The null and alternate hypotheses are: H 0 : ? d ? 0 H 1 : ? d > 0 The following
ID: 3429497 • Letter: T
Question
The null and alternate hypotheses are:
H0 : ?d ? 0
H1 : ?d > 0
The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.
Day
1
2
3
4
Day shift
10
12
13
18
Afternoon shift
8
9
12
16
At the .10 significance level, can we conclude there are more defects produced on the afternoon shift? Hint: For the calculations, assume the Day shift as the first sample.
(a)
State the decision rule. (Round your answer to 2 decimal places.)
Reject H0 if t >
(b)
Compute the value of the test statistic. (Round your answer to 3 decimal places.)
Value of the test statistic
(c)
What is the p-value?
p-value
(d)
What is your decision regarding H0?
H0
The null and alternate hypotheses are:
H0 : ?d ? 0
H1 : ?d > 0
The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.
Explanation / Answer
(a) The degree of freedom =n-1=4-1=3
Given a=0.1, the critical value is t(0.1, df=3) =1.64 (from student t table)
Reject H0 if t >1.64
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(b)
the value of the test statistic
t=mean difference/(s/vn)
=4.899
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(c) the p-value is 0.0081 (from student t table)
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(d) Reject Ho.
13.250 mean Day shift 11.250 mean Afternoon Shift 2.000 mean difference (Day shift - Afternoon Shift) 0.816 std. dev. 0.408 std. error 4 n 3 df