A small engineering company performed an experiment to compare the hardness of t
ID: 2922914 • Letter: A
Question
A small engineering company performed an experiment to compare the hardness of the cut edge of metal sheets after gas with two different fuel gases. Eight test plates of different steels and different thicknesses were used. for the test. Each plate was cut with both oxy-propane and oxy-natural gas. The first cut was made with oxy-propane on a random selection of 4 out of the 8 sheets, and the first cut was made with oxy-natural gas on the other 4 sheets. The results are given below (force applied (9.8N) /area of indentation (mm^-2)).
Plate
Oxy-propane cut
Vickers Hardness VH10
Oxy-natural gas cut
Vickers HardnessVH10
1
370
333
2
333
336
3
330
299
4
306
294
5
314
297
6
322
343
7
290
304
8
340
329
(a) What is the advantage (other than fewer sheets being required) of using both cutting methods on the same plates rather than taking 16 plates of different steels and thicknesses and randomly assigning 8 to each method.
(b) Construct a 90% confidence interval for the mean difference in hardness of the cut edge using the two methods.
(c) The company will change to oxy-natural gas provided they are reasonably confident that any increase in the mean hardness of the cut edge is less than 10.
What size sample would you recommend if the width of the 90% confidence interval for the mean is to be around 10? What is then the probability that the upper limit of such an interval will be less than 10 if the use of oxy-natural gas increases the mean hardness by 5?
(d) Why use plates of different steels and thicknesses rather than plates of the same steel and thickness?
Plate
Oxy-propane cut
Vickers Hardness VH10
Oxy-natural gas cut
Vickers HardnessVH10
1
370
333
2
333
336
3
330
299
4
306
294
5
314
297
6
322
343
7
290
304
8
340
329
Explanation / Answer
(a) The advantage is here that we have to test lesser number of plates and there are no biasness of sampling as both the experiment is done on the same plate.
(b) Here is the table of difference between hardness between both the cut edge technique witht calculation of standard deviation of difference.
Here Average of difference between Hardness level xd = 8.75
Standard deviation of the difference between hardness level sd = 20.42
90% confidence interval for difference = xd +- t7, 0.10 (sd /sqrt(n)) [ dF = 7, alpha = 0.10
= 8.75 +- 1.8946 * (20.42/sqrt(8))
= 8.75 +- 13.68
= (-4.93, 22.43)
(c) Width of the confidence interval = 10 for 90% confidenceinterval
10 = Z90* (s/sqrt(n)) = 1.645 * (20.42 / sqrt(n) )
sqrt(n) = 3.36
n = 12
Now it is asked
Pr(Difference < 10; 5 ; 5.274) = ?
Z = (10 -5)/ 5.274 = 0.95
Pr(Difference < 10; 5 ; 5.274) = (0.95) = 0.8289 or 82.89%
(d) We will use plates of different steels and thickness to bring more diversity in sample and in the results. It bring more randomness and removes exculsiviity and biasness.
Plate Oxy-propane cut Vickers Hardness VH10 Oxy-natural gas cut Vickers HardnessVH10 Difference(d) 1 370 333 37 2 333 336 -3 3 330 299 31 4 306 294 12 5 314 297 17 6 322 343 -21 7 290 304 -14 8 340 329 11 Average 8.75 Sd.Dev. 20.42