Download Zar Chapter 17 Exercises KEY

Zar Chapter 17 Exercises KEY
17.1
1. Model
Yi  0  1 X i  ei
a. Random variables of interest
Explanatory measurement: Xi = Environmental temperature (C) of the i-th
randomly sampled bird
Response Measurement: Yi = Oxygen consumption (ml/g/hr) of the i-th
randomly sampled bird at environmental temperature Xi
b. Parameters of interest
Intercept: 0  E Yi | X i  0 = the mean oxygen consumption (ml/g/hr) when
temperature = 0 C.
Slope: 1  d E Yi | X i  dX i  change in mean oxygen consumption per unit
change in temperature (ml/g/hr/C).
c. Assumptions
i. Location: E  ei   0  E Yi | X i   0  1 X i
ii. Dispersion: Var  ei    2
iii. Shape: ei distributed Normally
2. Hypotheses
H0: 1  0 versus HA: 1  0
3. Formulate
a. Test Criterion
F  MSM MSE
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b. Estimators
ˆ1 
S xy
S xx

  X  X Y  Y 
 X  X 
i
i
2
i
ˆ0  Y  ˆ1 X
4. Design
  0.05
n8
5. Perform the study, gather data, and compute
Using JMP data table SLR_Zar_Exercise_17_1.JMP
-18
-15
-10
-5
0
5
10
19
O2
Consumption
(ml/g/hr)
5.2
4.7
4.5
3.6
3.4
3.1
2.7
1.8
5.5
5
O2 Consumption
(ml/g/hr)
Temp
(C)
4.5
4
3.5
3
2.5
2
1.5
-20
-15
-10
-5
0
5
10
15
20
Temp (C)
Analysis of Variance
Source
Model
Error
C. Total
DF
1
6
7
Sum of
Squares
8.7
0.2
8.9
Mean
Square
8.745
0.028
F Ratio
309
P
<0.0001
Parameter Estimates
Term
Intercept
Temp (C)
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Estimate
3.4714
Std Error
0.06012
t Ratio
57.7
P
<0.0001
-0.0878
0.00499
-17.6
<0.0001
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6. Conclusion
a. The best estimate of the intercept is ˆ0  3.47 ml/g/h.
The best estimate of the slope is ˆ1  0.0878 ml/g/h/°C.
Thus, the prediction equation is
yˆ  3.47  0.0878 x
b. There is highly significant statistical evidence that the change in mean oxygen
consumption per unit change in temperature is different from 0 ml/g/hr/C
(F = 309, P < 0.0001).
c. There is highly significant statistical evidence that the change in mean oxygen
consumption per unit change in temperature is different from 0 ml/g/hr/C
(t = −17.6, P < 0.0001).
d. (not assigned) The standard error of estimate is the root mean square error,
ˆ  sY  X  MSE  0.02831  0.168
We need this for 17.2.
e. The coefficient of determination is R2 = MSModel/MSTotal = 8.7/8.9 = 0.98 =
98% of the variation in mean oxygen consumption (ml/g/hr) is explained by
the variation in temperature (°C).
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17.2
a. What is the mean rate of oxygen consumption in the population for birds at
15°C?
yˆ  3.47  0.0878 x  3.47  0.0878 15  2.16
This is computed in JMP file SLR_Zar_Exercise_17_2.JMP
5.5
O2 Consumption
(ml/g/hr)
5
4.5
4
3.5
3
2.5
2
1.5
-20
-15
-10
-5
0
5
10
15
20
Temp (C)
b. What is the 95% confidence interval for this mean rate?
As explained in Zar (1999) Example 17.5A,
 1  X  X 2 
 1 15   1.752 
i
  0.0283  
  0.1026
sYˆ  MSE  
i
S xx
1135.5
n

8





and the 1     95% confidence limits are
Yˆi  tn 2,1a 2 sYˆ  2.16  t6, 0.975  0.1026  2.16  2.447  0.1026
i
 2.16  0.251  1.91, 2.41
We are 95% confident that the mean oxygen consumption of all birds at 15°C
is between 1.91 and 2.41 ml/g/hr.
c. If we randomly chose one additional bird and measured its oxygen
consumption at 15°C, what we predict its oxygen consumption would be?
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yˆ  3.47  0.0878 x  3.47  0.0878 15  2.16
d. We can be 95% confident of this value lying between what limits?
As explained in Zar (1999) Example 17.5C,
 1  X  X 2 
 1 15   1.752 
i
  0.0283 1  
  0.1971
sYˆ  MSE 1  
i 1
S xx
1135.5
 n

 8





 
and the 1     95% confidence limits are
Yˆi  tn 2,1a 2 sYˆ  2.16  t6, 0.975  0.1971  2.16  2.447  0.197
i
 2.16  0.482  1.68, 2.64
We are 95% confident that the oxygen consumption of one randomly selected
birds at 15°C would be between 1.68 and 2.64 ml/g/hr.
JMP Computation of Confidence and Prediction Intervals
The confidence intervals of Zar Exercises 17.2 b and 17.2 d can be calculated in JMP
using the Fit Model Platform as follows.
First, add the value(s) of Xi for which you wish to estimate μY|X or predict Yˆi | X i . In the
present example, that value is Xi = 15.
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JMP
> Analyze > Fit Model
> Pick Role Variables > Y = O2 Consumption (ml/g/hr)
> Construct Model Effects > Add Temp (C)
> Run
> Fit Least Squares Hotspot
> Save Columns > Prediction Formula or Predicted Values
> Save Columns > Mean Confidence Limit Formula
> Save Columns > Indiv Confidence Limit Formula
The Save Column commands add columns to the JMP Data Table as follows.
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17.3
Computations in /data/Examples/SLR_Zar_Exercise_17_3.JMP and
/data/Examples/SLR_Zar_Exercise_17_3_Stacked.JMP
Impulse freq
(number/sec)
225 230 239
22
251
259
265
23
266
273
280
25
27
28
30
287
301
307
324
295
310
313
330
302
317
325
338
340
320
Impulse freq
(per sec)
Temp
(C)
20
300
280
260
240
220
18
20
22
24
26
28
30
32
Temp (C)
Anova (SLR)
Source
DF
Sum of
Squares
Mean
Square
F Ratio
P
R2
Model
1
21576.9
21576.9
311.0
<0.0001
0.942
Error
19
1318.4
69.4
Total
20
22895.2
0.058
1.000
Parameter Estimates
Term
Intercept
Temp (C)
Estimate
44.27
Std Error
13.91
t Ratio
3.2
Prob>|t|
0.0049
9.73
0.55
17.6
<.0001
a. The best estimate of the intercept is ˆ0  44.3 impulses/sec.
The best estimate of the slope is ˆ1  9.73 impulses/sec/°C.
Thus, the prediction equation is
yˆ  44.3  9.73x
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b. There is highly significant statistical evidence that impulse rate (number/sec) is
linearly related to temperature (C) (P < 0.0001).
c. The standard error of estimate is the root mean square error,
ˆ  sY  X  MSE  69.4  8.33
d. The coefficient of determination is R2 = 94.2% of the variation in impulse frequency
(per sec) is explained by the variation in temperature (°C).
e. (Not assigned) Test H0: The population regression is linear [i.e., for lack of fit].
Yij      0  1 X i       i    0  1 X i    eij
Yij
 
   0  1 X i       i    0  1 X i    eij

 eij
 Regression i    Lack of Fit i 
 

 
i
 eij
Anova SLR and Lack of Fit
SS
MS
F
P
R2
21576.9
21576.9
311.0
<.0001
0.94
(21 – 2) = 19
1318.4
69.4
Lack of Fit
7–2=5
513.0
Pure Error
21 – 7 = 14
805.3
(21 – 1) = 20
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Source
Regression
Residual, Error
df
1
0.06
102.6 1.8
57.5
0.18
0.02
0.04
= within groups
Total
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1-Way Anova
Source
Lack of Fit
df
Between group
SLR
Source
df
Source
df
Regression
1
Regression
1
Lack of Fit
(k − 2)
(k – 1)
means
Residual,
Residual,
Within groups,
(N – k)
Pure Error
(N – k)
Error
(N – 1)
Total
(N – 1)
Total
(N – 2)
Error
Total
(N – 1)
Both one-way Anova and SLR model the relationship among the group means as a
function of the explanatory variable. Here’s how those two models differ:
One-way Anova
SLR
explanatory variable treated as categorical
Explanatory variable is treated as a
measurement
More flexible, i.e., assumes less, i.e.,

Less flexible, i.e., assumes more, i.e.,

more degrees of freedom for the
model, fewer for residual

model, more for residual

allows any relationship among the
group means

assumes that the group means fall
on a straight line

less restrictive model
Explanatory variable can be qualitative
fewer degrees of freedom for the
more restrictive model
Explanatory variable must be quantitative
(nominal or ordinal) or quantitative
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