STAT 496, Spring 2013

```STAT 496, Spring 2013
Homework Assignment #2, Due by Friday, February 8
1. The three fundamental principles of a designed experiment are: control of outside
variables, randomization, and replication. Consider each of the following
scenarios. Does each adhere to all three principles? If so, describe how. If not,
indicate how you would change the design to incorporate the missing principle(s).
If a principle is partially adhered to but could be improved, comment on what you
would do to improve the experiment. Be specific and be sure to comment on all
three principles.
a) An experiment is performed to evaluate the effect of drill speed on the size of
drilled holes. A single 0.25 inch drill bit in a vertical head milling machine is
used for the experiment. All drilling is done by the same operator. A single
bar steel with a Brinell hardness of 400 is used. Four speeds are evaluated:
200, 250, 300 and 350 rpms. The operator picks a speed at random from the
four and will drill 20 holes. The process is repeated, picking a speed at
random from those remaining and drilling 20 holes until all 4 speeds have
been used. After all of the holes are drilled, each of the holes is measured
using a special caliper. The order in which the 80 holes are measured is done
at random.
b) A chemist wants to compare a new and simpler assay method for determining
the concentration of a mixture with the standard method. She prepares a batch
of solution, divides it into 30 specimens, and then selects 15 at random. She
asks her technician to analyze these 15 specimens using the new method and
the remaining 15 using the standard method. The technician finishes analyzing
the 15 specimens using the new method in the morning. He takes a break for
lunch and does the remaining 15 specimens using the standard method in the
afternoon.
c) An industrial psychologist does a study to determine whether incentives and
the size of the incentive, will improve productivity. Thirty workers, who all do
the same assembly job, agree to participate in the study. The thirty workers
are divided, at random, into three groups of 10. Each person in one group is
told that if he/she increases the number of correctly assembled products by 5%
he/she will receive a 5% increase in their pay. Each person in a second group
is told that if she/he increases the number of correctly assembled products by
5% she/he will receive a 10% increase in her/his pay. Each person in the third
group is told that the company’s goal is to increase the number of correctly
assembled parts by 5% but is given no individual pay incentive.
1
2. A manufacturer wishes to compare two foundries that supply parts for the
manufacturer’s production lines. An important quality characteristic is the %
elongation for the heat of metal that is used to make the parts.
a) If the chances of making errors are set at Alpha=0.05 and Beta=0.10 and a
difference in mean % elongation of 1 standard deviation is to be detected, how
many heats of metal should be evaluated for each foundry?
b) If resources limit the number of heats to 10 from each foundry, give two
combinations of Alpha and Beta and the size of the difference in means that
can be detected. Explain briefly the trade-offs between the two combinations.
c) Ten heats are randomly selected from those available at Foundry A and ten
heats are randomly selected from those available at Foundry B. All heats are
evaluated and % elongation values are obtained. Explain briefly why this is
an observational study and not an experiment.
Data for this study are given in the table on the next page. Output from JMP is
provided at the end of the assignment.
1. Foundry A
44
49
44
47
49
42
51
48
46
44
y1 = 46.4
2. Foundry B
58
53
61
60
60
63
64
61
59
55
y 2 = 59.4
s1 = 2.8752
s 2 = 3.3731
s12 = 8.27
s 22 = 11.38
d) Comment on the side-by-side box plots for the two foundries. Be sure to
e) Is there a statistically significant difference between the % elongation means
f) The JMP output reports the observed difference in means (Difference
13.0000) for the 20 heats evaluated. It also reports a 95% Confidence Interval
for the difference in foundry means (Upper CL Dif 15.9446, Lower CL Dif
10.0554) which can be interpreted as the likely values for the difference in
means if all heats available at Foundry A were compared to all heats available
at Foundry B. Explain why this difference and the confidence interval
represent enumerative purposes.
g) If we wanted to say that in the future that heats produced at Foundry B will
have a mean elongation between 10 and 16 percentage points higher than the
mean elongation for future heats produced at Foundry A (an analytic purpose)
what must be true about the two foundries?
2
JMP Output for Homework 2
Oneway Analysis of % Elongation By Foundry
65
% Elongation
60
55
50
45
40A
B
Foundry
Oneway Anova
t Test
B-A
Assuming equal variances
Difference
13.0000
Std Err Dif
1.4016
t Ratio
9.275204
DF
18
Upper CL Dif
15.9446
Prob &gt; |t|
&lt;.0001
*
Lower CL Dif
10.0554
Prob &gt; t
&lt;.0001
*
Prob &lt; t
1.0000
0.95
Confidence
-15
-10
-5
0
5
10
15
Means and Std Deviations
Std Err
Level
Number
Mean
Std Dev
Mean
Lower 95%
Upper 95%
A
10
46.4000
2.87518
0.9092
44.343
48.457
B
10
59.4000
3.37310
1.0667
56.987
61.813
3
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