courses.aplia.com recorded in seconds. The results of the study are presented in

Question

courses.aplia.com recorded in seconds. The results of the study are presented in the following data table. All scores are times necessary to complete the paper-folding task, Noise Type Participantclassical MusicWhite Noise Nature Sounds Participant Totals 3.22 3.27 3.47 3.43 3.35 3.42 3.31 3.93 3.91 4.16 3.46 3.35 .69 3.74 3.82 P. 10.10 P= 9.93 P 11.09 P= 11.08 P-11.33 k=3 G 53.53 x2 192.1709 T=16.74 T=18.73 T=18.06 ss-0.0441 ss=0.5286 ss=0.1575 The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population means are equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for the missing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on the order in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between your answers and the answer choices.) ANOVA Table Source Between treatments Within treatments df MS 0.0609 0.1372 Between subjects Error Total 14 FDistribution

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu1 = mu2 = mu3 Vs H1 : Atleast one of the mean is differ than 0.

mu1, mu2 and mu3 are three population means.

Assume alpha = level of significance = 5% = 0.05

This we can done using EXCEL.

steps :

Enter all the data in EXCEL sheet --> Data --> Data Analysis --> Anova : Single Factor --> ok --> Input range : select all tha data range --> Grouped by : columns --> Alpha : 0.05 --> Output Range : select one empty cell --> ok

Output is :

Here test statistic F = 3.370261

P-value = 0.068

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : All the three means are equal.

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Here also we have to test the hypothesis that,

H0 : There are no differences among the aggressive bahaviour means across the three time points.

H1 : Atleast one aggressive bahaviour means is different from another.

The same procedure as a) part.

Output is :

Test statistic F = 8.87595

P-value = 0.0028

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion :   Atleast one aggressive bahaviour means is different from another.

Measure of effect size :

n2 = SSbetween / SStotal

= 2.7352 / 5.04645 = 0.5420

The percentage of variance accounted for by the treatement effect is 54.20%.

Q.6) The Scheffe's test can't be conducted.

For Q.7) The population from which the samples were taken are normally distibuted.

Anova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 5 16.74 3.348 0.01102 Column 2 5 18.73 3.746 0.13213 Column 3 5 18.06 3.612 0.03937 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 0.410093 2 0.205047 3.370261 0.068928 3.885294 Within Groups 0.73008 12 0.06084 Total 1.140173 14

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