Question

Why do we need time n when we calculate SSTR? In other words, why n times (mean of a factor level - mean of total)^2  Could you give me some intuitive explanation or proof?

Here are my lecture notes.

Store IDs Total Mean ni S1 S2 S3 S4 S5(Y) 73 67 78 136 14.6 13.4 19.5 27.2 D1 11 17 16 1415 PackageD2 121015 19 11 design D3 23 20 18 17 Miss D4 27 33 22 26 28 Total 354 Y 18.63 19 Table Data summary packaging of breakfast cereals 1.4 Comparison of factor level means Want to che(k for deviations from the null hypothesis H0 : /11-: hypothesis is Ha : not all μs a flr, i.e., the alternative re equal Idea 1: A baseline value for comparison is the overall mean: . Idea 2: Calculate deviations from the overall mean for each factor level: 2 Under H0 : μι flr, these deviations are all zero. . Idea 3: Use the weighted sum of the above deviations as an overall measurement of the deviation from H0 : μ1 = = μ㎡ The weight of thei-th treatment group is its sample size ni, i.e., the more data, the more importance

Estimators Estimate the population means by their sample counterparts an Thus, Ξ 14-. However, Σί i ni (Y; is a statistic to measure the deviation from Ho: μ! an unbiased estimator of Σ¡l ni(μί-μ)2. In fact Y 2 to that of σ2 to decide whether Nevertheless, we can compare the magnitude of Στι ni the deviation is large or not i 1.5 Decomposition of Total Sum of Squares Write Yi -Y: deviation of the response from the overall mean; » Yi- Y.. : deviation of the i-th factor level mean from the overall mean; . Yii-Yi : deviation of the response from the corresponding factor level mean (residual) Then the ANOVA decomposition of the sum of squares: r i This can be expressed as SST。= SSE + SST R _ Σ. ¡ Ση 1(Xy-Y.)2 where SST0-L: 1 Σ is the ETTor Sum of Squares and SSTRーム1 ni(Yi (Xy-Y.)2 is the Total Sum of Squares; SSE r Y. 2 is the Treatment Sum of Squares

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We start understanding this problem from starting but you can see from page 2 also...

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