Question

A manager of an industrial plant asserts that workers on average do not complete a job using Method A in the same amount of time as they would using Method B. Five workers are randomly selected. The time to completion is recorded for Method A and Method B (in minutes). Is there evidence at the 5% level of Significance that Method A is faster than Method B? Method A: 12, 16, 15, 11 and 10. Method B: 6, 10, 8, 9 and 12. A. Yes, there is evidence at the 5% level of Significance that two method differ. B. No, there is no evidence at the 5% level of significance that method A is faster.

Answer #1

*Method A*

*N*_{1}: 5

*df*_{1} = *N* - 1 = 5 - 1 = 4

*M*_{1}: 12.8

*SS*_{1}: 26.8

*s*^{2}_{1} =
*SS*_{1}/(*N* - 1) = 26.8/(5-1) = 6.7

*Method B*

*N*_{2}: 5

*df*_{2} = *N* - 1 = 5 - 1 = 4

*M*_{2}: 9

*SS*_{2}: 20

*s*^{2}_{2} =
*SS*_{2}/(*N* - 1) = 20/(5-1) = 5

a) null and alternate hypothesis

H0:

H1:

b) level of significance = 0.05

c) test statistics

*s*^{2}_{p} =
((*df*_{1}/(*df*_{1} +
*df*_{2})) * *s*^{2}_{1}) +
((*df*_{2}/(*df*_{2} +
*df*_{2})) * *s*^{2}_{2}) =
((4/8) * 6.7) + ((4/8) * 5) = 5.85

*s*^{2}_{M1} =
*s*^{2}_{p}/*N*_{1}
= 5.85/5 = 1.17

*s*^{2}_{M2} =
*s*^{2}_{p}/*N*_{2}
= 5.85/5 = 1.17

*t* = (*M*_{1} -
*M*_{2})/√(*s*^{2}_{M1}
+ *s*^{2}_{M2}) =
3.8/√2.34 = 2.48

d) p value = 0.0189

E) since p value is less than level of significance so we reject H0

So there sufficient evidence at the 5% level of significance that method A is faster.

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