The following sample information shows the number of defective units produced on
ID: 3431199 • Letter: T
Question
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.
At the .05 significance level, can we conclude there are more defects produced on the day shift?
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 11 10 14 17 Afternoon shift 8 11 12 15Explanation / Answer
(1) The degree of freedom =n-1=4-1=3
Given a=0.05, the critical value is t(0.05, df=3)=2.353 (from student t table)
Reject H0 if t >2.35
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(2)
test statistic is
t= mean difference/(s/vn)
=1.732
13.000 mean Day shift 11.500 mean Aternoon shift 1.500 mean difference (Day shift - Aternoon shift) 1.732 std. dev. 0.866 std. error 4 n 3 df