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Assignment six = 20 points:
1.
In many educational institutions, the Athletic Department does a better job of
teaching teamwork than do the other units. What skills can AET faculty learn about
teaching teamwork by examining the analogy of a successful athletic team?
In any successful athletic team they one key ingredient that the team has is
trust. If you do not trust your teammate to be where you are supposed to be, or
run the right play, then you are not going to be very successful. Of course trust
is a two way street and is earned not given. But the most important skill is
leadership; leadership is having the ability to draw the line between being a
friend and a leader and getting your team to the right spot. I played basketball
here at MSU and when we stepped between the lines I had no friends, I did
everything I could to lead and make sure my team was ready for the next
game.
2.
The tensile strength of a fiber used in manufacturing cloth is of interest to the
purchaser. Previous experience indicates that the standard deviation of tensile
strength is 2 psi. A random sample of eight fiber specimens is selected, and the
average tensile strength is found to be 127 psi.

Test the hypothesis that the mean tensile strength equals 125 psi versus the
alternative that the mean exceeds 125 psi. Use α=0.05.
One-Sample Z
Test of mu = 125 vs > 125
The assumed standard deviation = 2
95% Lower
N Mean SE Mean
Bound Z P
8 127.000 0.707 125.837 2.83 0.002
Critcal value=1.895 (found in book n-1)
Z value is 2.83
So since 2.83 is greater than 1.895 reject the hypothesis
(a) What is the P-value for this test?
0.002
(b) Discuss why a one-sided alternative was chosen in part
Because the test was set to be greater than
(c) Construct a 95% lower confidence interval on the mean tensile strength.
95% lower= 127-2.83(2SQrt8)=125 (124.999999)
3.
A machine is used to fill containers with a liquid product. Fill volume can be
assumed to be normally distributed. A random sample of 10 containers is selected,
and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98,
11.96, 12.02, 12.05, and 11.99.
(a) Suppose that the manufacturer wants to be sure that the mean net
content exceeds 12 oz. What conclusions can be drawn from the data (use
α=0.01)?
One-Sample T: C1
Test of mu = 12 vs > 12
99.99%
Variable N Mean StDev SE Mean Lower Bound T
C1
10 12.0150 0.0303 0.0096
11.9575 1.57 0.076
P
Since the critical value, 2.82, is less than the calculated value, 1.57, you
can accept the hypothesis
(b) Construct a 95% two-sided confidence interval on the mean fill volume.
CI = 12.015 - (1.5655)(.0303/SQRT10) = 11.99
C2= 12.015 + (1.5655)(.0303/SQRT10) = 12.03
Not sure if this was done correctly
4.
Using the data bellow (Table 7.1 at your text) answer the following question
(Using Minitab).
a. Calculate mean, mode, median, standard deviation and variance for the data in
Table 7.1.
b. Draw the Histogram for width and analyze the graph.
c.
5.
Using correlation coefficient formula or Minitab, find the correlation
coefficient between width and gauge. What does this mean?
Determine the trial central line and control limits for a p chart using the following
data, which are for the payment of dental insurance claims. Plot the values on graph
paper and determine if the process is stable. If there are any out -of-control points,
assume an assignable cause and determine the revised central line and control
limits.
P Chart of C1
0.07
1
0.06
Proportion
0.05
0.04
UCL=0.04016
0.03
_
P=0.01747
0.02
0.01
0.00
LCL=0
1
3
5
7
9
11
13
15
17
19
21
23
25
Sample
UCL=0.04016
Pbar=0.01747 and
LCL=0
There are points out of control so we revisit the data and take out any point that are
out of control
P Chart of C1
0.04
UCL=0.03676
Proportion
0.03
0.02
_
P=0.01542
0.01
0.00
LCL=0
1
3
5
7
9
11
13
15
17
19
21
23
Sample
As you can see this makes a change the values and makes the process in control.
6.
Determine the trial limits and revised control limits for a u chart using the data in
the table for the surface finish of rolls of white paper. Assume any out -of-control
points have assignable causes.
Test Results for U Chart of C1
TEST 1. One point more than 3.00 standard deviations from center line.
Test Failed at points: 2, 6, 21
* WARNING * If graph is updated with new data, the results above may no
* longer be correct.
U Chart of C1
1
UCL=5.028
Sample Count Per Unit
5
4
_
U=3.304
3
2
LCL=1.579
1
1
1
0
1
4
7
10
13
16
Sample
19
22
25
28
Points 2, 6, and 21 need to be taken out and the data needs to be revisited.
U Chart of C1
5.5
UCL=5.169
Sample Count Per Unit
5.0
4.5
4.0
_
U=3.416
3.5
3.0
2.5
2.0
LCL=1.663
1.5
1
3
5
7
9
11
13 15
Sample
17
19
21
23
25
7.
An np chart is to be established on a painting process that is in
statistical
control. If 35 pieces are to be inspected every 4
hours, and the fraction
nonconforming is 0.06, determine the
central line and control limits.
Equations taken from CH. 7 power point
npbar = 35(0.06) = 2.1
UCL = 2.1 + 3 SQRT (2.1 (1 - 0.06)) = 2.1 + 3 SQRT (1.974) = 2.1 + 3(1.405) = 6.315
LCL = 2.1 – 3 SQRT (2.1 (1 – 0.06)) = 2.1 – 3 SQRT (1.974) = 2.1 – 3(1.405) = -2.115
8.
A quality technician has collected data on the count of rivet nonconformities in
four meters travel trailers. After 30 trailers, the total count of non-conformities is 316.
Trial control limits have been determined and a comparison with the data shows no outof-control points. What is the recommendation for the central line and the revised control
limits for a count of nonconformities chart?
Since there are no out of control points this could mean that there no real threats
within the units themselves and they can still be processed and sold to the general
population. If any of the points had been out of control then I would recommend
discarding that particular point and revisiting the data.