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For a one-way ANOVA,

a. identify one purpose of one-way ANOVA tables.

b. construct a generic one-way ANOVA table.

Short Answer

Expert verified

(a) Simply put, we utilize a one-way ANOVA to organize and summarize the values required for doing a one-way ANOVA.

(b) The general form of the One-way ANOVA table is

SourcedfSSMS=SS/dfF-statisticTreatmentk-1SSTRMSTR=SSTRk-1F=MSTRMSEErrorn-kSSEMSE=SSEn-kTotaln-1SST

Step by step solution

01

Part(a) Step 1: Definition

The statistical approach of analysis of variance, or ANOVA, divides observed variance data into multiple components for use in additional tests. For three or more groups of data, a one-way ANOVA is used to learn more about the relationship between the dependent and independent variables.

02

Part(a) Step 2: Explanation

The error mean square is the variance within the sample means. That is, the (n-k)divides the error sum squares.

MSE is calculated using the following formula:

MSE=SSEn-k

The squared difference between the measured dependent variable and its mean is S S T (Sum of Squares Total).

The sum of the discrepancies between the predicted value and the dependent variable's mean is called S S R (Sum of Squares due to Regression).

The discrepancy between the actual and anticipated values is called S S E (Sum of Squares Error).

The F-statistic is the ratio of deviation between sample means and variation within the sample. That is, the Error Mean square is divided by the Treatment Mean square.

F-statistic is calculated using the formula:-

F=MSTRMSE

Two mean squares are formed, one for treatments and the other for an error, using the sums of squares S S T R and S S E previously computed for the one-way ANOVA. M S T R and M S E are the abbreviations for these mean squares. An ANOVA table is a common format for displaying these results. The statistics used to test hypotheses about population means are also shown in the ANOVA table.

03

Part(b) Step 1: Definition

The statistical approach of analysis of variance, or ANOVA, divides observed variance data into multiple components for use in additional tests. For three or more groups of data, a one-way ANOVA is used to learn more about the relationship between the dependent and independent variables.

04

Part(b) Step 2: Explanation

The ANOVA table demonstrates how the sum of squares, and thus the mean sum of squares, is distributed per source of variation.

The general form of the table is

SourcedfSSMS=SS/dfF-statisticTreatmentk-1SSTRMSTR=SSTRk-1F=MSTRMSEErrorn-kSSEMSE=SSEn-kTotaln-1SST

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Most popular questions from this chapter

a. Obtain individual normal probability plots and the standard deviations of the samples.

b. Perform a residual analysis.

c. Use your results from parts (a) and (b) to decide whether conducting a one-way ANOVA test and the data is reasonable. If so, also do parts d-f.

d. Use a one-way ANOVA test to decide, at the 5%significance level, whether the data provide sufficient evidence to conclude that a difference exists among the means of the populations from which the samples were taken.

e. Interpret your results from part (d).

Weight Loss and BMI. In the paper "Voluntary Weight Reduction in Older Men Increases Hip Bone Loss: The Osteoporotic Fractures in Men Study" (Journal of Clinical Endocrinology & Metabolism, Vol. 90, Issue 4. Pp. 1998-2004), K. Ensrud et al. reported on the effect of voluntary weight reduction on hip bone loss in older men. In the study, 1342 older men participated in two physical examinations an average of 1.8years apart. After the second exam, they were categorized into three groups according to their change in weight between exams: weight loss of more than 5%, weight gain of more than , and stable weight (between 5%loss and5% gain). For purposes of the hip bone density study, other characteristics were compared, one such being body mass index (BMI). On the Weissstats site, we provide the BMI data for the three groups, based on the results obtained by the researchers.

SAMPLE 1SAMPLE 2SAMPLE 3
1104
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2

In Exercise \(13.42-13.47\) we provide data from independent simple random samples from several populations. In each case,

a. compute SST, SSTR and SSE by using the computing formulas given in Formula \(13.1\) on page \(535\).

b. compare your results in part (a) for SSTR and SSE with those you obtained in Exercises \(13.24-13.29\) where you employed the defining formulas.

c. construct a one-way ANOVA table.

d. decide at the \(5%\) significance level, whether the data provide sufficient evidence to conclude that the means of the populations from which the samples were drawn are not all the same.

many species of cuckoos are brood parasites. The females lay their eggs in the nests if similar bird species that then raise the young cuckoos at the expense of their own young. The question might be asked, "Do the cuckoos lay the same size eggs regardless of the size of the bird whose nest they use"? Data on the lengths, in millimeters of cuckoo eggs found in the nests of six bird species- Meadow Pipit, Tree pipit, Hedge Sparrow, Robin, Pied Wagtail and wren are provided on the WeissStats site. These data are collected by the late O. Latter in \(1902\) and used by L. Tippett in his text The Methods of Statistics.

a. Obtain individual normal probability plots and the standard deviations of the sample.

b. Perform a residual analysis

c. use your results from part (a) and (b) to decide whether conducting a one-way ANOVA test on the data is reasonable. If so, also do parts (d) and (e).

d. use a one-way ANOVA test to decide, at the\(5%\) significance level whether, the data provide sufficient evidence to conclude that a difference exists among the means of the populations from which the samples were taken.

e. Interpret your results from part (d).

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SAMPLE 2
SAMPLE 13
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