Question
In: Statistics and Probability
A manufacturer wants to compare the number of defects on the day shift with the number...
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 5 8 7 6 9 7
Evening Shift 8 10 7 11 9 12 14 9
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.
- State the null and alternate hypotheses.
- What is the level of significance?
- What is the test statistic?
- What is the decision rule?
- 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).
- What is your conclusion? Explain.
- Does the decision change at the 99% confidence level?
Solutions
orchestra answered 1 year agoRelated Solutions
A manufacturer wants to compare the number of defects on the day shift with the number...
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)...
A manufacturer wants to compare the number of defects on the day shift with the number...
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)...
A manufacturer wants to compare the number of defects on the day shift with the number...
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 5 8 7 6 9 7
Evening Shift 8 10 7 11 9 12 14 9
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.
State...
A manufacturer wants to compare the number of defects on the day shift with the number...
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 5 8 7 6
9 7
Evening Shift 8 10 7 11 9 12 14 9
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.
State...
Number of defective monitors manufactured in day shift and afternoon shift is to be compared. A...
Number of defective monitors manufactured in day shift and
afternoon shift is to be compared. A sample of the production from
six day shifts and eight afternoon shifts revealed the following
number of defects.
Day
4
5
8
6
7
9
Afternoon
9
8
10
7
6
14
11
5
Is there a difference in the mean number of defects per shift?
Choose an appropriate significance level.
(a) State the null hypothesis and the
alternative hypothesis.
(b) What is the...
Number of defective monitors manufactured in day shift and afternoon shift is to be compared. A...
Number of defective monitors manufactured in day shift and
afternoon shift is to be compared. A sample of the production from
six day shifts and eight afternoon shifts revealed the following
number of defects.
Day 4 5 8 6 7 9
Afternoon 9 8 10 7 6 14 11 5
Is there a difference in the mean number of defects per shift?
Choose an appropriate significance level.
(a) State the null hypothesis and the alternative
hypothesis.
(b) What is the...
The table below shows the number of graveyard shift, swing shift, and day shift workers (variable called JOB SHIFT) by "coffee drinking" and "non-coffee drinking" (variable called AWAKE)
The table below shows the number of graveyard shift, swing shift, and day shift workers (variable called JOB SHIFT) by "coffee drinking" and "non-coffee drinking" (variable called AWAKE). Test the null hypothesis that the two variables are unrelated by doing the following:
(Answer: Chi-square = 10.08)
State the null and research hypotheses.
Report the degrees of freedom.
Report the critical value (use alpha = .05)
Calculate the test statistic
In two or three sentences, briefly explain the results
(this...
Motorola used the normal distribution to determine the probability of defects and the number of defects...
Motorola used the normal distribution to determine the
probability of defects and the number of defects expected in a
production process. Assume a production process produces items with
a mean weight of 7 ounces.
a. The process standard deviation is 0.10, and
the process control is set at plus or minus 2 standard deviation s
. Units with weights less than 6.8 or greater than 7.2 ounces will
be classified as defects. What is the probability of a defect (to...
Motorola used the normal distribution to determine the probability of defects and the number of defects...
Motorola used the normal distribution to determine the
probability of defects and the number of defects expected in a
production process. Assume a production process produces items with
a mean weight of 14 ounces.
The process standard deviation is 0.2, and the process control
is set at plus or minus 0.5 standard deviation. Units with weights
less than 13.9 or greater than 14.1 ounces will be classified as
defects. What is the probability of a defect (to 4 decimals)?
In...
Motorola used the normal distribution to determine the probability of defects and the number of defects...
Motorola used the normal distribution to determine the
probability of defects and the number of defects expected in a
production process. Assume a production process produces items with
a mean weight of 12 ounces.
a. The process standard deviation is 0.10 ounces, and the
process control is set at plus or minus 1.75 standard deviations.
Units with weights less than 11.825 or greater than 12.175 ounces
will be classified as defects. What is the probability of a defect
(to 4...
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