topic8 - Statistics 512 Applied Linear Models Topic 8 Topic...

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Statistics 512: Applied Linear Models Topic 8 Topic Overview This topic will cover More on Multiple Comparisons / Confidence Intervals Two-way ANOVA with unequal numbers of observations in the cells Three-way ANOVA example Estimation of Factor Level Means (19.8) Still assume n constant across cells here. Point Estimates are ˆ μ i. = ¯ Y i.. μ .j = ¯ Y .j. ,andˆ μ i,j = ¯ Y i,j. These have associated variances (estimate by plugging in MSE): s 2 { ¯ Y i.. } = MSE/bn , s 2 { ¯ Y .j. } = MSE/an ,and s 2 { ¯ Y i,j. } = MSE/n . These may be used with t -critical values to form confidence intervals. The degrees of freedom are those associated with the MSE: ( n - 1) ab . It is not really appropriate to look at ˆ μ i. or ˆ μ .j when there is serious interaction. Computation Means can be obtained from proc means in SAS MSE for the model can be obtained from SAS as well Construct the CI using these values and the appropriate critical value. The critical value can be from the t -distribution. Or it may be Tukey, Bonferroni, or Scheffe adjusted as is appropriate. Contrasts We can look at contrasts of means (on the same factor) including multiple comparisons as we did in One-way ANOVA. When there is no interaction, for factor A the contrast L = c i μ i. is estimated by ˆ L = c i ¯ Y i.. . An unbiased estimator of the variance is s 2 { ˆ L } = bn c 2 i . Using a t -critical value (use error d.f.) we may construct a CI for the contrast. Factor B is analogous: the contrast L = c j μ .j is estimated by ˆ L = c j ¯ Y .j. .A n unbiased estimator of the variance is s 2 { ˆ L } = an c 2 j . 1
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If there is interaction , we may consider contrasts of the form ˆ L = c i,j ¯ Y i,j. . For these, s 2 { ˆ L } = MSE n c 2 i,j and CI’s may be obtained using an appropriate critical value Multiple Comparisons The multiple comparison procedures with no interaction are the same as for one-way ANOVA. Can use LSD, Tukey, Bonferroni, or Scheffe in SAS as appropriate in the means state- ment. Example Recall the bread sales example ( knnl860.sas ) Shelf height ( A ) has 3 levels Shelf width ( B ) has 2 levels There are 2 observations at each level (total 12 observations) Find a 95% CI for the mean sales using the middle wide shelf So we want a 95% CI for μ 2 , 2 . From SAS we have the means output and also =10 . 333. We also had ˆ μ 2 , 2 = ¯ Y 2 , 2 . = 69. There are ( n - 1) ab = 1(3)(2) = 6 degrees of freedom, and if this is the only interval of interest we may use the t -distribution so that the critical value is 2.447. Hence the CI is given by ˆ μ 2 , 2 ± 2 . 447 ± p MSE/n ² =69 ± 2 . 447 ± p 10 . 333 / 2 ² = (63 . 44 , 74 . 56). Find a 95% CI for the difference in sales between the middle shelf and the top shelf Here we are averaging across width, looking at the contrast μ 2 .
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This note was uploaded on 03/15/2012 for the course STAT 512 taught by Professor Staff during the Fall '08 term at Purdue.

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topic8 - Statistics 512 Applied Linear Models Topic 8 Topic...

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