Clark Heter is an industrial engineer at Lyons Products. He would like to determ
ID: 3126005 • Letter: C
Question
Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. A sample of 57 day-shift workers showed that the mean number of units produced was 348, with a population standard deviation of 23. A sample of 67 night-shift workers showed that the mean number of units produced was 353, with a population standard deviation of 33 units.
The decision rule is to reject H0: d n if z < . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
The test statistic is z = . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
At the .02 significance level, is the number of units produced on the night shift larger?Explanation / Answer
1.
Formulating the null and alternative hypotheses,
Ho: u1 - u2 >= 0
Ha: u1 - u2 < 0
At level of significance = 0.02
As we can see, this is a left tailed test.
Hence, it is a ONE TAILED TEST. [ANSWER]
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2.
Now, the critical value for z is, by table/technology,
zcrit = -2.05
Hence, we reject Ho is z < -2.05. [ANSWER]
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3.
Calculating the means of each group,
X1 = 348
X2 = 353
Calculating the standard deviations of each group,
s1 = 23
s2 = 33
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 57
n2 = sample size of group 2 = 67
Also, sD = 5.053160704
Thus, the z statistic will be
z = [X1 - X2 - uD]/sD = -0.989479712 [ANSWER, TEST STATISTIC]
where uD = hypothesized difference = 0
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4.
As z > -2.05, WE DO NOT REJECT THE NULL HYPOTHESIS. [ANSWER]