Question

Wenton Powersports produces dune buggies. They have three assembly lines, “Razor,” “Blazer,” and “Tracer,” named after the particular dune buggy models produced on those lines. Each assembly line was originally designed using the same target production rate. However, over the years, various changes have been made to the lines. Accordingly, management wishes to determine whether the assembly lines are still operating at the same average hourly production rate.

Razor				Blazer				Tracer
	11				10				9
	10				8				9
	8				9				10
	10				9				9
	9				13				8
	9				10				7
	13				11				8
	11				8				9

Production data (in dune buggies/hour) for the last eight hours are shown in the table above (in "unstacked" format - each group in a separate column). The Excel file contains the same data, but in a "stacked" format (the numeric values for all 3 groups are in a single column). Take a look at the difference.

To read the Excel file into R and view it:
>library(openxlsx)
>anyname <- read.xlsx(file.choose())
>attach(anyname)

>print(anyname)

a. Specify the competing hypotheses to test whether there are some differences in the mean production rates across the three assembly lines.

	H0: μRazor = μBlazer = μTracer. HA: Not all population means are equal.
	H0: μRazor ≤ μBlazer ≤ μTracer. HA: Not all population means are equal.
	H0: μRazor ≥ μBlazer ≥ μTracer. HA: Not all population means are equal.

b-1. Using the two R commands shown below, run a one-way ANOVA to test the above hypotheses. Use the column names in the Excel file for the "y" numeric variable and "factor" categorical variable. "name4results" is a name you make up to store the ANOVA results.

> name4results <- aov(y ~ factor)

> summary(name4results)

Fill in an ANOVA table below. (Round "Sum Sq" to 2 decimal places, "Mean Sq" and "F value" to 3 decimal places, and "p-value" to 4 decimal places.)

b-2. At the 5% significance level, what is the conclusion to the test?

b-3. At the 10% significance level, what is the conclusion to the test?

Answer 1

a)

option A) H₀: μ_Razor = μ_Blazer = μ_Tracer. H_A: Not all population means are equal. is correct

b)

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Razor	8	81	10.125	2.410714286
Blazer	8	78	9.75	2.785714286
Tracer	8	69	8.625	0.839285714
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	9.75	2	4.875	2.423076923	0.113017418	3.466800112
Within Groups	42.25	21	2.011904762
Total	52	23

b-2

p-value = 0.1130

p-value > 0.05

we fail to reject the null hypothesis

b-3

p-value > 0.10

we fail to reject the null hypothesis