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Unformatted text preview: STAT 350 Assignment 10 Solutions (50 points) 35. The following ANOVA table was created: Source DF SS MS F P brand 4 53231 13308 95.57 0.000 level 3 116218 38739 278.20 0.000 Error 12 1671 139 Total 19 171120 (a) (3) The Fratio for ' brand ' is F = 95.57. For df 1 = 4 and df 2 = 12, the value F = 95.57 has a Pvalue smaller than .001 (From Table VIII). Since this Pvalue is smaller than α = .01, we can conclude that there is a difference in power consumption among the 5 brands (b) (4) The Fratio for ' humidity ' is F = 278.20. For df 1 = 3 and df 2 = 12, the Pvalue associated with F = 278.20 is less than .001 (from Table VIII), so the null hypothesis H : average humidity is the same is rejected. We conclude that humidity levels do affect power consumption, so it was wise to use humidity as a blocking factor. 38. The following output was created Source df SS MS F pvalue Wood species 2 2216 1108 9.61 0.030 Wood grade 2 90755 45377 393.45 0.000 Error 4 461 115 Total 8 93432 (a) (3) The null hypothesis of interest is H : There are no differences between the mean bending parameters of the three species of wood. The Fratio for “wood species” is F = 9.61. For df 1 = 2 and df 2 = 4, the pvalue reported by Minitab is 0.030. Since the pvalue is less than α = .05, we reject H . We have sufficient evidence to claim that there are differences between the mean bending parameters of the three species of wood....
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This note was uploaded on 09/21/2011 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue.
 Spring '08
 Staff
 Statistics

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