Lecture note 8 Spring Lecture note 8. Analysis of Variance (ANOVA)
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1 Lecture note 8 Analysis of Variance (ANOVA) 1
2 Overview of ANOVA Analysis of variance (ANOVA) is a comparison of means. ANOVA allows you to compare more than two means simultaneously. Proper experimental design efficiently uses limited data to draw the strongest possible inferences. 2
3 The Goal: Explaining Variation ANOVA = ANalysis Of VAriance ANOVA compares the means of several groups. The groups are sometimes called "treatments" 3
4 The Goal: Explaining Variation ANOVA seeks to identify sources of variation in a numerical dependent variable Y (the response variable). Variation in Y about its mean is explained by one or more categorical independent variables (the factors) or is unexplained (random error). 4
5 Each possible value of a factor or combination of factors is a treatment. We test to see if each factor has a significant effect on Y using (for example) the hypotheses: H 0 : μ 1 = μ 2 = μ 3 = μ 4 H 1 : Not all the means are equal The test uses the F distribution. If we cannot reject H 0, we conclude that observations within each treatment have a common mean μ. 5
6 One Factor ANOVA One-Factor ANOVA ANOVA can have more than one factor. In business problems, one factor may suffice. We will illustrate ANOVA using one factor. 6
7 Example: Defect Rates (c = 4 locations) 7
8 One Factor Examples A one-factor ANOVA could test the hypothesis that the defect rate is affected by the manufacturing plant location: Defect rate = f(plant location) A one-factor ANOVA would test the hypothesis that the burst strength of PVC pipe is affected by the brand of pipe: Length of stay = f(brand of pipe) 8
9 One-Factor ANOVA (Completely Randomized Design) Data Format A one-factor ANOVA only compares the means of c groups (treatments or factor levels). Consider the format for a one-factor ANOVA with c treatments, denoted A 1, A 2,, A c 9
10 10
11 Grade Point Averages of 25 Business Students Freshman Sophomore Junior Senior Question: Do the mean semester GPAs differ significantly for these 4 groups of students (each observation is one student)? Or are we just seeing random variation in the samples? 11
12 Emergency Room Waiting Time in Minutes Hospital A Hospital B Hospital C Hospital D Question: Do the mean waiting times differ significantly for these 4 horpital (each observation is one patient). Or are we just seeing random variation in the samples? 12
13 Data Format Sample sizes within each treatment do not need to be equal (i.e., balanced). The total number of observations is equal to n = n 1 + n n c Hypothesis to Be Tested H 0 : μ 1 = μ 2 = = μ c H 1 : Not all the means are equal ANOVA tests all means simultaneously and so does not inflate the type I error. 13
14 One-Factor ANOVA as a Linear Model An equivalent way to express the one-factor model is to say that treatment j came from a population with a common mean (m) plus a treatment effect (A j ) plus random error (e ij ): y ij = μ + A j + e ij j = 1, 2,, c and i = 1, 2,, n Random error is assumed to be normally distributed with zero mean and the same variance for all treatments. 14
15 ANOVA Assumptions Analysis of Variance assumes that the - observations on Y are independent, - populations being sampled are normal, - populations being sampled have equal variances. ANOVA is somewhat robust to departures from normality and equal variance assumptions. 15
16 Group Means The mean of each group is calculated as The overall sample mean (grand mean) can be calculated as 16
17 Partitioned Sum of Squares For a given observation y ij, the following relationship must hold (y ij y ) = (y j y ) + (y ij y j ) where (y ij y ) = deviation of an observation from the grand mean (y j y ) = deviation of the column mean from the grand mean (between treatments) (y ij y j ) = deviation of the observation from its own column mean (within treatments). 17
18 Partitioned Sum of Squares This relationship is true for sums of squared deviations, yielding partitioned sum of squares: Simply put, SST = SSA + SSE 18
19 Partitioned Sum of Squares SSA and SSE are used to test the hypothesis of equal treatment means by dividing each sum of squares by it degrees of freedom to adjust for group size. These ratios are called Mean Squares (MSA and MSE). The resulting test statistic is F = MSA/MSE. 19
20 20
21 Test Statistic The F distribution describes the ratio of two variances. The F statistic is the ratio of the variance due to treatments (MSA) to the variance due to error (MSE). 21
22 Test Statistic When F is near zero, then there is little difference among treatments and we would not expect to reject the hypothesis of equal treatment means. Decision Rule F cannot be negative has no upper limit. For ANOVA, the F test is a right-tailed test. Use Appendix F or Excel to obtain the critical value of F for a given α. 22
23 Decision Rule 23
24 Steps in Performing One-Factor ANOVA Step 1: State the hypotheses H 0 : μ 1 = μ 2 = = μ c H 1 : Not all the means are equal Step 2: State the decision rule The degrees of freedom for the critical value F are Numerator d.f. = c 1 (between treatments) Denominator d.f. = n c (within treatments) For the given α, obtain the critical value from Appendix F or Excel. 24
25 Step 3: Perform the Calculations Use Excel(or STATA) to perform the calculations. For example, here are the results of an ANOVA: 25
26 Step 4: Make the Decision Reject H 0 if the critical value F exceeds the test statistic F a or if the p-value given by Excel(or STATA) is < α. 26
27 STATA Practice anova command Data file: wage.dta 1. Using su command, find the sample mean for each variable 2. Rename the three variables as wage_ba, wage_bs and wage_bba 3. Stack the three variables and make them one variable 27
28 4. Using anova command, test the following hypothesis H0: μ_ba= μ_bs= μ_bba Questions 1. In one-factor ANOVA, the unexplained variation is measured by the a. sum of squares for treatments b. sum of squares for error c. total sum of squares d. degrees of freedom 28
29 2. Which of the following is a correct interpretation of the null hypothesis for analysis of variance for one factor? a. There is no difference between the mean values of the random variable at the various levels of the test factor. b. The factor being tested had no effect on the random variable x. c. There is no variance amongst the mean values of x for each of the different factor levels. d. All of the above. 29
30 3. Which of the following statements is false regarding a single-factor ANOVA? a. Between-sample variation and within-sample variation are compared in an ANOVA test. b. The data values from repeated samplings are called replicates. c. SS(total) = SS(factor) + SS(error) d. None of the above 4. (T or F) To partition the sum of squares for the total in single-factor ANOVA is to separate the numerical value of SS(total) into two values, SS(factor) and SS(error), such that the sum of these two values is equal to SS(total). 30
31 5. The following ANOVA table shows results of independent samples collected to test the effect a factor had on a variable. Find the critical value for F at α= 0.05 and determine if H0 can be rejected. df ss factor error total
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