One step in the manufacture of large engines requires that holes of very precise
ID: 2929776 • Letter: O
Question
One step in the manufacture of large engines requires that holes of very precise dimensions be drilled. The tools that do the drilling are regularly examined and are adjusted to ensure that the holes meet the required specifications. Part of the examination involves measurement of the diameter of the drilling tool. A team studying the variation in the sizes of the drilled holes selected this measurement procedure as a possible cause of variation in the drilled holes. They decided to use a designed experiment as one part of this examination. Some of the data are given in the table below. The diameters in millimeters (mm) of five tools were measured by the same operator at three times (8:00 A.M., 11:00 A.M., and 3:00 P.M.). Three measurements were taken on each tool at each time. The person taking the measurements could not tell which tool was being measured, and the measurements were taken in random order.
Multiply each measurement by 0.04 to convert from millimeters to inches. Plot the means and run a two-way ANOVA using the transformed measurements. Summarize what parts of the analysis have changed and what parts have remained the same.
Explanation / Answer
Below code is carried out in R for performing the listed steps.
Machine<-read.csv("Machine.csv",header=TRUE)
str(Machine)
Machine$Time<-as.factor(Machine$Time) -- converting numerical to categorical
Machine$Tool<-as.factor(Machine$Tool) -- converting numerical to categorical
Machine$MeanDia<-rowMeans(Machine[,3:5])*0.04 -- > Converting millimeters to inches
summary(aov(formula=MeanDia~Time+Tool,data=Machine)) -- Performing anova
Df Sum Sq Mean Sq F value Pr(>F)
Time 2 1.013e-07 5.060e-08 5.703 0.0289 *
Tool 4 1.918e-06 4.796e-07 54.012 7.9e-06 ***
Residuals 8 7.100e-08 8.900e-09
Null Hypo : Mean Dia produced by each machine is equal
Alt. Hypo : Mean Dia produced by each machine is not equal
From the above we could figure out that both the independent factors ie. Time & Tool are significant
Though Time is significant at 95% C.I and not so at 99%, whereas Tool is highly significant
From the above we can conclude that the sample data represent that their is a gap in mean dia size produced by each tool. Hence we can reject the null hypothesis.
Below code helps in calculating the mean of Dia sizes / Tool
Machine$ToolMean<-with(Machine,ave(MeanDia,Tool,FUN = mean))
Below code helps in calculating the mean of Dia sizes / Time
Machine$TimeMean<-with(Machine,ave(MeanDia,Time,FUN = mean))
Plotting of the means
plot(Machine$Time,Machine$TimeMean)
plot(Machine$Tool,Machine$ToolMean)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.