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Must figure out which test to use one way nova, block design, factorial
2. Four levels of treatment are applied to six different supplies of lumber. The data is presented on worksheet B. We are interested in if there are differences in the treatments and the supplies of lumber are considered not to be a factor. Create the ANOVA table and test the hypothesis that all four treatments are the same or are not the same.
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Null hypothesis Ho 14. = μ,-,us-u-u-u Alternative hypothesis H1 : μ.-Ha # μ. # μ. # μ. # μ6. For Columns: Null hypothesis Ho :从,-从,-As = μ.4 Alternative hypothesis H1 14,1 #从2 #从3 #从4 Grand Total-7-ΣΣΧ.-2135 Raw sum of squares -190561 x2 = 190561 72 (2135) M 6x4 Correction Factor(CF)-_--189926.0417 Total Sum of Squares (TSS)-ΣΣΧ2-CF-190561-189926, 0417-634.9583 Row Sum of Squares (RSS)--Σ7-CF, where T,? be the square of row total -189926.0417 356.2083 Column Sum of Squares (CS.S) T-CF,where T be the square ofj column total (1140092)-189926 -189926.0417 89.7916 Error Sum of Squares (E.SS)-T.S.S- R.S.S-C.S.s 634.9583-356.2083-89.7916- 188.9584 ANOVA table Source Rows Columns 4-1- Error (6-1)(4-1)-15 188.9584 188.9572/9 12.597:2 Total DF MS 072 356.2083 356.2083/5-71.2417 5.6554F6132.90 89.791689.7916/3-29.9305 2.3760 Feus) 3.29 (5.15) 24-1- 23 634.9583 The value of the test stati stic corresponding to Rows is 5.6554 greater than the critical value 51,-2.90 at 5% level of significance. We reject the null hypothesis corresponding to rows There is a significant difference in the supplies of lumber The value of the test stati stic corresponding to Columns is 2.3760 less than the critical value 315,-3.29 at 5% level of significance. We fail to reject the null hypothesis corresponding to columns There is no significant difference in the treatments

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