Question

A manufacturer wants to compare the number of defects on the day shift with the number on the evening shift. A sample of production from recent shifts showed the following defects:

Day Shift​​: 6​ 9​ 8​ 7​ 10​ 8

Evening Shift: ​​9​ 11​ 8​ 12​ 10​ 13​ 15​ 10

The objective is to determine whether the mean number of defects on the night shift is greater than the mean number on the day shift at the 95% confidence level.

a) State the null and alternate hypotheses.

b) What is the level of significance?

c) What is the test statistic?

d) What is the decision rule?

e) Use the Excel Data Analysis pack to analyze the problem. Include the output with your answer. (Note: You may calculate by hand if you prefer).

f) What is your conclusion?  Explain.

g) Does the decision change at the 99% confidence level?

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud > 0
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 5
We have d = 5
pooled variance = calculate value of Sd= √S^2 = sqrt [ 388-(40^2/8 ] / 7 = 5.182
to = d/ (S/√n) = 2.729
critical Value
the value of |t α| with n-1 = 7 d.f is 1.895
we got |t o| = 2.729 & |t α| =1.895
make Decision
hence Value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 2.7289 ) = 0.01469
hence value of p0.05 > 0.01469,here we reject Ho
ANSWERS
---------------
a.
null, H0: Ud = 0
alternate, H1: Ud > 0
b.
level of significance =0.05
c.
test statistic: 2.729
d.
critical value: reject Ho, if to > 1.895
decision: Reject Ho
e.
p-value: 0.01469
f.
we have enough evidence to support the claim that hether the mean number of defects on the night shift is greater than the mean number on the day shift

g.
level of significance =0.01
Given that,
null, H0: Ud = 0
alternate, H1: Ud > 0
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.998
since our test is right-tailed
reject Ho, if to > 2.998
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 5
We have d = 5
pooled variance = calculate value of Sd= √S^2 = sqrt [ 388-(40^2/8 ] / 7 = 5.182
to = d/ (S/√n) = 2.729
critical Value
the value of |t α| with n-1 = 7 d.f is 2.998
we got |t o| = 2.729 & |t α| =2.998
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 2.7289 ) = 0.01469
hence value of p0.01 < 0.01469,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud > 0
test statistic: 2.729
critical value: reject Ho, if to > 2.998
decision: Do not Reject Ho
p-value: 0.01469
we do not have enough evidence to support the claim that hether the mean number of defects on the night shift is greater than the mean number on the day shift