Stat 3411
Spring 2009
Exam 3
Name __________________________________
(1) Out of 235 nickel-cadmium cell made at a plant under a trial set of operating conditions, 9 of the 235 cells
had shorts. Give an unsimplified numerical expression for a 95% confidence interval for the fraction of cells
that have shorts using these operating conditions.
9
pˆ 
For proportions  2  p 1  p   pˆ 1  pˆ 
235
2
SE pˆ 
n
pˆ  1.96

pˆ 1  pˆ 
n
pˆ 1  pˆ 
235
(2) A sample of 5 specimens of cold-rolled steel had a mean Rockwell hardness of 19.8 and a standard
deviation of 3.5. Give an unsimplified numerical expression for a 95% prediction interval for the hardness of
the next specimen of cold –rolled steel specimen produced under these conditions.
For predicting the next value there are 2 sources of uncertainty or variance
2

Estimating the mean Var  X  

Variability of individual values Var  X new    2

All together Var  X new  X    2 
19.8  2.776 3.52 
n
2
n
 S2 
S2
n
2
3.5
5
(3) According to the Encyclopedia of Optical Engineering, light intensity in water should decrease
exponentially with depth.
Light Intensity  I 0e  K *depth
I0 is the light intensity at the surface, and K is sometimes called the "extinction coefficient".
The following are values of light intensity measurements in Lake Superior taken at different depths.
Depth
5
6
7
8
9
10
11
12
13
Light
538.4
425.1
351.2
285.0
236.3
195.6
163.5
133.3
112.5
Light Intensity
600
400
200
0
4
6
8
10
12
14
Depth
Given these data, how would you estimate values for I0 and K for fitting these data with the equation
Light Intensity  I 0e  K *depth ? You don't need to do anything with these numbers. Just describe the steps you
would take to find estimated values of the parameters I0 and K for these data.
Y = Light intensity
X = Depth
Ln(Y) = Ln(I0) – K*X = b0 + b1X
Use Excel to fit ln(Y) versus X
Ln  I 0   b0
I 0  eb0
 K  b1 K  b1
(4) We have collected data on the effects of diameter (X1) and feed rate (X2) on thrust required (Y) to drill an
aluminum alloy and fit the regression model Ln(Y) = b0 + b1Ln(X1) + b2Ln(X2) using Excel.
X1
0.250
0.406
0.406
0.250
0.225
0.318
0.450
0.318
0.318
0.318
0.318
0.318
X2
0.006
0.006
0.013
0.013
0.009
0.005
0.009
0.017
0.009
0.009
0.009
0.009
Y
230
375
570
375
280
225
580
565
400
400
380
380
(a) We plan to use the Excel Data Analysis Regression tool to fit the model Ln(Y) = b0 + b1Ln(X1) + b2Ln(X2)
in order to derive a model for predicting Ln(Y). What assumptions need to hold in order for the resulting model
we get from Excel to be perfectly valid for predicting Ln(Y)?
1. The equation fitted is a reasonable representation of the relationship.
2. The residuals are
a. Normal
b. Equal Variance
c. Independent
(b) The fitted model results in the following information.
ANOVA
df
Regression
2
SS
1.088
9
0.038
  y  yˆ 
2
Residual
2
1.126
 y
Total
11
i
i
 y
Summarize how well the required Ln(thrust) is predicted by Ln(diameter) and LN(feed rate) for these data.
R2 
1.088
 0.966  96.6%
1.126
The diameter and feed rate variables explain 96.6% of the variation in required thrust. A very good fit.
(c) What do the following plots tell us about the degree to which the required assumptions for the usual Excel
regression analysis are met for these data? For each plot tell what that plot tells you about assumptions for the
analysis.
0.1
0.05
Residuals
Ln(X1) Residual Plot
0
-1.5
-1.3
-1.1
-0.9
-0.7
-0.05
-0.5
-0.1
Ln(X1)
-0.15
This plot plots looks fine. The fitted equation is not consistently missing points in some range of Ln(X1) values.
Variances seem fairly constant.
Ln(X2) Residual Plot
0.05
0
-5.5
-5
-4.5
-0.05
Ln(X2)
-4
Residuals
0.1
-0.1
-0.15
The equation is consistently missing values in the midrange of Ln(X2) values on the high side. The relationship
with X2 could be improved. Variances still seem fairly constant.
0.1
Residual
0.05
0
5
5.5
6
6.5
-0.05
Predicted
-0.1
-0.15
This one isn't too bad. The fitted plane isn't consistently missing high or low predicted values.
(d) What else would you need to do to verify other assumptions of the analysis?
Normality: Normal plot of residuals
Independence: Checking how the data were collected. Were 2 holes drilled into the piece of aluminum?
(e) The Excel output includes
Intercept
Ln(X1)
Ln(X2)
Coefficients
10.267
0.994
0.677
Standard
Error
0.287
0.095
0.056
t Stat
35.833
10.507
12.061
P-value
5.081E-11
2.368E-06
7.372E-07
Lower
95%
9.618
0.780
0.550
Upper
95%
10.915
1.208
0.804
Write the fitted equation for predicting Y (not Ln(Y)) from X1 and X2. Write the equation in its simplest form.
Y  e10.267 X 10.994 X 20.677  28, 767 X 10.994 X 20.677
(f) What is the average effect on required thrust (Y) due to increasing the Feed Rate (X2) by 3%? I am looking
for the general effect on Y (not Ln(Y)) no matter what initial values we have for X1 and X2 .
0.994
X 20.677
Ynew 28, 767 X 1 1.03* X 2 

 1.030.677  1.0202
0.994
0.677
Ynew
28, 767 X 1 X 2
0.677
2.02% increase
or approximately 0.677 *3%  2.03%
(5) According to the Annis model, the ratio of flight times for folded vs clipped construction paper helicopters
should be 1.19. That is, flight times of helicopters without a paper clip would be 19% longer than for
helicopters with a paper clip. The data collected from the class were as follows.
Helicopter
1
2
3
4
5
Log Scale
Folded
0.414
0.352
0.286
0.370
0.364
6
7
0.389
0.288
0.281
0.301
0.108
-0.013
8
9
10
11
12
13
0.399
0.284
0.392
0.393
0.341
0.412
0.311
0.322
0.306
0.313
0.309
0.324
0.088
-0.038
0.086
0.080
0.033
0.088
14
15
0.346
0.350
0.268
0.358
0.078
-0.007
0.359
0.0019
0.044
0.296
0.0010
0.033
0.063
0.0022
0.049
Mean
Variance
St Deviation
Log Scale Log Scale
Clipped Difference
0.300
0.113
0.300
0.052
0.227
0.059
0.276
0.093
0.242
0.122
(a) Find a 95% confidence interval for the average difference in Ln(time) for folded and clipped helicopters.
Note: Use whichever numbers above make most sense to use here. You won't use all of the summary
information given above.
With these paired data
0.049
15
0.063  2.145*0.013
0.063  2.145
0.036 to 0.090
(b) Find a confidence interval for the (geometric) average ratio of times for folded and clipped helicopters.
100.036 to 100.090
1.09 to 1.23
(c) Based on this confidence interval do we have strong evidence for rejecting the Annis model?
Why or why not?
Since the ratio 1.19 is in this interval, the Annis model is not rejected.