The supervisor of a soap production line finds that the amount of scrap produced
ID: 3207511 • Letter: T
Question
The supervisor of a soap production line finds that the amount of scrap produced per hour is proportional to the speed of the line. He requests you to see if this is true. The following is his data.
Speed (cycles/min)
Scrap/hour
100
218
125
248
220
360
205
351
300
470
255
394
225
332
175
321
270
410
170
260
155
241
190
331
140
275
290
425
265
367
a) Develop a (linear) model to verify his observation. Write down the model and comment on his claim. Further, the supervisor asks you if it is possible to give him limiting (high) values for the average scrap per hour taken over an 8 hour shift at each speed level so that he can decide if the line needs to be 'reset'. Resetting the machines on the line is an expensive proposition and so he wants you to be 95% confident that resetting is really required. (Hint: A shift of 8 hours means that you have 8 samples)
Solve this problem, showing all the intermediate steps.
(Hint: you can compute the width of the CI)
b) On a particular shift (of 8 hours) he ran the line at a speed of 190 cycles/min and obtained, on the average, 339 pieces of scrap per hour. Should he reset the line? Justify your answer with proper analysis (Hint: you need to check whether the number of pieces of scrap is outside the CI)
Speed (cycles/min)
Scrap/hour
100
218
125
248
220
360
205
351
300
470
255
394
225
332
175
321
270
410
170
260
155
241
190
331
140
275
290
425
265
367
Explanation / Answer
we shall analyse this using the open source statistical package R
Please see the complete R snippet below
# read the data into R dataframe
data.df<- read.csv("C:\Users\586645\Downloads\Chegg\scrap.csv",header=TRUE)
str(data.df)
# fit the linear regression model
fit <- lm(data.df$Scrap.hour~ data.df$Speed..cycles.min.)
# summarise the results
summary(fit)
# take a random sample
sample8<-data.df[sample(nrow(data.df),size=8),]
# get the confidence interval
predict(fit,sample8,interval="confidence")
The results are
Call:
lm(formula = data.df$Scrap.hour ~ data.df$Speed..cycles.min.)
Residuals:
Min 1Q Median 3Q Max
-34.500 -14.403 5.496 16.046 28.418
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 97.96533 20.51430 4.775 0.000362 ***
data.df$Speed..cycles.min. 1.14539 0.09578 11.959 2.18e-08 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 22.19 on 13 degrees of freedom
Multiple R-squared: 0.9167, Adjusted R-squared: 0.9103
F-statistic: 143 on 1 and 13 DF, p-value: 2.177e-08
The model is signifcant as the p value is less than 0.05, the regression equation is
Y = 97.96 + 1.14*X and the confidence interval is given as
> predict(fit,sample8,interval="confidence")
fit lwr upr
1 212.5041 187.3792 237.6289
2 241.1388 220.3583 261.9192
3 349.9506 337.2210 362.6801
4 332.7697 320.3897 345.1498
5 441.5815 418.4684 464.6947
6 390.0391 373.9942 406.0841
7 355.6775 342.6679 368.6870
8 298.4081 284.4974 312.3188
9 407.2199 389.0421 425.3978
10 292.6812 278.2691 307.0932
11 275.5004 259.2785 291.7222
12 315.5889 302.7923 328.3856
13 258.3196 239.9387 276.7004
14 430.1277 408.7331 451.5222
15 401.4930 384.0585 418.9275
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