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?

Question

contestada

Respuesta :

fichoh fichoh · Answer 1 · 2021-02-16T08:24:43+01:00

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)