(Round all intermediate calculations to at least 4 decimal places.) A lead inspe
ID: 3061273 • Letter: #
Question
(Round all intermediate calculations to at least 4 decimal places.) A lead inspector at ElectroTech, an electronics assembly shop, wants to convince management that it takes longer, on a per-component basis, to inspect large devices with many components than it does to inspect small devices because it is difficult to keep track of which components have already been inspected. To prove her point, she has collected data from the last 25 devices. The data are shown in the accompanying table Number of ComponentsInspection Time on Device 32 13 9 17 15 (seconds) 84 49 30 60 51 24. 42 7 12 19 71 42 63 26 80 48 31 62 52 60 72 102 59 30 12 10 19 19 25 16 13 21 12 23 67 46 70 lick here for the Excel Data FileExplanation / Answer
a)
As the scatter plot is upward sloping implies the time increases with the number of parts
Hence Yes
b-1
By plotting a scatter plot and adding a trend line by right click on the plot
we get the R squared value for linear as R² = 0.9362
Now we repeat the same procedure but this time we select polynomial of order 2 to get the R squared value for Quadratic equation= R² = 0.9782
Repeat the same procedure but this time we select polynomial of order 3 to get the R squared value for Cubic equation= R² = 0.9889
b-2
As the Rsquared of Cubic equation is highest we say that Cubic model has best fit
c
From the excel we get the equation of cubic as
y = 0.0023x3 - 0.2068x2 + 7.5627x - 21.848
Substituting 35 as x
We get 88.129 =88.13 seconds
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.