Exam 12: Experimental Design and Analysis of Variance

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

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df error = 15
(4 − 1)(6 − 1) = 15

ANOVA table ANOVA table    Post hoc analysis Tukey simultaneous comparison t-values    The Excel/MegaStat 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. Use the information above and determine a Tukey simultaneous 95 percent confidence interval for μ<sub>1</sub> − μ<sub>2</sub>. The mean and sample sizes for brand 1 and brand 2 are as follows:   = 2.95,   = 2.28, n<sub>1</sub> = 4, and n<sub>2</sub> = 5. Post hoc analysis Tukey simultaneous comparison t-values ANOVA table    Post hoc analysis Tukey simultaneous comparison t-values    The Excel/MegaStat 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. Use the information above and determine a Tukey simultaneous 95 percent confidence interval for μ<sub>1</sub> − μ<sub>2</sub>. The mean and sample sizes for brand 1 and brand 2 are as follows:   = 2.95,   = 2.28, n<sub>1</sub> = 4, and n<sub>2</sub> = 5. The Excel/MegaStat 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. Use the information above and determine a Tukey simultaneous 95 percent confidence interval for μ1 − μ2. The mean and sample sizes for brand 1 and brand 2 are as follows: ANOVA table    Post hoc analysis Tukey simultaneous comparison t-values    The Excel/MegaStat 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. Use the information above and determine a Tukey simultaneous 95 percent confidence interval for μ<sub>1</sub> − μ<sub>2</sub>. The mean and sample sizes for brand 1 and brand 2 are as follows:   = 2.95,   = 2.28, n<sub>1</sub> = 4, and n<sub>2</sub> = 5. = 2.95, ANOVA table    Post hoc analysis Tukey simultaneous comparison t-values    The Excel/MegaStat 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. Use the information above and determine a Tukey simultaneous 95 percent confidence interval for μ<sub>1</sub> − μ<sub>2</sub>. The mean and sample sizes for brand 1 and brand 2 are as follows:   = 2.95,   = 2.28, n<sub>1</sub> = 4, and n<sub>2</sub> = 5. = 2.28, n1 = 4, and n2 = 5.

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[.035, 1.305]
[(2.95 − 2.28) ± (4.11/ [.035, 1.305] [(2.95 − 2.28) ± (4.11/   )(   )] = [.67 ± (2.9066)(.2185)] = [.67 ± .635] = [.035, 1.305] )( [.035, 1.305] [(2.95 − 2.28) ± (4.11/   )(   )] = [.67 ± (2.9066)(.2185)] = [.67 ± .635] = [.035, 1.305] )] = [.67 ± (2.9066)(.2185)] = [.67 ± .635] = [.035, 1.305]

Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments. Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments.    Test H<sub>0</sub>: there is no difference between treatment effects at α = .05. Test H0: there is no difference between treatment effects at α = .05.

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Since our calculated F value (1.91) is less than the critical F value (2.422) we cannot reject the null hypothesis; there is no evidence of difference between treatments. Since our calculated F value (1.91) is less than the critical F value (2.422) we cannot reject the null hypothesis; there is no evidence of difference between treatments.   <sub> </sub>=   = 1.91 F = 1.91 F<sub>.05,5,45</sub> = 2.422 1.91 < 2.422, so do not reject the null hypothesis. = Since our calculated F value (1.91) is less than the critical F value (2.422) we cannot reject the null hypothesis; there is no evidence of difference between treatments.   <sub> </sub>=   = 1.91 F = 1.91 F<sub>.05,5,45</sub> = 2.422 1.91 < 2.422, so do not reject the null hypothesis. = 1.91
F = 1.91
F.05,5,45 = 2.422
1.91 < 2.422, so do not reject the null hypothesis.

In one-way ANOVA, the numerator degrees of freedom equals the number of samples being compared.

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Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments. Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments.    What is the calculated F statistic for treatment? What is the calculated F statistic for treatment?

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In a one-way analysis of variance with three treatments, each with five measurements, in which a completely randomized design is used, what are the degrees of freedom for error?

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

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A company that fills one-gallon containers of water has four machines. The quality control manager needs to determine whether the average fill for these machines is the same. For a sample of 19 one-gallon containers, we have the following data of fill measures (x) in quarts. Machine 1 Machine 2 Machine 3 Machine 4 N 4 6 5 4 (3 - 2) ± 4.53 A company that fills one-gallon containers of water has four machines. The quality control manager needs to determine whether the average fill for these machines is the same. For a sample of 19 one-gallon containers, we have the following data of fill measures (x) in quarts. Machine 1 Machine 2 Machine 3 Machine 4 N 4 6 5 4 (3 - 2) ± 4.53   = (-2.377, 4.377) 4.03 4.0017 3.974 4.005 S 0.0183 0.0117 0.0182 0.0129 And the following partial ANOVA table.    Determine the degrees of freedom for the treatment, error, and total, and state the critical value of the F statistic at α = .05. = (-2.377, 4.377) 4.03 4.0017 3.974 4.005 S 0.0183 0.0117 0.0182 0.0129 And the following partial ANOVA table. A company that fills one-gallon containers of water has four machines. The quality control manager needs to determine whether the average fill for these machines is the same. For a sample of 19 one-gallon containers, we have the following data of fill measures (x) in quarts. Machine 1 Machine 2 Machine 3 Machine 4 N 4 6 5 4 (3 - 2) ± 4.53   = (-2.377, 4.377) 4.03 4.0017 3.974 4.005 S 0.0183 0.0117 0.0182 0.0129 And the following partial ANOVA table.    Determine the degrees of freedom for the treatment, error, and total, and state the critical value of the F statistic at α = .05. Determine the degrees of freedom for the treatment, error, and total, and state the critical value of the F statistic at α = .05.

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

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When we compute 100(1 − α) confidence intervals, the value of α is called the

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ANOVA table ANOVA table   Post hoc analysis Tukey simultaneous comparison t-values   The Excel/MegaStat 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 .05, the null hypothesis for the ANOVA F test is rejected. Analysis of the Tukey simultaneous confidence intervals shows that at the significance level (experimentwise) of .05, we would conclude that Post hoc analysis Tukey simultaneous comparison t-values ANOVA table   Post hoc analysis Tukey simultaneous comparison t-values   The Excel/MegaStat 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 .05, the null hypothesis for the ANOVA F test is rejected. Analysis of the Tukey simultaneous confidence intervals shows that at the significance level (experimentwise) of .05, we would conclude that The Excel/MegaStat 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 .05, the null hypothesis for the ANOVA F test is rejected. Analysis of the Tukey simultaneous confidence intervals shows that at the significance level (experimentwise) of .05, we would conclude that

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Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments. Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.    What is the treatment mean square? What is the treatment mean square?

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The ________ units are the entities (objects, people, etc.) to which the treatments are assigned.

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  Consider the randomized block design with 4 blocks and 3 treatments given above. What is the value of the F statistic for blocks? Consider the randomized block design with 4 blocks and 3 treatments given above. What is the value of the F statistic for blocks?

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Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments. Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments.    What is the treatment mean square? What is the treatment mean square?

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Looking at four different diets, a researcher randomly assigned 20 equally overweight individuals into each of the four diets. What are the degrees of freedom for the individual confidence intervals?

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Find a Tukey simultaneous 95 percent confidence interval for μ1 − μ2, where Find a Tukey simultaneous 95 percent confidence interval for μ<sub>1</sub> − μ<sub>2</sub>, where   = 33.98,   = 36.56, and MSE = .669. There were 15 observations total and 3 treatments. Assume that the number of observations in each treatment is equal. = 33.98, Find a Tukey simultaneous 95 percent confidence interval for μ<sub>1</sub> − μ<sub>2</sub>, where   = 33.98,   = 36.56, and MSE = .669. There were 15 observations total and 3 treatments. Assume that the number of observations in each treatment is equal. = 36.56, and MSE = .669. There were 15 observations total and 3 treatments. Assume that the number of observations in each treatment is equal.

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In one-way ANOVA, the total sum of squares is equal to ________.

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When computing individual confidence intervals using the t statistic, for all possible pairwise comparisons of means, the experimentwise error rate will be

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A vitamin-water manufacturer wants to compare the effects on sales of three water colors: green, blue, and red. Four regions are selected for the test, with the following ANOVA results. A vitamin-water manufacturer wants to compare the effects on sales of three water colors: green, blue, and red. Four regions are selected for the test, with the following ANOVA results.    Compute the mean square and F to test the null hypothesis that there is no interaction at α = .01. Compute the mean square and F to test the null hypothesis that there is no interaction at α = .01.

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