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.  

35

36

48

49

45

Catalyst
Temperature			1			2
3 1	32		35		36
350 K	33	38	34
30	39	37
47		49	41	44
360 K		48	43
	45	46	40

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.

69

52 0.7746 6.06 0.000

Catalyst*Temp -1.7238 -0.8619 0.7746 -1.11 0.2

79

Analysis of Variance for Rate (coded units)

Source DF Seq SS Adj SS Adj MS F P

Main Effects 2 519.085 521.

91

2 260.96 18.38 0.000

2-Way Interactions 1 17.5

78

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:

0.20

97

104

109

0.25

109

93

93

104

104

109

113

103

104

115

100

104

Depth of Cut (in)
Feed Rate (in/min)	0.15	0.18		0.20		0.25
79	84	87						104
69	73			93				109
65	78		97	101
	103
91		113		115
	100
119
0.30	116
107	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.

1

2

3

Car
Gasoline grade		1		2		3		4		5
0.545	0.685	0.530	0.445	0.550
0.510	0.644		0.497		0.339	0.522
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:

1

2

4

5

1

–

–

–

2

–

–

0.497

0.339

–

3

–

–

0.498

0.410

–

Gasoline Grade	Cars
	3
						–				–

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