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For this assignment all work /computer output can be copied and pasted or typed into this document. Only 1c,d,and e require hand calculations. Instructions are

For this assignment all work /computer output can be copied and pasted or typed into this document. Only 1c,d,and e require hand calculations. Instructions are for R , 2. Consider the independent samples A, B,C

A B C 2.6 3.3 5.8 3.2 7.3 5.6 4 5.7 6.7 4.5 6.3 7.8 4.6 4.8 7 4.8 5.3 7.2 5 5 8 5.5 4.7 8.6 6.3 7.6 9.4 ( 10 pts) a. Create a normal quantile quantile plot for each sample. Note: There is a module video on Normal QQ plots if you need to review this . example using R: copy and paste each line individually and Press after each line A=c(2.6, 3.2, 4, 4.5, 4.6, 4.8, 5, 5.5, 6.3) B=c(3.3, 7.3, 5.7, 6.3, 4.8, 5.3, 5, 4.7, 7.6) C=c(5.8, 5.6, 6.7, 7.8, 7, 7.2, 8, 8.6, 9.4) combined=cbind(A,B,C) qqnorm(A) x=seq(-4,4,by=.1) lines(x,sd(A)*x+mean(A)) qqnorm(B) lines(x,sd(B)*x+mean(B)) qqnorm(C) lines(x,sd(C)*x+mean(C)) boxplot(combined) Does one-way ANOVA procedure seem justified? That is, do the samples appear normal with approximately equal variance? Provide all 4 plots and conclusion. ************************************************************************* ( 10 pts) b. Regardless of your answer to part a, perform a one-way ANOVA for a difference in means. Use R . State hypotheses and conclusion. download HW5Q1.csv. change R working directory to this location Q2=read.csv('HW4Q2.csv',header=T) anova=aov(Q2$Response~Q2$Treatment) summary(anova) ***************************************************************** ( 10 pts) c. Use the Bonferroni multiple comparisons procedure to detect which population means differ and how they differ. Use 95% confidence ( alpha = .05) . The number of comparisons are 3 C 2 = 3. Compute for all 3 possibilities To get the critical t value for each of the 3 intervals, use R and type qt(1-.05/(2*3),24) . Show the confidence interval for each difference and state whether or not it indicates a significant difference in population means. Does the boxplot from part a support your answer? ( 10 pts) d. Use Sheffes Honest Significant Difference Test to detect which population means significantly differ. Use 95% confidence ( alpha = .05) . Compute HSD= ((k-1)F_alpha 2MSE/n) Here k=3, n=9, and the critical F is the ANOVA critical F and MSE is the ANOVA MSE. for critical value , qf( .95 , dfnumerator, dfdenominator) Complete the table filling in the differences in sample means for A,B,C Any differences whose absolute value exceeds HSD are significantly different at alpha =.05

A B C A x B x x C x x x

Report HSD value, differences table, and conclusions. ***************************************************************** ( 10 pts) e. Use Tukeys Honest Significant Difference Test to detect which population means significantly differ. Use 95% confidence ( alpha = .05) . Compute HSD= MSE is same as above and n= 9. q is the critical value for the studentized range distribution, q. To obtain this critical value, type qtukey(.95,3, 24 ) Complete the table filling in the differences in sample means for A,B,C Any differences whose absolute value exceeds HSD are significantly different at alpha =.05. A B C A x B x x C x x x

Report HSD value, differences table, and conclusion. Check your work by TukeyHSD(anova=aov(Q2$Response~Q2$Treatment)) This gives confidence intervals for the difference in means using this method. If starting from scratch: Q2=read.csv('HW4Q2.csv',header=T) anova=aov(Q2$Response~Q2$Treatment) TukeyHSD(anova)

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