response variable, denoted σ 2, is the same for all of the populations. s The observations must be independent. Slide 6 Analysis of Variance: Testing for the Equality of k Population Means
s s s s BetweenTreatments Estimate of Population Variance WithinTreatments Estimate of Population Variance Comparing the Variance Estimates: The F Test The ANOVA Table Slide 7 An Introduction to Analysis of Variance
s s Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means using data obtained from observational or experimental studies. We want to test the following hypotheses. H0: µ 1 = µ 2 = µ 3 =
. . . = µ . . . k Ha: Not all population means are equal
s s If H0 is rejected, we cannot conclude that all population means are different. Rejecting H0 means that at least two population means have different values. Slide 8 Notations and Formulas
Let xij = observation i for treatment j nj = number of observations for treatment j xj = sample mean for treatment j = sample variance for treatment j s2 j Then
nT = n1 + n2 + ..... + nk = total number of observations n1 x1 + n2 x2 + ..... + nk xk x= = sample mean of all observations nT x1 + x2 + .... + xk . If n1 = n2 = ..... = nk = n, then x = k Slide 9 BetweenTreatments Estimate of Population Variance
s A betweentreatments estimate of σ square due to treatments (MSTR).
k 2 2 is called the mean MSTR =
s n j ( x j − x )2 ∑
j =1 k −1 s The numerator of MSTR is called the sum of squares due to treatments (SSTR). The denominator of MSTR represents the degrees of freedom associated with SSTR. Slide 10 WithinTreatments Estimate of Population Variance
s The estimate of σ 2 based on the variation of the sample observations within each treatment is called the mean square due to error (MSE).
MSE = ( n j −1) s 2 ∑ j
j =1 k nT − k s s The numerator of MSE is called the sum of squares due to error (SSE). The denominator of MSE represents the degrees of freedom associated with SSE. Slide 11 Example: Reed Manufacturing
s Analysis of Variance • Mean Square due to Treatments = x SSTR = MSTR = • Mean Square Error SSE = MSE = Slide 12 Test for the Equality of k Population Means
s Hypotheses H0: µ 1 = µ 2 = µ 3 = Test Statistic F = MSTR/MSE Rejection Rule • Using test statistic: Reject H0 if F > Fα where the value of Fα is based on an F distribution with k where the value of is based on an 1 numerator degrees of freedom and nT k denominator degrees of freedom. • Using pvalue: Reject H0 if pvalue < α Slide 13 . . . = µ . . . k Ha: Not all population means are equal
s s Sampling Distribution of MSTR/MSE
s The figure below shows the rejection region associated with a level of significance equal to α where Fα denotes the critical value. Do Not Reject H0 Reject H0 MSTR/MSE Fα Critical Value Slide 14 Example: Reed Manufacturing
s Analysis of Variance • Test Statistic: • Conclusion: Slide 15 Example: Reed Manufacturing
s Analysis of Variance • F Test If H0 is true, the ratio MSTR/MSE should be near 1 since both MSTR and MSE are estimating σ 2. If Ha is true, the ratio should be significantly larger than 1 since MSTR tends to overestimate σ 2. • Rejection Rule Assuming α = .05, F.05 = 3.89 (2 d.f. numerator, 12 d.f. denominator). Reject H0 if F > 3.89 Slide 16 The ANOVA Table
Source of Sum of Degrees of Mean Variation Squares Freedom Squares F Treatments SSTR k 1 MSTR MSTR/MSE Error SSE nT k MSE Total SST nT 1 SST divided by its degrees of freedom nT 1 is simply the overall sample variance that would be obtained if we treated the entire nT observations as one data set. SST = ∑∑ ( xij − x ) 2 = SSTR + SSE
j =1 i =1 k nj Slide 17 Example: Reed Manufacturing
s Analysis of Variance • ANOVA Table Source of Sum of Degrees of Mean Variation Squares Freedom Square F Treatments Error Total Slide 18 ...
View
Full
Document
 Summer '08
 Priya

Click to edit the document details