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Chapter 13: Q.13.40 (page 542)

Fill in the missing entries in the partially completed one-way ANOVA tables.

Short Answer

Expert verified

The completed ANOVA table is

sourcedfSSMS=SS/dfF=statisticTreatment437.849.461.464error20135.206.76total24173.04

Step by step solution

01

Given Information

Given data:

02

Explanation

Determine the total number of degrees of freedom. totaldf=errordf+treatmentdf

=20+4

=24

MSE (mean square error):

MSE=SSEn-k

SSE=MSE(n-k)

=6.76(20)

=135.2

03

Explanation

Calculate the treatment's mean squares:

MSTR=SSTRk-1

=37.844

⇒MSTR=9.46

k-1=3

Look for theF-static.

F-statistic=MSTRMSE

=9.466.76

=1.464

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Most popular questions from this chapter

Rock Sparrows breeding in northern Italy are the subject of a long-term ecology and conservation study due to their wide variety of breeding patterns. Both males and females have a yellow patch on their breasts that is thought to play a significant role in their sexual behavior. A Pilastro et al. conducted an experiment in which they increased or reduced the size of a female's breast patch by dying feathers at the edge of a patch and then observed several characteristics of the behavior the male. Their results were published in the paper "Male Rock Sparrow Adjust Their Breeding Strategy According to Female Ornamentation: Parental or Mating Investment? Eight mating pairs were observed in each of three groups: a reduced patch size group, a group control and an enlarged patch size group. The data on the WeissStats site, based on the results reported by the researchers, give the number of minutes per hour that makes sang in the vicinity of the nest after the patch size manipulation was done on the females.

Suppose that a one-way ANOVA is being performed to compare the means of three populations and that the sample sizes are 10, 12, and 15. Determine the degrees of freedom for the F-statistic.

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Using the Fa-notation, identify the F-value having area 0.975 to its left.
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