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(1) (45 points) A study of children exposed to lead had 3 groups.
1) Control – normal blood lead levels past 2 years
2) Currently exposed – elevated blood lead level this year
3) Previously exposed – elevated lead level last year, normal this year
The measurement of a finger-wrist tapping score for each child gave the following results:
Group
1
2
3
n
63
22
21
mean
55.1
47.6
49.4
Standard deviation
10.9
7.0
10.1
(a) Find the values for the following ANOVA table
Source
Group
Error
Total
df
___
___
___
SS
________
________
________
MS
_______
_______
F
____
(b) Are the two contrasts given below orthogonal? Explain.
(1)
(2)
(3)
Currently vs
previously exposed
0
1
-1
Exposed vs
unexposed
-2
1
1
(2) (35 points)
Suppose that shear strengths of plywood boards have a population standard deviation of
12 psi. We will compare the usual glue to a new glue, using n boards with the new glue
and n boards with the old glue. We are going to test H 0 : 1  2 against the two-sided
alternative H a : 1  2 . Assume the new glue has the same standard deviation as the
usual glue.
a) We want an 80% chance of rejecting H 0 : 1  2 vs H a : 1  2 if in fact the new
glue gives a bonding strength of 4 psi’s greater than the old glue. We also want a 90%
chance that we will not reject H 0 : 1  2 if in fact the glues are equivalent. How
many boards do we need to include in our experiment?
b) If we use n = 80 boards for each glue, find the approximate probability of being able to
reject H 0 : 1  2 vs H a : 1  2 if the new glue adds 4 psi’s to the bonding strength.
c) After seeing the computation in part (a) the company says that this is too many boards
and they cannot afford to do this big of an experiment. Are there any alternatives you
can suggest to the company rather than using this many boards?
(3) (35 points)
A study on mercury contents in the wing muscles of Australian waterfowl gave the
following results.
Species
Shelduck
Shoveler
Blue-billed
Number of birds
6
3
18
Sample mean
9
10
15
Sample
standard deviation
4
5
5
a) Find the grand mean y..
b) Find the pooled variance.
c) Write a general formula for the sum of squares for treatments in a one-way ANOVA.
Use this formula to find the sum of squares for treatment (species) in the data above.
d) Fill in the following ANOVA table.
Source
Species
Error
Total
df
__
__
__
Sum of Squares
__________
__________
__________
Mean squares
________
________
________
F
___
e) Give an approximate range for the p-value for the F computed above. In words, what is
the null hypothesis tested by this F-test? Is the null hypothesis rejected at the 0.05 level?
What does this say about these bird species?
(4)
An experiment is carried out to test the effectiveness of four new drugs on hypertension.
The standard drug (drug 1) is used along with the four new drugs (drug 2, drug 3, drug 4,
and drug 5). The resulting SAS output is given below.
a) Give values for the blanks labeled ?_________?.
Dependent Variable: RESPONSE
Source
DF
Model
Error
??
??
Sum of
Squares
?_______?
?_______?
Root MSE
?_______?
Level of
TREAT
1
2
3
4
5
Parameter
2 vs 1
3 vs 1
4 vs 1
5 vs 1
Estimate
?_______?
10.0000000
2.5000000
3.7500000
N
4
4
4
4
4
Mean
Square
?_______?
?_______?
F value
?_______?
RESPONSE Mean
17.100000
-----------RESPONSE----------Mean
SD
21.5000000
1.29
15.7500000
2.50
11.5000000
2.38
19.0000000
3.16
17.7500000
1.70
T for H0:
Parameter=0
?____?
6.14
1.54
2.30
PR > |T|
Std Error of
Estimate
?_______?
0.0001
0.1454
0.0360
Show your calculations.
b) Is treatment effect significant at the 0.01 level?
c) Suppose the only comparisons of interest are to compare each drug to the standard
drug. Which drugs are significantly different from the standard drug at the 0.05 level
controlling the experimentwise error rate appropriately?
(5) (10 points)
Seven random people (labeled person 2-9 below) had their blood serums bilirubin levels
measured several times each. The following SAS commands create normal plots and
summary statistics for the 7 people.
options linesize = 80;
goptions device = ps
gsfmode = append
gsfname = grafout;
filename grafout ‘bil.ps’;
data;
infile ‘serum.2.dat’;
input person bil;
proc sort;
by person bil;
proc means mean sd cv;
var bil;
by person;
proc rank normal = blom out = bilrank;
var bil;
ranks bilscore;
by person;
proc print data = bilrank;
axis1 label = (‘NORMAL SCORE’);
symbol1 interpol = join value = dot l = 1 c = black;
symbol2 interpol = join value = circle l = 3 c = black;
symbol3 interpol = join value = star l = 5 c = black;
symbol4 interpol = join value = square l = 7 c = black;
symbol5 interpol = join value = circle l = 9 c = black;
symbol6 interpol = join value = star l = 11 c = black;
symbol7 interpol = join value = circle l = 13 c = black;
proc gplot;
plot bil*bilscore = person
/haxis = axis1;
a) The lines on the normal plot are reasonably straight. What does that indicate?
b) The lines on the plot are not as parallel as we would like. What does that indicate?
(6)
Four labs are each sent 5 samples of a homogeneous material to analyze with the
following partial results for the 20 total values.
Source
labs
error
total
df
SS
85.93
34.38
120.31
If we sent 10 samples to another lab and found the sample standard deviation of these 10
samples, about what would you expect the standard deviation to be? (Give a numerical
value.)
(7) (33 points)
The effect of salinity on growth of rainbow trout was examined in the paper “Growth,
Training and Swimming Ability of Young Trout Maintained under Different Salinity
Conditions” (J. Marine Biological Assoc of U.K., 1982) with the following results for
weight gain.
Treatment
Fresh water
Brackish
Sea water
n
12
12
8
mean
8.08
7.86
6.47
st. dev.
1.79
1.76
1.34
a) What is the estimate of the pooled standard deviation, σ?
b) Give a general formula for the treatment or model sum of squares. Find the numerical
value of these data.
c) Test the null hypothesis that all water groups give the same population weight gain.
Give a p-value as near as you can from the table. Do you reject the null hypothesis at
the 0.05 level? What does this tell you about the trout?
d) Test the significance of the comparison “sea water vs others”. That is test
  2
H0 : 1
 3.
2
(8)
(Data courtesy of Dr. Jean Regal) Guinea pigs were given 4 treatments
1. Antigen challenge/ diet1
2. Antigen challenge/ diet 2
3. No antigen challenge/ diet 1
4. No antigen challenge/ diet 2
The resulting data for white blood cell counts (wbc) were as follows:
Treat
1
2
n
7
8
mean
y1. = 1.64
y2. = 1.53
st. deviation
0.09
0.18
3
4
8
8
y3. = 1.01
y4. = 0.93
0.15
0.17
a) Test the null hypothesis of no treatment effect H 0 : 1  2  3  4 at the 0.05 level.
b) Test the null hypothesis of no diet effect, H 0 :
( 1  3 ) 2  4

vs a two-sided
2
2
alternative at the 0.05 level.
c) Give a 99% confidence interval for the average difference between diets 1 and 2,
(   3 )  2   4
H0 : 1

.
2
2
d) Another study is being planned using a new diet and no antigen challenge. About
how many animals need to be used so that the sample mean of the exponential
animals’ wbc count is no more than 0.1 away from the true population mean? (with
90% confidence)
e) A new experiment is being planned with these same 4 treatments as above. About
many animals would be needed in each of the 4 groups in order to test the null
hypothesis that diet 1 is the same as diet 2 against the one-sided alternative that diet
1 gives greater wbc counts? We want a 99% chance of claiming that diet 1 increases
wbc counts if in fact diet 1 adds 0.1 to the average wbc count. In addition, we want
only a 5% chance of claiming diet 1 increases wbc counts if in fact animals on diet 1
and diet 2 have the same population wbc counts.