Problem 1. 4.8 (pg 169)
Summary of Fit
0.088012
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum
Wgts)
0.074194
798.2601
2821.296
135
Analysis of
Variance
Source
Model
Error
C. Total
Parameter
Estimates
Term
Intercept
EXCHRATE
PRICE
DF
Sum of
Squares
2
132
8117338
4058669
84112922
637219
92230260
134
Mean Square
F Ratio
6.3693
Prob > F
0.0023
Estimate
Std Error
t Ratio
Prob>|t|
3361.9317
1.8691544
-2413.837
633.1943
4.223047
846.4798
5.31
0.44
-2.85
<.0001
0.6588
0.0051
(a) From the result of JMP above, the estimated regression equation is:
SHIPMENT=3361.9317+1.8691544* EXCHRATE -2413.837* PRICE
(b)
Hypothesis
H0: EXCHRATE = PRICE = 0
Ha: At least one of the slopes is not zero
Decision Rule
F(0.05, 2, 135-2-1) = 3.06
Reject H0 if the sample F-statistic f > 3.06
Test-Statistic
Sample F-ratio = 6.3693
Decision
Since 6.3693> 3.06, we reject H0 and conclude that some of the predictors coefficients are
significantly different from 0.
(c)
Hypothesis
H0: EXCHANGE =0
Ha: EXCHANGE is not equal to 0 in the full model.
Decision Rule
t(0.025, 135-2-1) = 1.98
Reject H0 if the sample t-ratio t > 1.98 or t<-1.98
Test-Statistic
Sample t-ratio = 0.44
Decision
We retain H0 and conclude that the coefficient of EDUC is not significantly different from 0 when
taking account of PRICE.
(d)
The R-square value is 0.088012, so 8.8% of variation has been explained.
(e)
From the result of jump, PRICE=-2413.837 and SE( )= 846.4798. Also t(0.025, 135-2-1) = 1.98 and then the
95% CI for the regression coefficient of PRICE is
(-2413.837-846.4798*1.98, -2413.837+846.4798*1.98) = (-4089.867, -737.807).
PRICE
2. Problem 4.8 (pg 169)
(a)
From the result of JMP, the estimated regression equation is:
Longevity = 3.2438212+0.4508583* Mother + 0.4111835* Father +0.016553* Gmothers+0.0868583* Gfathers
(b and c)
Using JMP it is easy to compute the 95% prediction interval for the longevity of the individual man;
it is (67.76, 79.72); and the 95% confidence interval for the average longevity of the men in this
group is (70.95, 76.53).
(d)
Hypothesis
H0: gmother=gfather=0
Ha: At least one of them is zero
For the JMP result in the beginning of the solution, in the full model =674.2430.
When excluding the factors of gmothers and/or gfathers, using JMP we can get:
Analysis of
Variance
DF
Source
Model
Error
C. Total
Sum of
Squares
2
97
99
Mean Square
1917.0052
958.503
686.7548
7.080
2603.7600
F Ratio
135.3828
Prob > F
<.0001
So the F-statistic is
=()/2/95=(686.7548674.2430)/2674.2430/95=0.88
Testing at level .05, F(.05, 2, 100-4-1) = 3.09. We reject H0 if the sample F-statistic f > 3.09
Decision
Since 0.88< 3.09, we retain H0 and conclude that under level .05, gmothers and gfathers are not
useful for predicting longevity once mother and father have been taken into account.
(e) i.e. (h)
We fit a model with only gmother and father into account
Summary of Fit
0.356257
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum
Wgts)
0.342984
4.156913
72.32
100
Analysis of
Variance
Source
Model
Error
C. Total
DF
Sum of Squares
Mean Square
2
97
927.6075
1676.1525
463.804
17.280
99
2603.7600
H0: gmother=gfather=0
Ha: At least one coefficient is not equal to 0.
Decision Rule
F(.05, 2, 100-2-1) = 3.09
F Ratio
26.8406
Prob > F
<.0001
Reject H0 if the sample F-statistic f > 3.09
Test-Statistic
Sample F-ratio = 26.8406
Decision
Since 26.8406> 3.09, we reject H0 and conclude that some of the predictors coefficients are
significantly different from 0 and this model is useful for predicting mens longevity, which
means that if the values of mother and father arent known, but the values of gmothers and
gfathers are known, then these values of gmothers and gfathers are useful for predicting
longevity.
Explanation
Why are gmothers and gfathers insignificant when mother and father are known, but
become significant when mother and father are unknown? From the former result, we can
conclude that gmothers and gfathers are useful to describe the longevity of men. However, the
latter implies that, mother and father can cover some of the information in gmothers and
gfathers, and thus gmothers and gfathers are insignificant when mother and father are
known. This is understandable because the the ages at death of the mother and father are highly
related to the age at death of the grandparents.
3. Problem 17 (4.14)
(0) We run a forward stepwise regression model where we set the probability to enter to be 0.05. Our
results indicate that we should include all three variables, ERA, BA, an HR, into predicting WINS
(aka our Y)
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
0.896931
0.885039
5.00226
80.83333
30
Analysis of Variance
Source
Model
Error
C. Total
DF
3
26
29
Sum of Squares
5661.5790
650.5876
6312.1667
Mean Square
1887.19
25.02
F Ratio
75.4195
Prob > F
<.0001*
Parameter Estimates
Term
Intercept
HR
BA
ERA
Estimate
-21.88065
0.0975914
606.30599
-16.89736
Std Error
28.92622
0.035722
100.7689
1.758246
t Ratio
-0.76
2.73
6.02
-9.61
Prob>|t|
0.4562
0.0112*
<.0001*
<.0001*
The model tells us that to be a successful baseball team, you should have higher home runs, higher
average and batting, have low average ERA.
(a)
Our R2 is 0.896931, which is very high. This indicates that the regression model we built is a good fit
to the data and be an excellent predictor of WINS
(b)
The predicted number of wins for Team A, A, is 68.2285357 while the predicted number of wins for
Team B, B, is 91.4148184. The predicted difference is simply 91.4148184 68.2285357 = 23.18628.
The prediction interval is simply
B A +/- 2 * RMSE * sqrt(2) = 23.18628 +/- 2*5.00226 *sqrt(2) = (9.03775, 37.33481)
OR (slightly more accurate one where we actually obtain the critical t value and use Sp of each team
instead of approximating Sp by RMSE)
B A +/- 2.055529 * sqrt(Sp(A)2 + Sp(B)2)
= 23.18628 +/- 2.055529 *7.580458
= (7.604429, 38.76813)
4. Problem 4.8 (pg 169)
Log10(Cost/SqFt) = 1.3050734 - 0.0206945*Log10(SqFt) + 0.0319822*(Log10(SqFt)-4.18431)^2
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
0.14286
0.135067
0.039938
1.227443
223
Analysis of Variance
Source
Model
Error
C. Total
DF
2
220
222
Sum of Squares
0.05848530
0.35090492
0.40939021
Mean Square
0.029243
0.001595
F Ratio
18.3337
Prob > F
<.0001*
Parameter Estimates
Term
Intercept
Log10(SqFt)
(Log10(SqFt)-4.18431)^2
Estimate
1.3050734
-0.020695
0.0319822
Std Error
0.022416
0.005164
0.008855
t Ratio
58.22
-4.01
3.61
Prob>|t|
<.0001*
<.0001*
0.0004*
(a)
For a house with 2,500 sqFt, we get $17.96933/SqFt with a 95% prediction interval of
(1.17514488, 1.33391885) in log scale. After unlogging it, we get ($14.9635/SqFt,
$21.57341/SqFt). The 50% prediction interval is
+/- t * RMSE = 1.25453186 +/- (0.6756066)*0.039938 = (1.227549, 1.281514)
After unlogging, we get ($16.88686/SqFt, $19.12115/SqFt) as our interval.
OR (slightly more accurate one where we use Sp instead of approximating Sp by RMSE)
+/- t * sqrt(Sp) = 1.25453186 +/- 0.6757066 * 0.04028149 = (1.227313, 1.281750)
(b)
There are two possible residual plots. Both are correct for the purposes of studying normality and
homoscedasticity. Based on the plots, the residuals are normally distributed because it fits inside the
butterfly lines and the regression looks homoscedastic because there is no spreading in the
residual plots.
Residual by Predicted Plot
Log10(Cost/
SqFt) Res idual
0.15
0.10
0.05
0.00
-0.05
-0.10
1.1 1.15 1.2 1.25 1.3 1.35 1.4
R es id ual
Log10(Cost/SqFt) Predicted
0.10
0.00
-0.10
10
20
Log10(SqFt)^2
2.33
0.98
1.64
0.95
1.28
0.9
0.67
0.0
-0.67
0.8
0.5
0.2
-1.28
0.1
-1.64
0.05
0.02
-2.33
-0.1
-0.05
(c) (Optional)
0
0.05
0.1
0.15
No rm al Qu a ntil e Pl ot
And the QQ plot is
As the number of square feet increases, the marginal utility of renting the office space decreases.
Marginal utility, which is the change in utility for every Sqft increase in office space, goes down
and using a quadratic term takes into account the presence of marginal utility.
5. Problem 4.8 (pg 169)
(a)
H0: %<=20 = %>=50 = 0 Ha: at least one of the two betas
F=36.618, p-value<0.0001, so we reject the null hypothesis and conclude that class sizes are
useful to predict Grad.Rate.
(b)
Bs GradRate will be higher than As by 1.2095*0.15=0.181425.
(c)
The 95% confidence will be
(d)
Grad.Rate=-0.973+1.325Fresh.Ret_0.000397SAT75
(e)
Ha: at least one of the two betas
F=501.22, p-value<0.0001, so we reject the null hypothesis and conclude that the overall model
under Analysis II useful for predicting Grad.Rate.
(f)
So we reject the null hypothesis and conclude that SAT75 is a useful predictor of Grad.Rate after
controlling for Fresh.Ret.
(g)
Analysis of Variance
Source
DF
Model
4
Error
168
C. Total
172
Sum of Squares
F-Ratio
4.684
270.975
0.72
6
5.410
SST=5.410 from Analysis II SST; SSR=SST-SSE=5.410-0.726=4.684
F=MSR/MSE=(4.684/4)/(0.726/168)=270.975
(h)
Ha: at least one the two betas
So we reject the null hypothesis and conclude that there be statistically significant information
about class sizes contained in the two variables %<=20 and %>=50 after controlling for
Fresh.Ret and SAT75.
(i)
We choose the three variables based on how backward stepwise regression choose these three
variables. Backward stepwise regression chooses the t with the lowest ratio to go from a four
variable model to a three variable model. Hence, we get Fresh.Ret, SAT75, %<=20
(j)
Analysis of Variance
Source
DF
Sum of Squares
F-Ratio
Model
3
4.68397
363.43
Error
169
0.72603
C. Total
172
5.410
SST=5.410 from Analysis II;
From ANOVAIII, for %>=50 t-ratio=0.08
If we consider two regression models: Analysis III and Analysis IV
Partial F ratio=
But partial F ratio=(t ratio)^2 where the t ratio comes from the variable we take out.
We solve the equation and get SSEr=0.72603, which is SSE in Analysis IV
SSR=SST-SSE=5.410-0.72603=4.68397
F=MSR/MSE=(4.68397/3)/(0.72603/169)=363.43
Note that F is very high (probably higher than the critical value of F) and hence, the overall
model is useful.
(k)
Three-variable model R^2=4.68397/5.410=0.8658
The Best Two-variable model R^2=4.625/5.410=0.8549 (Fresh Ret. And SAT 75)
To test if %<=20 is significant in the three-variable model
So we reject the null hypothesis and conclude that three-variable model is significantly better
than the two-variable model.