Analysis Of Variance Compiled by T.O. Antwi-Asare, U.G

Transcription

1 Analysis Of Variance Compiled by T.O. Antwi-Asare, U.G 1

2 ANOVA Analysis of variance compares two or more population means of interval data. Specifically, we are interested in determining whether differences exist between the population means. The procedure works by analyzing the sample variances.

3 The assumptions underlying the analysis of variance technique are the same as those used in the t test when comparing two different means. We assume that the samples are randomly and independently drawn from Normally distributed populations which have equal variances. We deal with variable within the interval scale or ratio scale

4 To formalise this we break down the total variance of all the observations into 1. the variance due to differences between treatments or factors, and. the variance due to differences within treatments (also known as the error variance).

5 we have to work with three sums of squares: The total sum of squares measures (squared) deviations from the overall or grand average using all the observations. It ignores the existence of the different factors. The between sum of squares is based upon the averages for each factor and measures how they deviate from the grand average. The within sum of squares is based on squared deviations of observations from their own factor mean.

6 Total sum of squares=between Sum of Squares + Within Sum of Squares The larger - the between sum of squares relative to the within sum of squares, the more likely it is that the null is false.

7 One Way Analysis of Variance Example 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 Emphasize the convenience of using the product. Emphasize the quality of the product. Emphasize the product s low price.

8 One Way Analysis of Variance Example: An experiment was conducted as follows: 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.

9 Problem assumptions The data are interval The problem objective is to compare sales in the three cities. We hypothesize that the three population means are equal

10 One Way Analysis of Variance Weekly sales Convenience 59 Quality 804 Price

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

12 Notation Independent samples are drawn from k populations (treatments). First observation, first sample Second observation, second sample Sample size Sample mean 1 k X 11 x 1.. X n1,1 n1 x1 X 1 x.. X n, n x X 1k x k.. X nk,k X is the response variable. The variables value are called responses. n x k k

13 Terminology 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 marketing strategies.

14 The rationale of the test statistic Two types of variability are employed when testing for the equality of the population means

15 x 15 x x x x 1 10 x 15 7 A small variability within the samples makes it easier Treatment 1 Treatment Treatment 3 to draw a conclusion about the population means. The 1 sample means are the same as before, but the larger within-sample Treatment 1 Treatment Treatment variability 3 makes it harder to draw a conclusion about the population means.

16 The rationale behind the test statistic Part 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 hypothesis is true, at least some of the sample means would differ. Thus, we measure variability between sample means.

17 Variability between sample means The variability between the sample means is measured as the sum of squared distances between each treatment mean and the grand mean. This sum is called the Sum of Squares for Treatments-SST or Between Sum of Squares BSS In our example treatments are represented by the different advertising strategies.

18 NOTE: Here SST Total Sum of Squares TSS = BSS It is the Between Sum of Squares

19 Sum of squares for treatments (SST) or Between Sum of Squares BSS BSS or SST k j1 n (x x) j j There are k treatments The size of sample j Note: When the sample means are close to one another, their distance from the grand mean is small, leading to a small SST. Thus, large SST indicates large variation between sample means, which supports H 1. The mean of sample j or Factor j or treatment j

20 Sum of squares for treatments (SST) or BSS Solution continued Calculate SST or BSS x x x X The grand mean is calculated by n x 1 1 n 1 n n x n n k k x SST k k j1 n j (x j x) = 0( ) + 0( ) + 0( ) = 57,51.3

21 The rationale behind test statistic Part II 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, SST must be judged relative to the within samples variability.

22 Within samples variability SSE or WSS (Within Sum of Squares) or ESS 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 SSE or WSS 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).

23 For example: SSE or WSS (n 1-1)s 1 + (n -1)s + (n 3-1)s (n k 1)s k = k j=1 n j 1 s j k = no. of treatments SSE k j1 n j i1 ( x ij x j )

24 k j=1 n j 1 s j = SSE or WSS where x is the column j mean j SSE k j1 n j i1 ( x ij x j )

25 Sum of squares for errors (SSE) Solution Continued: Calculate SSE s 1 10, s 7,38,11 s 3 8,670.4 SSE k j1 n j i1 ( x ij x j ) Or, SSE (n 1-1)s 1 + (n -1)s + (n 3-1)s 3 = (0-1)10, (0-1)7, (0-1)8,670.4 = 506,983.50

26 The mean sum of squares To perform the test we need to calculate the mean squares as follows: Calculation of MST - Mean Square for Treatments Calculation of MSE Mean Square for Error MST k SST MSE SSE 1 n k 57, , , ,894.45

27 Calculation of the test statistic F MST MSE 8, , Required Conditions: 1. The populations tested are normally distributed.. The variances of all the populations tested are equal. 3.3 with the following degrees of freedom: v 1 =k -1 and v =n-k

28 The F test rejection region And finally the Decision Rule H 0 : m 1 = m = =m k H 1 : At least two means differ Test statistic: Reject H 0 if: F>F a,k-1,n-k F MST MSE

29 The F test H o : m 1 = m = m 3 H 1 : At least two means differ Test statistic F= MST/ MSE= 3.3 R.R. : F Fa k1 nk F0.05,31,603 MST F MSE 8, , Since 3.3 > 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.

30 ANOVA Anova: Single Factor SUMMARY Groups Count Sum Average Variance Convenience Quality Price ANOVA Source of Variation SS df MS F P-value F crit Between Groups Within Groups Total(TSS)

31 k n 1 s = SSE j=1 j j BSS or SST k j1 n (x x) j j

32 Question The reaction times of three groups of sportsmen were measured on a particular task, with the following results (time in milliseconds): Racing drivers Tennis players Boxers Test whether there is a difference in reaction times between the three groups.

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34 Introduction ANOVA is the technique where the total variance present in the data set is spilt up into non- negative components where each component is due to one factor or cause of variation. Factors of variation Assignable Can be many Non-assignable Error or Random variation

35 Utility ANOVA is used to test hypotheses about differences between two or more means. The t-test can only be used to test differences between two means. When there are more than two means, it is possible to compare each mean with each other mean using t-tests. However, conducting multiple t-tests can lead to severe inflation of the Type I error type. ANOVA is used to test differences among several means for significance without increasing the Type I error rate using an F test

36 The ANOVA Procedure: This is the ten step procedure for analysis of variance: 1.Description of data.assumption: Along with the assumptions, we represent the model for each design we discuss. 3. Hypothesis 4.Test statistic 5.Distribution of test statistic 6.Decision rule

37 7.Calculation of test statistic: The results of the arithmetic calculations will be summarized in a table called the analysis of variance (ANOVA) table. The entries in the table make it easy to evaluate the results of the analysis. 8.Statistical decision 9.Conclusion 10.Determination of p value

38 ONE-WAY ANOVA- Completely Randomized Design (CRD) One-way ANOVA: It is the simplest type of ANOVA, in which only one source of variation, or factor, is investigated. It is an extension to three or more samples of the t test procedure for use with two independent samples In another way t test for use with two independent samples is a special case of oneway analysis of variance.

39 Experimental design used for one-way ANOVA is called Completely randomised design. This tests the effect of equality of several treatments of one assignable cause of variation. Based on two principles- Replication and randomization. Advantages: Very simple: Reduces the experimental error to a great extent. We can reduce or increase some treatments. Suitable for laboratory experiments. Disadvantages: Design is not suitable if the experimental units are not homogeneous. Design is not so much efficient and sensitive as compared to others. Local control is completely neglected.

40 Hypothesis Testing Steps: 1. Description of data: The measurements( or observation) resulting from a completely randomized experimental design, along with the means and totals. Available Subjects Random numbers

41 Table of Sample Values for the CRD Treatment 1 3 K x 11 x 1 x 13 x 1k x 1 x x 3. X k.... x x n1 x 1 n n3 3 x nkk Total T.1 T. T.3 T.k T.. Mean x.1 x. x.3 x.k x..

42 Table of Sample Values for the Randomized Complete Block Design Treatments Blocks 1 3 k Total Mean 1 x 11 x 1 x x 1k T 1. x 1. x 1 x x 3 x k T. x.... n x n1 x n x n3. x nk Tn. X n. Total T.1 T. T.3 T.k T.. Mean x.1 x. x.3 x.k x..

43 x ij = the i th observation resulting from the j th treatment (there are a total of k treatment) T.j = x ij = total of the j th treatment x.j = T.j/nj = mean of jth treatment T.. = T.j = x ij = total of all observations x.. = T../N, N = n j

44 . Assumption: The Model The one-way analysis of variance may be written as follows: x ij = m j e ij ; i=1, n j, j= 1,.k The terms in this model are defined as follows: 1. m represents the mean of all the k population means and is called the grand mean.. j represents the difference between the mean of the j th population and the grand mean and is called the treatment effect. 3. e ij represents the amount by which an individual measurement differs from the mean of the population to which it belongs and is called the error term.

45 Assumptions of the Model The k sets of observed data constitute k independent random samples from the respective populations. Each of the populations from which the samples come is normally distributed with mean m j and variance j. Each of the populations has the same variance. That is 1 = = k =, the common variance. The j are unknown constants and j = 0, since the sum of all deviations of the m j from their mean, m, is zero. The (errors) e ij have a mean of 0, since the mean of x ij is m j The e ij have a variance equal to the variance of the x ij, since the e ij and x ij differ only by a constant. The e ij are normally (and independently) distributed.

46 3. Hypothesis: We test the null hypothesis that all population or treatment means are equal against the alternative that the members of at least one pair are not equal. We may state the hypothesis as follows H 0 : µ 1 = µ =..= µ k H A : not all µ j are equal If the population means are equal, each treatment effect is equal to zero, so that alternatively, the hypothesis may be stated as H 0 : τ j = 0, j=1,,.,k H A : not all τ j =0

47 4. Test statistic: Table: Analysis of Variance Table for the Completely Randomized Design Source of variation Among sample Within samples Sum of square d.f Mean square Variance ratio SSA k n j1 ). j.. k-1 MSA=SSA/(k-1) N-k Total N-1 k n j SST ( x ij x..) j1 i1 j ( k n j SSW j1 i1 x ( x ij x x. j) MS due to Treatment MSW=SSW/(N-k) MS due to error V.R=MSA/ MSW=F The Total Sum of squares(tss): It is the sum of the squares of the deviations of individual observations taken together.

48 The Within Groups of Sum of Squares: The first step in the computation call for performing some calculations within each group. These calculation involve computing within each group the sum of squared deviations of the individual observations from their mean. When these calculations have been performed within each group, we obtain the sum of the individual group results. The Among Groups Sum of Squares: To obtain the second component of the total sum of square, we compute for each group the squared deviation of the group mean from the grand mean and multiply the result by the size of the group. Finally we add these results over all groups. Total sum of square is equal to the sum of the among and the within sum of square. TSS=SSA+SSW

49 The First Estimate of σ : Within any sample n j j1 ( x n ij j 1. j) Provides an unbiased estimate of the true variance of the population from which the sample came. Under the assumption that the population variances are all equal, we may pool the k estimate to obtain k n j j1 i 1 k j1 ( x ( ij n j x x 1). j)

50 The Second Estimate of σ : The second estimate of σ may be obtain from the familiar formula for the variance of sample means,. If we solve this equation for σ x n, the variance of the population from which the samples were drawn, we have An unbiased estimate of provided by k j1 ( x k x 1 n x, computed from sample data, is. jx.. ) If we substitute this quantity into equation we obtain the desired estimate of σ n k j1 ( x k. jx.. ) 1

51 When the sample sizes are not all equal, an estimate of σ based on the variability among sample means is provided by The Variance Ratio: k j1 n j ( x k. jx.. ) 1 What we need to do now is to compare these two estimates of σ, and we do this by computing the following variance ratio, which is the desired test statistic: V.R = Among groups mean square Within groups mean square

52 6. Distribution of Test statistic: F distribution we use in a given situation depends on the number of degrees of freedom associated with the sample variance in the numerator and the number of degrees of freedom associated with the sample variance in the denominator. we compute V.R. in situations of this type by placing the among groups mean square in the numerator and the within groups mean square in the denominator, so that the numerator degrees of freedom is equal to the number of groups minus 1, (k-1), and the denominator degrees of freedom value is equal to k ( n 1) n j j1 j1 k j k N k

53 7. Significance Level: Once the appropriate F distribution has been determined, the size of the observed V.R. that will cause rejection of the hypothesis of equal population variances depends on the significance level chosen. The significance level chosen determines the critical value of F, the value that separates the nonrejection region from the rejection region. 8. Statistical decision: To reach a decision we must compare our computed V.R. with the critical value of F, which we obtain by entering Table G with k-1 numerator degrees of freedom and N-k denominator degrees of freedom. If the computed V.R. is equal to or greater than the critical value of F, we reject the null hypothesis. If the computed value of V.R. is smaller than the critical value of F, we do not reject the null hypothesis.

54 9. Conclusion: When we reject H 0 we conclude that not all population means are equal. When we fail to reject H 0, we conclude that the population means may be equal. 10. Determination of p value

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