Exam 14: Analysis of Variance

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Number of observation for particular combination of treatments is called a(n)________________.

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In a two-factor ANOVA,there are 4 levels for factor A,5 levels for factor B,and 3 observations for each combination of factor A and factor B levels.The total number of observations in this experiment equals:

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Motorcycle Repair Cost Motorcycle insurance appraisers examine motorcycles that have been involved in accidental collisions and estimate the cost of repairs.An insurance executive claims that there are significant differences in the estimates from different appraisers.To support his claim he takes a random sample of six motorcycles that have recently been damaged in accidents.Three appraisers then estimate the repair costs of all six motorcycles.The data are shown below.  Motorcycle Repair Cost  Motorcycle insurance appraisers examine motorcycles that have been involved in accidental collisions and estimate the cost of repairs.An insurance executive claims that there are significant differences in the estimates from different appraisers.To support his claim he takes a random sample of six motorcycles that have recently been damaged in accidents.Three appraisers then estimate the repair costs of all six motorcycles.The data are shown below.   ​ ​ -{Motorcycle Repair Cost Narrative} Can we infer at the 5% significance level that the executive's claim is true? ​ ​ -{Motorcycle Repair Cost Narrative} Can we infer at the 5% significance level that the executive's claim is true?

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H0: μ1 = μ2 = μ3 vs.H1: At least two means differ Conclusion: Reject the null hypothesis.Yes,the insurance executive's claim is true,according to this data.

We can use the F-test to determine whether μ1 is greater than μ2.

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The sum of squares for treatments,SST,achieves its smallest value (zero)when all the sample means are equal.

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In one-way ANOVA,the test statistic is defined as the ratio of the mean square for error (MSE)and the mean square for treatments (MST),namely,F = MSE / MST.

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A randomized block design with 4 treatments and 5 blocks produced the following sum of squares values: SS(Total)= 2000,SST = 400,SSE = 200.The value of MSB must be 350.

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In the two-factor ANOVA where a is the number of factor A levels,b is the number of factor B levels,and r is the number of replicates,the number of degrees of freedom for interaction is:

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GMAT Scores A recent college graduate is in the process of deciding which one of three graduate schools he should apply to.He decides to judge the quality of the schools on the basis of the Graduate Management Admission Test (GMAT)scores of those who are accepted into the school.A random sample of six students in each school produced the following GMAT scores.Assume that the data are normally distributed with equal variances for each school. GMAT Scores  A recent college graduate is in the process of deciding which one of three graduate schools he should apply to.He decides to judge the quality of the schools on the basis of the Graduate Management Admission Test (GMAT)scores of those who are accepted into the school.A random sample of six students in each school produced the following GMAT scores.Assume that the data are normally distributed with equal variances for each school.   ​ ​ -{GMAT Scores Narrative} Can he infer at the 10% significance level that the GMAT scores differ among the three schools? ​ ​ -{GMAT Scores Narrative} Can he infer at the 10% significance level that the GMAT scores differ among the three schools?

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In a one-way ANOVA,error variability is computed as the sum of the squared errors,SSE,for all values of the response variable.This variability is the:

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Number of observation for a particular combination of treatments is called a replicate.

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The distribution of the test statistic for analysis of variance is the:

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LSAT Scores A recent college graduate is in the process of deciding which one of three Law schools he should apply to.He decides to judge the quality of the schools on the basis of the Law School Admission Test (LSAT)scores of those who are accepted into the school.A random sample of six students in each school produced the following LSAT scores.An F-test using ANOVA concluded that average LSAT scores differ for at least two of the schools. LSAT Scores  A recent college graduate is in the process of deciding which one of three Law schools he should apply to.He decides to judge the quality of the schools on the basis of the Law School Admission Test (LSAT)scores of those who are accepted into the school.A random sample of six students in each school produced the following LSAT scores.An F-test using ANOVA concluded that average LSAT scores differ for at least two of the schools.   ​ ​ -{LSAT Scores Narrative} Use Tukey's method with α = 0.05 to determine which population means differ. ​ ​ -{LSAT Scores Narrative} Use Tukey's method with α = 0.05 to determine which population means differ.

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Which of the following components in an ANOVA table is not additive?

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The alternative hypothesis of ANOVA is that ____________________ population means are different.

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In one-way ANOVA,the amount of total variation that is unexplained is measured by the:

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The test statistic in the analysis of variance (ANOVA)technique follows a normal distribution.

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In ANOVA the populations are classified according to one or more criterion,called ____________________.

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In a two-factor ANOVA,always test for ____________________ first.

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TV News Viewing Habits A statistician employed by a television rating service wanted to determine if there were differences in television viewing habits among three different cities in New York.She took a random sample of five adults in each of the cities and asked each to report the number of hours spent watching television in the previous week.The results are shown below. (Assume normal distributions with equal variances.) TV News Viewing Habits  A statistician employed by a television rating service wanted to determine if there were differences in television viewing habits among three different cities in New York.She took a random sample of five adults in each of the cities and asked each to report the number of hours spent watching television in the previous week.The results are shown below. (Assume normal distributions with equal variances.)   ​ ​ -{TV News Viewing Habits Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value. ​ ​ -{TV News Viewing Habits Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.

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