Exam 10: Experimental Design and Analysis of Variance

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At the 5% level of significance,is there a difference in the average starting salaries among the four disciplines?

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Based on the results of a two-factor factorial experiment,the ANOVA table showed that SSE = 5.5.If we ignore one of the factors and perform a one-way ANOVA with the remaining factor using the same data,the SSE will always be larger than 5.5.

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ANOVA table Source SS df MS F p -value Treatment 6.000 3 1.9998 18.85 3.4E-05 Error 1.486 14 0.1061 Total 7.485 17 Post hoc analysis Tukey simultaneous comparison t-values ( ( d.f. =14) =14) Brand 3 nbsp; Brand 2 nbsp; Brand 4 nbsp; Brand 1 1.402.282.582.95 Brand 3 1.40 Brand 2 2.28 Brand 4 2.58 Brand 1 2.95 4.27 5.38 1.35 7.09 3.07 1.63 The Excel/Mega-Stat output given above summarizes the results of a one-way analysis of variance in an attempt to compare the performance characteristics of four brands of vacuum cleaners. The response variable is the amount of time it takes to clean a specific size room with a specific amount of dirt. -At a significance level of 0.05,the null hypothesis for the ANOVA F test is rejected.Analysis of the Tukey's simultaneous confidence intervals shows that at the significance level (experimentwise)of 0.05 we would conclude that,in regards to average time,

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A two-way ANOVA for a two-factor factorial design has 4 levels of factor 1, 3 levels of factor 2, and 5 observations per cell. -In testing the significance of interaction,the numerator and denominator degrees of freedom for the ANOVA F test are:

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Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 groups (treatments). Source Sum of Squares Treatments 15.93 Blocks 42.09 Error 23.84 Total 81.86 -What is the block mean square?

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A _____ design,is an experimental design that compares p groups by using b blocks,where each block is used exactly once to measure the effect of each treatment.

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ANOVA table Source SS df MS F p -value Treatment 6.000 3 1.9998 18.85 3.4E-05 Error 1.486 14 0.1061 Total 7.485 17 Post hoc analysis Tukey simultaneous comparison t-values ( ( d.f. =14) =14) Brand 3 nbsp; Brand 2 nbsp; Brand 4 nbsp; Brand 1 1.402.282.582.95 Brand 3 1.40 Brand 2 2.28 Brand 4 2.58 Brand 1 2.95 4.27 5.38 1.35 7.09 3.07 1.63 The Excel/Mega-Stat output given above summarizes the results of a one-way analysis of variance in an attempt to compare the performance characteristics of four brands of vacuum cleaners. The response variable is the amount of time it takes to clean a specific size room with a specific amount of dirt. -At a significance level of 0.05,we would:

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When using a one -way ANOVA to analyze data from a completely randomized design,the calculated F statistic will decrease as _____.

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A one-way analysis of variance is a method that allows us to estimate and compare the effects of several treatments on a response variable.

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A machine manufacturer conducted a study to compare the performance of two types of similar machines A and B in terms of the hardness of the product manufactured.According to the production supervisor,the machine setting (high,medium,low)also affects the hardness of the product.The hardness readings (response variable)are given in a two-factor factorial design format with two observations per cell. Machine type Machine Setting A B Low 8 4 8 4 Medium 7 5 6 2 High 9 4 10 5 You are also given the following additional information: SS (machine type)= 48,SS (machine setting)= 8,SST = 64 -Complete the analysis of variance table and test to determine whether the means for factor B differ at a = .10.Interpret the results.

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Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 groups (treatments). Source SS Treatments 2,477.53 Blocks 3,180.48 Error 11,661.38 Total -Determine the degrees of freedom for error.

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Is there a significant difference in the annual sales of the five company categories?

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Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 groups (treatments). Source Sum of Squares Treatments 15.93 Blocks 42.09 Error 23.84 Total 81.86 -Determine the degrees of freedom for groups (treatments).

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In one-way ANOVA,a large value of F results when the within-treatment variability is large compared to the between-treatment variability.

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A company that fills containers of water has four machines.The quality control manager needs to determine whether the average fill for these machines is the same.Looking at 19 containers,we have the following data of fill measures in litres: Machine 1 Machine 2 Machine 3 Machine 4 4 6 5 4 4.03 4.0017 3.974 4.005 0.0183 0.0117 0.0182 0.0129 and the following partial ANOVA table (with "group" labeled as "treatment"): Source SS d.f. MS F Treatments 0.002359 Errors Total 0.010579 -Complete the ANOVA table and calculate F.

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