Anthony J Greene1 ANOVA: Analysis of Variance 1-way ANOVA

Anthony J Greene2 ANOVA I.What is Analysis of Variance 1.The F-ratio 2.Used for testing hypotheses among more than two means 3.As with t-test, effect is measured in numerator, error variance in the denomenator 4.Partitioning the Variance II.Different computational concerns for ANOVA 1.Degrees Freedom for Numerator and Denominator 2.No such thing as a negative value III.Using Table B.4 IV.The Source Table V.Hypothesis testing

Anthony J Greene3 M2M2 M3M3 M1M1

4 ANOVA Analysis of Variance Hypothesis testing for more than 2 groups For only 2 groups t 2 (n) = F (1,n)

Anthony J Greene5 BASIC IDEA As with the t-test, the numerator expresses the differences among the dependent measure between experimental groups, and the denominator is the error. If the effect is enough larger than random error, we reject the null hypothesis. M 1 = 1 M 2 = 5 M 3 = 1 Is the Effect Variability Large Compared to the Random Variability Grp 1 Grp 2 Grp 3 Effect V Random V =

Anthony J Greene6 BASIC IDEA If the differences accounted for by the manipulation are low (or zero) then F = 1 If the effects are twice as large as the error, then F = 3, which generally indicates an effect.

Anthony J Greene7 Sources of Variance

Anthony J Greene8 Why Is It Called Analysis of Variance? Aren’t We Interested In Means, Not Variance? Most statisticians do not know the answer to this question? If we’re interested in differences among means why do an analysis of variance? The misconception is that it compares  1 2 to  2 2. No The comparison is between effect variance (differences in group means) to random variance.

Anthony J Greene9 Learning Under Three Temperature Conditions T is the treatment total, G is the Grand total M2M2 M3M3 M1M1

Anthony J Greene10 Computing the Sums of Squares

Anthony J Greene11 How Variance is Partitioned This simply disregards group membership and computes an overall SS Variability Between and Within Groups is Included Keep in mind the general formula for SS M 1 = 1 M 2 = 5 M 3 = 1 Grp 1 Grp 2 Grp 3

Anthony J Greene12 How Variance is Partitioned Imagine there were no individual differences at all. The SS for all scores would measure only the fact that there were group differences. Grp 1 Grp 2 Grp 3 Keep in mind the general formula for SS M 1 = 1 M 2 = 5 M 3 = 1T 1 = 5 T 2 = 25 T 3 = 5

Anthony J Greene13 How Variance is Partitioned SS computed within a column removes the mean. Thus summing the SS’s for each column computes the overall variability except for the mean differences between groups. M 1 = 1 M 2 = 5 M 3 = 1 Grp 1 Grp 2 Grp 3 Keep in mind the general formula for SS

Anthony J Greene14 How Variance is Partitioned M 1 = 1 M 2 = 5 M 3 = 1 Grp 1 Grp 2 Grp

Anthony J Greene15 Computing Degrees Freedom df between is k-1, where k is the number of treatment groups (for the prior example, 3, since there were 3 temperature conditions) df within is N-k, where N is the total number of ns across groups. Recall that for a t-test with two independent groups, df was 2n-2? 2n was all the subjects N and 2 was the number of groups, k.

Anthony J Greene16 Computing Degrees Freedom

Anthony J Greene17 How Degrees Freedom Are Partitioned N-1 = (N - k) + (k - 1) N-1 = N - k + k – 1

Anthony J Greene18 Partitioning The Sums of Squares

Anthony J Greene19 Computing An F-Ratio

Anthony J Greene20 Consult Table B-4 Take a standard normal distribution, square each value, and it looks like this

Anthony J Greene21 Table B-4

Anthony J Greene22 Two different F-curves

Anthony J Greene23 ANOVA: Hypothesis Testing

Anthony J Greene24 Basic Properties of F-Curves Property 1: The total area under an F-curve is equal to 1. Property 2: An F-curve starts at 0 on the horizontal axis and extends indefinitely to the right, approaching, but never touching, the horizontal axis as it does so. Property 3: An F-curve is right skewed.

Anthony J Greene25 Finding the F-value having area 0.05 to its right

Anthony J Greene26 Assumptions for One-Way ANOVA 1.Independent samples: The samples taken from the populations under consideration are independent of one another. 2.Normal populations: For each population, the variable under consideration is normally distributed. 3.Equal standard deviations: The standard deviations of the variable under consideration are the same for all the populations.

Anthony J Greene27 Learning Under Three Temperature Conditions M 1 = 1 M 2 = 5 M 3 = 1

Anthony J Greene28 Learning Under Three Temperature Conditions M 1 = 1 M 2 = 5 M 3 = 1 Is the Effect Variability Large Compared to the Random Variability

Anthony J Greene29 Learning Under Three Temperature Conditions

Anthony J Greene30 Learning Under Three Temperature Conditions

Anthony J Greene31 Learning Under Three Temperature Conditions

Anthony J Greene32 Learning Under Three Temperature Conditions

Anthony J Greene33 Learning Under Three Temperature Conditions M2M2 M3M3 M1M1

Anthony J Greene34 Learning Under Three Temperature Conditions Σ X 2 = M2M2 M3M3 M1M1

Anthony J Greene35 Learning Under Three Temperature Conditions M2M2 M3M3 M1M1

36 Learning Under Three Temperature Conditions M2M2 M3M3 M1M1

Calculating the F statistic Ss total = X 2 -G 2 /N = 46 SS between = SS between = 30 SS total = Ss between + SS within Ss within = 16

Anthony J Greene38 Distribution of the F-Statistic for One-Way ANOVA Suppose the variable under consideration is normally distributed on each of k populations and that the population standard deviations are equal. Then, for independent samples from the k populations, the variable has the F-distribution with df = (k – 1, n – k) if the null hypothesis of equal population means is true. Here n denotes the total number of observations.

Anthony J Greene39 ANOVA Source Table for a one-way analysis of variance

Anthony J Greene40 The one-way ANOVA test for k population means (Slide 1 of 3) Step 1The null and alternative hypotheses are H o :  1 =  2 =  3 = …=  k H a : Not all the means are equal Step 2Decide On the significance level,  Step 3The critical value of F , with df = (k - 1, N - k), where N is the total number of observations.

Anthony J Greene41 The one-way ANOVA test for k population means (Slide 2 of 3)

Anthony J Greene42 The one-way ANOVA test for k population means (Slide 3 of 3) Step 4Obtain the three sums of squares, STT, STTR, and SSE Step 5Construct a one-way ANOVA table: Step 6If the value of the F-statistic falls in the rejection region, reject H 0 ;

Anthony J Greene43 Post Hocs H 0 :  1 =  2 =  3 = …=  k Rejecting H 0 means that not all means are equal. Pairwise tests are required to determine which of the means are different. One problem is for large k. For example with k = 7, 21 means must be compared. Post-Hoc tests are designed to reduce the likelihood of groupwise type I error.

Anthony J Greene44 Criterion for deciding whether or not to reject the null hypothesis

Anthony J Greene45 One-Way ANOVA controllow dosehigh dose A researcher wants to test the effects of St. John’s Wort, an over the counter, herbal anti-depressant. The measure is a scale of self-worth. The subjects are clinically depressed patients. Use α = 0.01

Anthony J Greene46 One-Way ANOVA controllow dosehigh dose G= T 1 =5T 2 =10T 3 =30 Compute the treatment totals, T, and the grand total, G

Anthony J Greene47 One-Way ANOVA controllow dosehigh dose G=45 346N=15 014k=3 117 T 1 =5T 2 =10T 3 =30 n 1 =5n 2 =5n 3 =5 Count n for each treatment, the total N, and k

Anthony J Greene48 One-Way ANOVA controllow dosehigh dose G=45 346N=15 014k=3 117 T 1 =5T 2 =10T 3 =30 n 1 =5n 2 =5n 3 =5 M 1 =1M 2 =2M 3 =6 Compute the treatment means

Anthony J Greene49 One-Way ANOVA controllow dosehigh dose G=45 346N=15 014k=3 117 T 1 =5T 2 =10T 3 =30 n 1 =5n 2 =5n 3 =5 M 1 =1M 2 =2M 3 =6 SS=6SS=8SS=10 Compute the treatment SSs (0-1) 2 =1 (1-1) 2 =0 (3-1) 2 =4 (0-1) 2 =1 (1-1) 2 =0 sum

Anthony J Greene50 One-Way ANOVA controllow dosehigh dose G=45 346N=15 014k=3 117  X 2 = 229 T 1 =5T 2 =10T 3 =30 n 1 =5n 2 =5n 3 =5 M 1 =1M 2 =2M 3 =6 SS=6SS=8SS=10 Compute all X 2 s and sum them

Anthony J Greene51 One-Way ANOVA controllow dosehigh dose G=45 346N=15 014k=3 117  X 2 = 229 T 1 =5T 2 =10T 3 =30 SS Total =94 n 1 =5n 2 =5n 3 =5 M 1 =1M 2 =2M 3 =6 SS=6SS=8SS=10 Compute SS Total SS Total =  X 2 – G 2 / N

Anthony J Greene52 One-Way ANOVA controllow dosehigh dose G=45 346N=15 014k=3 117  X 2 = 229 T 1 =5T 2 =10T 3 =30 SS Total =94 n 1 =5n 2 =5n 3 =5 SS Within =24 M 1 =1M 2 =2M 3 =6 SS 1 =6SS 2 =8SS 3 =10 Compute SS Within SS Within =  SS i

Anthony J Greene53 One-Way ANOVA controllow dosehigh dose G=45 346N=15 014k=3 117  X 2 = 229 T 1 =5T 2 =10T 3 =30 SS Total =94 n 1 =5n 2 =5n 3 =5 SS Within =24 M 1 =1M 2 =2M 3 =6 d.f. Within =12 SS 1 =6SS 2 =8SS 3 =10 d.f. Between =2 d.f. Total =14 Determine d.f.s d.f. Within =N-k d.f. Between =k-1 d.f. Total =N-1 Note that (N-k)+(k-1)=N-1

Anthony J Greene54 One-Way ANOVA controllow dosehigh dose G=45 346N=15 014k=3 117  X 2 = 229 T 1 =5T 2 =10T 3 =30 SS Total =94 n 1 =5n 2 =5n 3 =5 SS Within =24 M X1 =1M X2 =2M X3 =6 d.f. Within =12 SS 1 =6SS 2 =8SS 3 =10 d.f. Between =2 d.f. Total =14 Ready to move it to a source table

Anthony J Greene55 One-Way ANOVA Compute the missing values SourceSSdfMSF Between702 Within2412 Total9414

Anthony J Greene56 One-Way ANOVA Compute the missing values SourceSSdfMSF Between70235 Within24122 Total9414

Anthony J Greene57 One-Way ANOVA Compute the missing values SourceSSdfMSF Between Within24122 Total9414

Anthony J Greene58 One-Way ANOVA 1.Compare your F of 17.5 with the critical value at 2,12 degrees of freedom,  = 0.01: reject H 0 SourceSSdfMSF Between Within24122 Total9414

Anthony J Greene59 One-Way ANOVA Low Medium High Students want to know if studying has an impact on a 10-point statistics quiz, so they divided into 3 groups: low studying (0-5hrs./wk), medium studying (6-15 hrs./wk) and high studying (16+ hours/week). At α=0.01, does the amount of studying impact quiz scores?

Anthony J Greene60 One-Way ANOVA low medium high G= T 1 =12T 2 =30T 3 =54 Compute the treatment totals, T, and the grand total, G

Anthony J Greene61 One-Way ANOVA lowmedium high G=96 358N= k= T 1 =12T 2 =30T 3 =54 n 1 =6n 2 =6n 3 =6 Count n for each treatment, the total N, and k

Anthony J Greene62 One-Way ANOVA lowmedium high G=96 358N= k= T 1 =12T 2 =30T 3 =54 n 1 =6n 2 =6n 3 =6 M 1 =2M 2 =5M 3 =9 Compute the treatment means

Anthony J Greene63 One-Way ANOVA low medium high G=96 358N= k= T 1 =12T 2 =30T 3 =30 n 1 =6n 2 =6n 3 =6 M 1 =2M 2 =5M 3 =9 SS=10SS=8SS=10 Compute the treatment SSs (2-2) 2 =0 (4-2) 2 =4 (3-2) 2 =1 (0-2) 2 =4 (2-2) 2 =0 (1-2) 2 =1 sum

Anthony J Greene64 One-Way ANOVA lowmedium high G=96 358N= k=  X 2 =682 T 1 =12T 2 =30T 3 =54 n 1 =6n 2 =6n 3 =6 M 1 =2M 2 =5M 3 =9 SS=10SS=8SS=10 Compute all X 2 s and sum them

Anthony J Greene65 One-Way ANOVA lowmedium high G=96 358N= k=  X 2 = 682 T 1 =12T 2 =30T 3 =54 SS Total =170 n 1 =6n 2 =6n 3 =6 M 1 =2M 2 =5M 3 =9 SS=10SS=8SS=10 Compute SS Total SS Total =  X 2 – G 2 / N

Anthony J Greene66 One-Way ANOVA lowmedium high G=96 358N= k=  X 2 = 682 T 1 =12T 2 =30T 3 =54 SS Total =170 n 1 =6n 2 =6n 3 =6 SS Within =28 M 1 =2M 2 =5M 3 =9 SS 1 =10SS 2 =8SS 3 =10 Compute SS Within SS Within =  SS i

Anthony J Greene67 One-Way ANOVA lowmedium high G=90 358N= k=  X 2 = 682 T 1 =12T 2 =30T 3 =54 SS Total =170 n 1 =6n 2 =6n 3 =6 SS Within =28 M 1 =2M 2 =5M 3 =9 d.f. Within =15 SS 1 =10SS 2 =8SS 3 =10 d.f. Between =2 Determine d.f.s d.f. Within =N-k d.f. Between =k-1 d.f. Total =N-1 Note that (N-k)+(k-1)=N-1

Anthony J Greene68 One-Way ANOVA Fill in the values you have SourceSSdfMSF Between2 Within2815 Total17017

Anthony J Greene69 One-Way ANOVA Compute the missing values SourceSSdfMSF Between Within Total17017

Anthony J Greene70 One-Way ANOVA 1.Compare your F of with the critical value at 2,15 degrees of freedom,  = 0.01: reject H 0 SourceSSdfMSF Between Within Total17017