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A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The average thi
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A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The average thickness and the standard deviation of thickness of shift 1, based on a random sample of 30 gaskets, are 10.53mm and 0.14mm. The similar figures for shift 2, based on a random sample of 25 gaskets, are 10.55mm and 0.17mm. Let µ1-µ2 be the mean difference in thickness between shifts 1 and 2.
a) Find a 95% confidence interval for µ1-µ2.
b) Based on your answer to part a, are you convinced that the gaskets from shift 2 are, on average, wider than those from shift 1? Why or why not?
c) How would your answers to parts a and b change if the sample sizes were instead 300 and 250?
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Mathematics
4 years
2021-08-27T10:20:59+00:00
2021-08-27T10:20:59+00:00 1 Answers
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Answers ( )
Answer:
(−0.103371 ; 0.063371) ;
No ;
( -0.0463642, 0.0063642)
Step-by-step explanation:
Shift 1:
Sample size, n1 = 30
Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm
Shift 2:
Sample size, n2 = 25
Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17
Mean difference ; μ1 – μ2
Zcritical at 95% confidence interval = 1.96
Using the relation :
(m1 – m2) ± Zcritical * (s1²/n1 + s2²/n2)
(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)
Lower boundary :
-0.02 – 0.0833710 = −0.103371
Upper boundary :
-0.02 + 0.0833710 = 0.063371
(−0.103371 ; 0.063371)
B.)
We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.
C.)
For sample size :
n1 = 300 ; n2 = 250
(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)
Lower boundary :
-0.02 – 0.0263642 = −0.0463642
Upper boundary :
-0.02 + 0.0263642 = 0.0063642
( -0.0463642, 0.0063642)