Anova Guide PDF

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Summary

Test 2 on ANOVA....

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ANOVA GUIDE Transfer data from table to excel

Make sure you have headings for your columns Input data DOWN the column Double check all your numbers to ensure you’ve typed them correctly and you haven’t missed any Open R >library(Rcmdr) Import Data on R Commander Data -> Import data -> from excel file…

Make sure that checkbox is ticked Check Summary Statistics Statistics -> Summaries -> Numerical summaries…

->Summarise by groups…

Check if they match!!

Oh no, they don’t match!! Go back and check your spreadsheet

FIX IT

and repeat the previous steps

^That’s better

(define your parameters and state your null and alternate hypothesis) Conduct ANOVA Statistics - > Means -> One-Way ANOVA Table

Copy out the table from above

p-value=0.000326 alpha = 0.05 p-value < alpha at the 5% level of significance, there is sufficient evidence to reject the null hypothesis. Therefore, there is significant evidence to conclude that not all the mean satisfactions of the different cars are equal Tukey (Conduct when the null hypothesis is rejected)

Statistics - > Means -> One-Way ANOVA Table TIME

MAKE SURE THAT CHECKBOX IS TICKED THIS

You might need to scroll to find the right info. I circled the important info.

***Say which ones you can reject the null hypothesis for and which ones you can’t, and in which order they are of increasing mean

The ANOVA test has 3 assumptions: The error terms are independent. You cannot verify independence empirically but only through logic. It is logical that the error terms are independent. The error terms are normally distributed. We can verify through graphing a histogram of residuals. IN THE R SCRIPT BOX (top one of R commander) >hist(AnovaModel.1$residuals)

*Sketch* Is it SYMMETRICAL and does it have a BELL-CURVE shape?? If yes, the normality of the error terms is verified? (Pls answer these questions and not copy this bit out) The error terms have equal variance. This can be verified through Bartlett’s Test. Statistics->Variances->Bartlett’s Test

p-value = 0.3695 alpha = 0.05 p-value > alpha At the 5% level of significance, we fail to reject the null hypothesis. Therefore, there is significant evidence to suggest that all the variances of the satisfactions for the different car brands are equal. Hence the assumption is verified....