Assignment
1. (2.5 points) A biochemist is interested in the effects of two different catalysts and two different temperatures on the rate of a chemical reaction. She performs six replicates of a 22 design, making the runs in random order. Analyze the rate of chemical reaction data that follow and draw appropriate conclusions.
|
|
Catalyst |
|||
|
Temperature |
1 |
2 |
||
|
|
31 |
32 |
35 |
36 |
|
350 K |
33 |
38 |
34 |
35 |
|
|
30 |
36 |
39 |
37 |
|
|
47 |
49 |
41 |
44 |
|
360 K |
48 |
48 |
49 |
43 |
|
|
45 |
46 |
40 |
45 |
For the given data following is the output of analysis:
Fractional Factorial Fit: Rate versus Catalyst, Temp
Estimated Effects and Coefficients for Rate (coded units)
Term Effect Coef SE Coef T P
Constant 40.3381 0.7746 52.07 0.000
Catalyst -1.0095 -0.5048 0.7746 -0.65 0.522
Temp 9.3905 4.6952 0.7746 6.06 0.000
Catalyst*Temp -1.7238 -0.8619 0.7746 -1.11 0.279
Analysis of Variance for Rate (coded units)
Source DF Seq SS Adj SS Adj MS F P
Main Effects 2 519.085 521.912 260.96 18.38 0.000
2-Way Interactions 1 17.578 17.578 17.58 1.24 0.279
Residual Error 20 283.962 283.962 14.20
Pure Error 20 283.962 283.962 14.20
Total 23 820.625
Unusual Observations for Rate
Obs Rate Fit SE Fit Residual St Resid
13 47.0000 35.2857 1.4242 11.7143 3.36R
R denotes an observation with a large standardized residual
Estimated Coefficients for Rate using data in uncoded units
Term Coef
Constant -475.095
Catalyst 121.381
Temp 1.45619
Catalyst*Temp -0.344762
Using the analysis above, we can observe that the catalyst has a significant effect on reaction rate (p-value = 0.522 > 0.05. therefore, the rate of reaction is almost the same for both catalysts. However, it differs significantly due to the temperature (P-value = 0.000 < 0.05). Also, there is no significant change in reaction rate when observing both the catalyst and temperature.
2. (2.5 points) An engineer suspects that the surface finish of a metal part is influenced by the feed rate and the depth of cut. She selects three feed rates and four depths of cut. She then conducts a factorial experiment and obtains the following data:
|
|
Depth of Cut (in) |
|||
|
Feed Rate (in/min) |
0.15 |
0.18 |
0.20 |
0.25 |
|
|
79 |
84 |
87 |
104 |
|
0.20 |
69 |
73 |
93 |
109 |
|
|
65 |
78 |
97 |
101 |
|
|
97 |
103 |
104 |
109 |
|
0.25 |
91 |
109 |
113 |
115 |
|
|
93 |
93 |
100 |
104 |
|
|
104 |
109 |
113 |
119 |
|
0.30 |
103 |
104 |
115 |
116 |
|
|
107 |
100 |
104 |
112 |
a. Analyze the data and conclusions. Use = 0.05.
# R command
Part=c(79,69,65,84,73,78,87,93,97,104,109,101,97,91,93,103,109,93,104,113,100,
109,115,104,104,103,107,109,104,100,113,115,104,119,116,112)
Feed. Rate=factor(c(rep(0.20,12), rep(0.25, 12), rep(0.30, 12)))
Depth.of.cut=factor(c(rep(c(rep(0.15,3), rep(0.18,3), rep(0.20,3), rep(0.25,3)),3)))
model=aov(Part~Feed.Rate+Depth.of.cut+Feed.Rate*Depth.of.cut)
summary(model)
## End
Note: the p-values for the main effects of the feed rate and the depth of cut are 0.000 and 0.000. therefore, we can conclude that there is a statistically significant effect on the main effects of feed rate and the depth cut on the surface finish. Also, the p-value of the interaction effects of feed rate and the depth cut is 0.018, which is less than the p-value of 0.05. thus we can conclude that the interaction of the feed rate and the depth of cut has a significant effect on the surface finish of a metal part at the 0.05 significance level.
3. (2.5 points) the effect of three different grades of gasoline on fuel economy in car engines is being studied. Fuel economy is measured using brake-specific fuel consumption after the engine has been running for 15 minutes. Five different car engines are available for the study, and the experimenters conduct the following randomized complete block design.
|
|
Car |
||||
|
Gasoline grade |
1 |
2 |
3 |
4 |
5 |
|
1 |
0.545 |
0.685 |
0.530 |
0.445 |
0.550 |
|
2 |
0.510 |
0.644 |
0.497 |
0.339 |
0.522 |
|
3 |
0.523 |
0.605 |
0.498 |
0.410 |
0.520 |
(a) Analyze the data from this experiment.
The table provides the data on the effect of 3 different gasoline grades on fuel economy in the car engines of 5 cars. Here, the correction is supposed to be applied as the familywise type 1 error rate is greater than the per analysis error rate.
(b) Use the Bonferroni correction method to make comparisons among the three grades of gasoline to determine specifically which grades differ in brake-specific fuel consumption.
Assuming that the per analysis alpha rate is 1.5.
Divide the per analysis alpha rate by the number of statistical analyses performed.
For example, 1.5/3 = 0.5.
The observed values less than the corrected p-value (i.e., 0.5) are concluded to be statistically significant. Therefore, the Bonferroni correction method is:
|
Gasoline Grade |
Cars |
||||
|
|
1 |
2 |
3 |
4 |
5 |
|
1 |
- |
- |
- |
- |
- |
|
2 |
- |
- |
0.497 |
0.339 |
- |
|
3 |
- |
- |
0.498 |
0.410 |
- |
4. (2.5 points) suppose that we are testing H0: µ = µ0 versus H1: µ ≠ µ0. Calculate the P-value for the following observed values of the test statistic:
(a) Z0 = 4.9
H0: µ = µ0
H1: µ ≠ µ0
from the table at Z0 = 4.9
p=p(z>4.9) = 1-0.99999952
= 0.00000048
(c) Z0 = -3
From the table at Z0 = -3
P=p(z>-3) = 1-0.001349
=0.99865
(d) Z0 = 4.3
From the table at Z0 = 4.3
P=p(z> 4.3)
=0.000009
(e) Z0 = 3.9
From the table at Z0 = 3.9
P=p(z>3.9)=1-0.99995
=0.00005
(f) Z0 = -0.5
From the table Z0 = -0.5
P=p(z>-0.5) = 1-0.30854
=0.69146