ANLY500-FINAL-Question-3-_Anova.docx - ANLY500 FINAL...

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ANLY500 FINAL Question 3 - ANOVA Ruimin Wang Last Updated: 2020-08-11 #Capture an ANOVA model on the iris dataset. Set the dependent variable to 'Species'. Capture a summary. data = read.csv ( "iris_exams.csv" ) library (ez) ## Warning: package 'ez' was built under R version 4.0.2 ## Registered S3 methods overwritten by 'lme4': ## method from ## cooks.distance.influence.merMod car ## influence.merMod car ## dfbeta.influence.merMod car ## dfbetas.influence.merMod car data $ partno = 1 : nrow (data) options ( scipen = 999 ) ezANOVA ( data = data, dv = Petal.Width, between = Species, wid = partno, type = 3 , detailed = T) ## Warning: Converting "partno" to factor for ANOVA. ## Warning: Converting "Species" to factor for ANOVA. ## Coefficient covariances computed by hccm() ## $ANOVA ## Effect DFn DFd SSn SSd F ## 1 (Intercept) 1 297 430.7761 11.7455 10892.722 ## 2 Species 2 297 168.0546 11.7455 2124.737 ## p ## 1 0.000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000004223412 ## 2 0.000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000109555385349935824 2844783246820233166661776680306429819619489934933
## p<.05 ges ## 1 * 0.9734578 ## 2 * 0.9346747 ## ## $`Levene's Test for Homogeneity of Variance` ## DFn DFd SSn SSd F p p<.05 ## 1 2 297 0.9561276 3.982661 35.65078 0.00000000000001327717 * oneway.test (Petal.Width ~ Species, data = data) ## ## One-way analysis of means (not assuming equal variances) ## ## data: Petal.Width and Species ## F = 2908.6, num df = 2.00, denom df = 166.29, p-value < ## 0.00000000000000022 library (MOTE) omega.F ( dfm = 2 , dfe = 297 , Fvalue = 2124.737 , n = 300 , a = .05 ) ## $omega ## [1] 0.9340293 ## ## $omegalow ## [1] 0.9173461 ## ## $omegahigh ## [1] 0.946744 ## ## $dfm ## [1] 2 ## ## $dfe ## [1] 297 ## ## $F ## [1] 2124.737 ## ## $p ## [1] 0.000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000001095565 ## ## $estimate
## [1] "$\\omega^2$ = 0.93, 95\\% CI [0.92, 0.95]" ## ## $statistic ## [1] "$F$(2, 297) = 2124.74, $p$ < .001" #Provide an interpretation of the results in your own words. Support your response with results captured from running ANOVA.

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