Exam 16: Analysis of Variance

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Fill in the blanks (identified by asterisks) in the following partial ANOVA table. Source of Variation SS df MS F Treatments * * 79.95 * Error 17.0 * * Total 176.9 10

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Automobile insurance appraisers examine cars 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 cars that have recently been damaged in accidents. Three appraisers then estimate the repair costs of all six cars. From the data shown below, can we infer at the 5% significance level that the executive's claim is true? Estimated repair cost Car Appraiser 1 Appraiser 2 Appraiser 3 1 650 600 750 2 930 910 1010 3 440 450 500 4 750 710 810 5 1190 1050 1250 6 1560 1270 1450

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A randomised 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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One-way ANOVA is applied to independent samples taken from four normally distributed populations with equal variances. If the null hypothesis is rejected, then we can infer that: A. all popul ation means are equal B. all population means differ. C. at least two population means are equal. D. at least two population means differ.

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The sum of squares for treatments (SST) is the variation attributed to the differences between the treatment means, while the sum of squares for error (SSE) measures the variation within the samples.

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In one-way analysis of variance, within-treatments variation stands for the: A. sum of squares for error. B. sum of squares for treatments. C. total sum of squares. D. None of these choices are correct.

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Statistics practitioners use the analysis of variance (ANOVA) technique to compare two or more populations of interval data.

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The strength of a weld depends to some extent on the metal alloy used in the welding process. A scientist working in the research laboratory of a major automobile manufacturer has developed three new alloys. In order to test their strengths, each alloy is used in several welds. The strengths of the welds are then measured, with the results shown below. Can the scientist conclude at the 5% significance level that differences exist among the strengths of the welds with the different alloys? Strenghh of welds Alloy 1 Alloy 2 Alloy 3 15 17 25 23 21 27 16 19 24 29 25 31 28 23 19

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In one-way ANOVA, suppose that there are five treatments with n1=n2=n3=5n _ { 1 } = n _ { 2 } = n _ { 3 } = 5 and n4=n5=7n _ { 4 } = n _ { 5 } = 7 . Then the mean square for error, MSE, equals: A. /4. B. /29. C. /24 D. /5.

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When the effect of a level for one factor depends on which level of another factor is present, the most appropriate ANOVA design to use in this situation is the: A. one-way ANOVA. B. two-way ANOVA. C. randomised block design. D. matched pairs design.

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Consider the following partial ANOVA table: Source of Variation SS df MS F Treatments 75 * 25 6.67 Error 60 * 3.75 Total 135 19 The numbers of degrees of freedom for numerator and denominator, respectively, (identified by asterisks) are: A. 4 and 15. B. 3 and 16. C. 15 and 4. D. 16 and 3.

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Three tennis players, one a beginner, one experienced and one a professional, have been randomly selected from the membership of a large city tennis club. Using the same ball, each person hits four serves with each of five racquet models, with the five racquet models selected randomly. Each serve is clocked with a radar gun and the result recorded. Among ANOVA models, this setup is most like the single-factor analysis of variance: independent samples.

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In the randomised block design ANOVA, the sum of squares for error equals: A. SS(Total) - SST. B. SS(Total) - SSB C. SS(Total) - SST - SSB. D. SS(Total) - SS(A) - SS(B) - SS(AB)

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In one-way ANOVA, the term xˉ\overline { \bar { x } } refers to the: A. sum of the sample means. B. sum of the sample means divided by the total number of observations. C. sum of the population means. D. weighted mean of the sample means.

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Which of the following is a required condition for one-way ANOVA? A. The populations must all be normally distributed. B. The population variances must be equal. C. The samples for each treatment must be selected randomly and independently. D. All of these choices are correct.

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When the data are obtained through a controlled experiment in the single-factor ANOVA, we call the experimental design the completely randomised design of the analysis of variance.

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Given the significance level 0.025, the F-value for the numbers of degrees of freedom d.f. = (4, 8) is 5.05.

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A federal politician wants to investigate the difference in average weekly losses of workers due to accidents between South Australia (SA) and Western Australia (WA). The following table shows the average weekly losses of worker hours due to accidents in 2012 at five randomly selected manufacturing firms in each of SA and WA. SA WA 35 39 73 83 55 34 125 98 33 42 Assume that the weekly losses of worker hours are normally distributed. Is there a significant difference between the average weekly losses of workers hours due to accidents between SA and WA? Test at the 5% significance level. (Hint: Perform an F-test for one-way ANOVA to determine whether the population means differ.)

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The F-test of the analysis of variance requires that the populations be normally distributed with equal variances.

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If we want to conduct a test to determine whether a population mean is greater than another population mean, we: A. can use the analysis of variance. B. must use the independent-samples t -test for the difference between two means. C. must use the chi-squared test. D. can use the analysis of variance or can use the independent-samples t -test for the difference between two means.

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