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    Analysis of Variance.ppt

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    Analysis of Variance.ppt

    1、Analysis of VarianceChapter 1515.1 Introduction Analysis of variance compares two or more populations of interval data. Specifically, we are interested in determining whether differences exist between the population means. The procedure works by analyzing the sample variance. The analysis of varianc

    2、e is a procedure that tests to determine whether differences exits between two or more population means. To do this, the technique analyzes the sample variances15.2 One Way Analysis of Variance Example 15.1 An apple juice manufacturer is planning to develop a new product -a liquid concentrate. The m

    3、arketing manager has to decide how to market the new product. Three strategies are considered Emphasize convenience of using the product. Emphasize the quality of the product. Emphasize the products low price.One Way Analysis of Variance Example 15.1 - continued An experiment was conducted as follow

    4、s: In three cities an advertisement campaign was launched . In each city only one of the three characteristics (convenience, quality, and price) was emphasized. The weekly sales were recorded for twenty weeks following the beginning of the campaigns.One Way Analysis of Variance One Way Analysis of V

    5、arianceSee file Xm15 -01Weekly salesWeekly salesWeekly sales Solution The data are interval The problem objective is to compare sales in three cities. We hypothesize that the three population means are equalOne Way Analysis of Variance H0: m1 = m2= m3H1: At least two means differTo build the statist

    6、ic needed to test thehypotheses use the following notation: SolutionDefining the HypothesesIndependent samples are drawn from k populations (treatments).1 2 kX11x21.Xn1,1X12x22.Xn2,2X1kx2k.Xnk,kSample sizeSample meanFirst observation,first sampleSecond observation,second sampleX is the “response var

    7、iable”.The variables value are called “responses”.NotationTerminology In the context of this problemResponse variable weekly salesResponses actual sale valuesExperimental unit weeks in the three cities when we record sales figures.Factor the criterion by which we classify the populations (the treatm

    8、ents). In this problems the factor is the marketing strategy.Factor levels the population (treatment) names. In this problem factor levels are the marketing trategies.Two types of variability are employed when testing for the equality of the population meansThe rationale of the test statistic Graphi

    9、cal demonstration:Employing two types of variability20253017Treatment 1 Treatment 2 Treatment 31012199Treatment 1Treatment 2Treatment 32016151411109The sample means are the same as before,but the larger within-sample variability makes it harder to draw a conclusionabout the population means.A small

    10、variability withinthe samples makes it easierto draw a conclusion about the population means. The rationale behind the test statistic I 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). If the alternative hypot

    11、hesis is true, at least some of the sample means would differ. Thus, we measure variability between sample means. The variability between the sample means is measured as the sum of squared distances between each mean and the grand mean.This sum is called the Sum of Squares for TreatmentsSSTIn our ex

    12、ample treatments arerepresented by the differentadvertising strategies.Variability between sample meansThere are k treatmentsThe size of sample j The mean of sample jSum of squares for treatments (SST)Note: When the sample means are close toone another, their distance from the grand mean is small, l

    13、eading to a small SST. Thus, large SST indicates large variation between sample means, which supports H1. Solution continuedCalculate SST = 20(577.55 - 613.07)2 + + 20(653.00 - 613.07)2 + + 20(608.65 - 613.07)2 = 57,512.23The grand mean is calculated by Sum of squares for treatments (SST)Is SST = 57

    14、,512.23 large enough to reject H0 in favor of H1?See next.Sum of squares for treatments (SST) Large variability within the samples weakens the “ability” of the sample means to represent their corresponding population means. Therefore, even though sample means may markedly differ from one another, SS

    15、T must be judged relative to the “within samples variability”. The rationale behind test statistic II The variability within samples is measured by adding all the squared distances between observations and their sample means.This sum is called the Sum of Squares for Error SSEIn our example this is t

    16、he sum of all squared differencesbetween sales in city j and thesample mean of city j (over all the three cities).Within samples variability Solution continuedCalculate SSESum of squares for errors (SSE) = (n1 - 1)s12 + (n2 -1)s22 + (n3 -1)s32= (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24 = 50

    17、6,983.50Is SST = 57,512.23 large enough relative to SSE = 506,983.50 to reject the null hypothesis that specifies that all the means are equal?Sum of squares for errors (SSE) To perform the test we need to calculate the mean squares as follows:The mean sum of squares Calculation of MST - Mean Square

    18、 for Treatments Calculation of MSEMean Square for ErrorCalculation of the test statistic with the following degrees of freedom:v1=k -1 and v2=n-kRequired Conditions:1. The populations testedare normally distributed.2. The variances of all thepopulations tested areequal.And finally the hypothesis tes

    19、t:H0: m1 = m2 = =mkH1: At least two means differTest statistic: R.R: FFa,k-1,n-kThe F test rejection region The F testHo: m1 = m2= m3H1: At least two means differ Test statistic F= MST/ MSE= 3.23Since 3.23 3.15, there is sufficient evidence to reject Ho in favor of H1, and argue that at least one of

    20、 the mean sales is different than the others. Use Excel to find the p-value fx Statistical FDIST(3.23,2,57) = .0467The F test p- value p Value = P(F3.23) = .0467Excel single factor ANOVASS(Total) = SST + SSEXm15-01.xls15.3 Analysis of Variance Experimental Designs Several elements may distinguish be

    21、tween one experimental design and others. The number of factors. Each characteristic investigated is called a factor. Each factor has several levels.Factor ALevel 1Level2Level 1Factor BLevel 3Two - way ANOVATwo factorsLevel2One - way ANOVASingle factorTreatment 3 (level 1)ResponseResponseTreatment 1 (level 3)Treatment 2 (level 2)


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