Analysis of Variance ST 511

Introduction n Analysis of variance compares two or more populations of quantitative data. n Specifically, we are interested in determining whether differences exist between the population means. n The procedure works by analyzing the sample variance.

§1 One Way Analysis of Variance n The analysis of variance is a procedure that tests to determine whether differences exist between two or more population means. n To do this, the technique analyzes the sample variances

One Way Analysis of Variance: Example n A magazine publisher wants to compare three different styles of covers for a magazine that will be offered for sale at supermarket checkout lines. She assigns 60 stores at random to the three styles of covers and records the number of magazines that are sold in a one-week period.

One Way Analysis of Variance: Example n How do five bookstores in the same city differ in the demographics of their customers? A market researcher asks 50 customers of each store to respond to a questionnaire. One variable of interest is the customer’s age.

Idea Behind ANOVA – two types of variability 1.Within group variability 2.Between group variability

Treatment 1Treatment 2 Treatment Treatment 1 Treatment 2Treatment The sample means are the same as before, but the larger within-sample variability makes it harder to draw a conclusion about the population means. A small variability within the samples makes it easier to draw a conclusion about the population means.

Idea behind ANOVA: recall the two-sample t-statistic n Difference between 2 means, pooled variances, sample sizes both equal to n n Numerator of t 2 : measures variation between the groups in terms of the difference between their sample means n Denominator: measures variation within groups by the pooled estimator of the common variance. n If the within-group variation is small, the same variation between groups produces a larger statistic and a more significant result.

One Way Analysis of Variance: Example n Example 1 –An apple juice manufacturer is planning to develop a new product -a liquid concentrate. –The marketing manager has to decide how to market the new product. –Three strategies are considered F Emphasize convenience of using the product. F Emphasize the quality of the product. F Emphasize the product’s low price.

One Way Analysis of Variance n Example 1 - continued –An experiment was conducted as follows: F In three cities an advertisement campaign was launched. F In each city only one of the three characteristics (convenience, quality, and price) was emphasized. F The weekly sales were recorded for twenty weeks following the beginning of the campaigns.

One Way Analysis of Variance Weekly sales

One Way Analysis of Variance n Solution –The data are quantitative –The problem objective is to compare sales in three cities. –We hypothesize that the three population means are equal

H 0 :  1 =  2 =  3 H 1 : At least two means differ To build the statistic needed to test the hypotheses use the following notation: Solution Defining the Hypotheses

Independent samples are drawn from k populations (treatment groups). 12k X 11 x 21. X n1,1 X 12 x 22. X n2,2 X 1k x 2k. X nk,k Sample size Sample mean First observation, first sample Second observation, second sample X is the “response variable”. The variables’ values are called “responses”. Notation

Terminology n In the context of this problem… Response variable – weekly sales Responses – actual sale values Experimental unit – weeks in the three cities when we record sales figures. Factor – the criterion by which we classify the populations (the treatments). In this problems the factor is the marketing strategy. Factor levels – the population (treatment) names. In this problem factor levels are the 3 marketing strategies: 1) convenience, 2) quality, 3) price

Two types of variability are employed when testing for the equality of the population means The rationale of the test statistic 1.Within sample variability 2.Between sample variability

The rationale behind the test statistic – I n If the null hypothesis is true, we would expect all the sample means to be close to one another (and as a result, close to the grand mean). n If the alternative hypothesis is true, at least some of the sample means would differ. Thus, we measure variability between sample means. H 0 :  1 =  2 =  3 H 1 : At least two means differ

This sum is called the Sum of Squares for Groups SSG The variability between the sample means is measured as the sum of squared distances between each mean and the grand mean. In our example, treatments are represented by the different advertising strategies. Variability between sample means

There are k treatments The size of sample j The mean of sample j Sum of squares for treatment groups (SSG) Note: When the sample means are close to one another, their distance from the grand mean is small, leading to a small SSG. Thus, large SSG indicates large variation between sample means, which supports H 1.

Sum of squares for treatment groups (SSG) n Solution – continued Calculate SSG = 20( ) ( ) ( ) 2 = = 57, The grand mean is calculated by

Sum of squares for treatment groups (SSG) Is SSG = 57, large enough to reject H 0 in favor of H 1 ?

The rationale behind test statistic – II n Large variability within the samples weakens the “ability” of the sample means to represent their corresponding population means. n Therefore, even though sample means may markedly differ from one another, SSG must be judged relative to the “within samples variability”.

This sum is called the Sum of Squares for Error SSE Within samples variability n The variability within samples is measured by adding all the squared distances between observations and their sample means. In our example, this is the sum of all squared differences between sales in city j and the sample mean of city j (over all the three cities).

Sum of squares for errors (SSE) n Solution – continued Calculate SSE  (n 1 - 1)s (n 2 -1)s (n 3 -1)s 3 2 = (20 -1)10,775 + (20 -1)7, (20-1)8, = 506,983.50

Sum of squares for errors (SSE) Is SSG = 57, large enough relative to SSE = 506, to reject the null hypothesis that specifies that all the means are equal?

mean squares To perform the test we need to calculate the mean squares as follows: The mean sum of squares Calculation of MSG - Mean Square for treatment Groups Calculation of MSE Mean Square for Error

Calculation of the test statistic with the following degrees of freedom: v 1 =k -1 and v 2 =n-k Required Conditions: 1. The populations tested are normally distributed. 2. The variances of all the populations tested are equal.

And finally the hypothesis test: H 0 :  1 =  2 = …=  k H 1 : At least two means differ Test statistic: R.R: F>F ,k-1,n-k The F test rejection region

The F test H o :  1 =  2 =  3 H 1 : At least two means differ Test statistic F= MSG  MSE= 3.23 Since 3.23 > 3.15, there is sufficient evidence to reject H o in favor of H 1, and argue that at least one of the mean sales is different than the others.

The F test p- value n Use Excel to find the p-value f x Statistical F.DIST.RT(3.23,2,57) =.0469 p Value = P(F>3.23) =.0469

Excel single factor ANOVA SS(Total) = SSG + SSE

Multiple Comparisons n When the null hypothesis is rejected, it may be desirable to find which mean(s) is (are) different, and at what ranking order. n Two statistical inference procedures, geared at doing this, are presented: –“regular” confidence interval calculations –Bonferroni adjustment

Multiple Comparisons n Two means are considered different if the confidence interval for the difference between the corresponding sample means does not contain 0. In this case the larger sample mean is believed to be associated with a larger population mean. n How do we calculate the confidence intervals?

“Regular” Method n This method builds on the equal variances confidence interval for the difference between two means. n The CI is improved by using MSE rather than s p 2 (we use ALL the data to estimate the common variance instead of only the data from 2 samples)

Experiment-wise Type I error rate (the effective Type I error) n The preceding “regular” method may result in an increased probability of committing a type I error. n The experiment-wise Type I error rate is the probability of committing at least one Type I error at significance level . It is calculated by: experiment-wise Type I error rate = 1-(1 –  ) g where g is the number of pairwise comparisons (i.e. g = k C 2 = k(k-1)/2. n For example, if  =.05, k=4, then experiment-wise Type I error rate =1-.735=.265 n The Bonferroni adjustment determines the required Type I error probability per pairwise comparison (  * ), to secure a pre-determined overall .

Bonferroni Adjustment n The procedure: –Compute the number of pairwise comparisons (g) [g=k(k-1)/2], where k is the number of populations. –Set  * =  /g, where  is the true probability of making at least one Type I error (called experiment-wise Type I error). –Calculate the following CI for  i –  j

Bonferroni Method n Example - continued –Rank the effectiveness of the marketing strategies (based on mean weekly sales). –Use the Bonferroni adjustment method n Solution –The sample mean sales were , 653.0, –We calculate g=k(k-1)/2 to be 3(2)/2 = 3. –We set  * =.05/3 =.0167, thus t.0167/2, 60-3 = (Excel). –Note that s = √ = 94.31

Bonferroni Method: The Three Confidence Intervals There is a significant difference between  1 and  2.

Bonferroni Method: Conclusions Resulting from Confidence Intervals Do we have evidence to distinguish two means? n Group 1 Convenience: sample mean n Group 2 Quality: sample mean 653 n Group 3 Price: sample mean n List the group numbers in increasing order of their sample means; connecting overhead lines mean no significant difference 1 3 2

Bonferroni Method: Conclusions Resulting from Confidence Intervals Do we have evidence to distinguish two means? n Group 1 Convenience: sample mean n Group 2 Quality: sample mean 653 n Group 3 Price: sample mean n List the group numbers in increasing order of their sample means; connecting overhead lines mean no significant difference 1 3 2