Exam 12: Experimental Design and Analysis of Variance

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In a completely randomized ANOVA,other things equal,as the sample means get closer to each other,the probability of rejecting the null hypothesis:

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A company that fills gallon 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 one-gallon containers,we have the following data of fill measures (x)in quarts. F. F = 10.10 Source SS DF MS F Treatments .007077 3 0.002359 10.10 Errors .003502 15 .0002335 Total 0.010579 18

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 Machine 1  Machine 2  Machine 3  Machine 4 N4654xˉ4.034.00173.9744.005S0.01830.01170.01820.0129\begin{array} { c l l l l } & \text { Machine 1 } & \text { Machine 2 } & \text { Machine 3 } & \text { Machine 4 } \\N & 4 & 6 & 5 & 4 \\\bar { x } & 4.03 & 4.0017 & 3.974 & 4.005 \\S & 0.0183 & 0.0117 & 0.0182 & 0.0129\end{array}
And the following partial ANOVA table.  Source  SS  DF  MS  F  Treatments 0.002359 Errors  Total 0.010579\begin{array} { l c c c c } \text { Source } & \text { SS } & \text { DF } & \text { MS } & \text { F } \\\text { Treatments } & & & 0.002359 \\\text { Errors } & & & \\\text { Total } & 0.010579 & &\end{array}
Complete the ANOVA table and calculate F.
F = 10.10  Source  SS  DF  MS  F  Treatments .00707730.00235910.10 Errors .00350215.0002335 Total 0.01057918\begin{array} { l c c c c } \text { Source } & \text { SS } & \text { DF } & \text { MS } & \text { F } \\\text { Treatments } & .007077 & 3 & 0.002359 & 10.10 \\\text { Errors } & .003502 & 15 & .0002335 & \\\text { Total } & 0.010579 & 18 & &\end{array}
Feedback:DF treatment = 4 machines - 1 = 3
DF total = n - 1 = 19 - 1 = 18
DF error = DF total - DF treatment = 18 - 3 = 15
SS treatment = MS treatment × DF treatment = .002359 × 3 = .007077
SS error = SS total - SS treatment = .010579 - .007077 = .003502
MS error = SS error/DF error = .003502/15 = .0002335
F = MS treatment/MS error = .002359/.0002335 = 10.10

Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.

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 Source  Sum of Squares  Treatments 15.93 Blocks 42.09 Error 23.84 Total 81.86\begin{array} { l c } \text { Source } & \text { Sum of Squares } \\\text { Treatments } & 15.93 \\\text { Blocks } & 42.09 \\\text { Error } & 23.84 \\\text { Total } & 81.86\end{array}
What is the mean square error?
1.59
Feedback: 23.84(3)(5)=1.59\frac { 23.84 } { ( 3 ) ( 5 ) } = 1.59

Find Tukey's simultaneous 95 percent confidence interval for μC - μB,where

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What is the degrees of freedom treatment (between-group variation)of a completely randomized design (one-way)ANOVA test with 4 groups and 15 observations per each group?

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Consider the one-way ANOVA table.

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The dependent variable,the variable of interest in an experiment,is also called the ___________ variable.

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When computing confidence intervals using Tukey's procedure,for all possible pairwise comparison of means,the experimentwise error rate will be:

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Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments.

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Suppose you are a researcher investigating the annual sales differences among five categories of businesses.The study looks at a total of 55 companies equally divided among categories groups A,B,C,D,and E. Source SS df MS F Treatment 583.39 4 145.85 7.501 Error 972.18 50 19.4436 Total 1555.57 54

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In performing a one-way ANOVA,the _________ is the between-group variance.

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A sum of squares that measures the variability among the sample means is referred to as the:

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Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.

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The variable of interest in an experiment is referred to as the __________ variable.

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After rejecting the null hypothesis of equal treatments,a researcher decided to compute a 95 percent confidence interval for the difference between the mean of treatment 1 and mean of treatment 2 based on Tukey's procedure.At α = .05,if the confidence interval includes the value of zero,then we can reject the hypothesis that the two population means are equal.

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___________ refers to applying a treatment to more than one experimental unit.

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Consider the following calculations for a one-way analysis of variance from a completely randomized design with 20 total observations.The response variable is sales in millions of dollars and four treatment levels represent the four regions that the company serves. MSE = 101.25

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The ANOVA procedure for a two-factor factorial experiment partitions the total sum of squares into three components,SS first factor,SS second factor,and SSE.

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Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments.

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In a one way ANOVA table,the ___________ the value of MSE,the higher the probability of rejecting the hypothesis that all treatment means are equal.

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