HW2
2.6
Parameter Estimates
Variable DF Parameter Standard t Value Pr > |t| 95% Confidence Limits
Estimate
Error
Intercept
1
10.20000
0.66332
15.38 <.0001
8.67037
11.72963
transfer
1
4.00000
0.46904
8.53 <.0001
2.91839
5.08161
This table can help you do the data analysis.
a. t(0.975; 8) = 2.306, b1 = 4.0, s{b1} = 0.469, so we get 4±2.306(0.469). Then we get the interval
[2.918,5.082]
b. H0: β1 = 0, Ha: β1 ≠0.
t* = (4.0-0)/0.469=8.529. If |t*|≤2.306 conclude H0, otherwise Ha. Conclude Ha.
P-value=0.00003
c. b0 = 10.20, s{b0} = 0.663, so we get 10.20±2.306(0.663), the interval is 8.671≤β0≤11.729
d. H0: β0 ≤ 9, Ha: β0 ≥ 9.
t*= (10.20-9)/0.663 = 1.810. If t*≤ 2.306 conclude H0, otherwise Ha. Conclude H0. Pvalue=0.0053
e. H0: β1 = 0: δ= |2-0|/0.5 = 4, power =0.93 (or 0.94)
H0: β0 ≤ 9: δ= |11- 9|/0.75 = 2.67, power =0.50304
Obs
n sigb0 beta0 df alpha
1 10
0.75
11
delta
t_c
power
8 0.025 2.66667 2.75152 0.50304
2.8
a. H0: β1 = 3.0, Ha: β1≠3.0.
t*= (3.57-3.0)/0.3470 = 1.643 ;t(.975; 23) = 2.069. If |t*|≤2.069 conclude H0, otherwise Ha. Conclude
H0.
b. δ=|0.5|/0.35=1.43, power=0.16+((1.43-1)/(2-1))(0.48-0.16)=0.297(or0.3)
c. The test in part (a) is based on the slope=3, and the F* in the printout only explain the slope of the
model that is fitted to the original dataset. Therefore F* is irrelevant to the test in (a)
2.10
a. Prediction
b. Mean response
c. Prediction
2.13
Output Statistics
Obs ACT Dependent Predicted
Std Error 95% CL Mean
Variable
Value Mean Predict
121
.
28
3.2012
0.0706
Residual
3.0614 3.3410
.
a.
̂
̂
̂
𝑌
ℎ = 3.2012, s {𝑌ℎ } = .0706, t(.975; 118) = 1.9803, 3.2012± 1.9803(.0706); 3.0614 ≤E{𝑌ℎ } ≤3.3410
(or use the SAS output above)
b.
Output Statistics
Obs ACT Dependent Predicted
Std Error 95% CL Predict Residual
Variable
Value Mean Predict
.
121 28
3.2012
0.0706
1.9594
4.4431
b. s{pred} = .6271, 3.2012 ±1.9803(.6271), 1.9594 ≤ 𝑌ℎ(𝑛𝑒𝑤) ≤4.4430 (or get it directly from the table
above)
c. Yes, yes
d.
Obs
n alpha dfn dfd
1 120
0.05
t
w1
w
alphat
t_c
2 118 1.98027 6.14618 2.47915 0.014582 2.47915
W2 = 2F(.90; 2, 118) = 2(3.0731) = 6.1462, W = 2.4792, 3.2012±2.4792(0.07060),
3.0262 ≤β0 + β1Xh ≤ 3.3762, yes, yes
2.17
Greater, H0: β1 = 0
2.21
No, because R-sqr is only a relative reduction from SSTO and provides no information about absolute
precision for estimation a mean response or prediction a new observation. (KNNL 76)
2.25
a.
Analysis of Variance
Source
DF
Sum of
Squares
Mean F Value Pr > F
Square
Model
1 160.00000 160.00000
Error
8
Corrected Total
9 177.60000
17.60000
72.73 <.0001
2.20000
SSM+SSE=SST, DFm+DFe=DFt, DFm+DFe+1=number of the total observations
b.
H0: β1 = 0, Ha: β1 ≠ 0
F* = 160.00/2.20 = 72.727, F(.95; 1, 8) = 5.32. If F* ≤ 5.32 conclude H0, otherwise Ha.
Conclude Ha.
c.
t* = (4.00 - 0)/.469 = 8.529, (𝑡 ∗)2 = (8.529)*(8.529) = 72.7 = F*
(You also can use the t* shows in the table below, t*=8.53)
Parameter Estimates
Variable DF Parameter Standard t Value Pr > |t| 95% Confidence Limits
Estimate
Error
Intercept
1
10.20000
0.66332
15.38 <.0001
8.67037
11.72963
transfer
1
4.00000
0.46904
8.53 <.0001
2.91839
5.08161
d.
1.48324 R-Square 0.9009
Root MSE
Dependent Mean 14.20000 Adj R-Sq 0.8885
10.44535
Coeff Var
R-square = .9009, r = .9492, 90.09%
2.28
a.
Output Statistics
Obs age Dependent Predicted
Std Error 95% CL Mean Residual
Variable
Value Mean Predict
39
79.0000
60
84.9468
1.0552
82.8347
87.0590
-5.9468
̂
̂
𝑌
ℎ = 84.9468, s {𝑌ℎ } = 1.05515, t (.975 ; 58) = 2.00172,
84.9468 ± 2.00172(1.05515), 82.835≤E{𝑌̂
ℎ } ≤ 87.059 (also can get from the SAS output above)
b.
Output Statistics
Obs age Dependent Predicted
Std Error 95% CL Predict Residual
Variable
Value Mean Predict
61
60
.
84.9468
1.0552
68.4507 101.4430
.
S{Yh(new)} = 8.24101, 84.9468 ± 2.00172(8.24101), 68.451 ≤ Yh(new) ≤101.443 (also can get from the
SAS output above)
c.
Obs
n alpha dfn dfd
1 60
0.05
2
t
w1
w
alphat
t_c
58 2.00172 6.31186 2.51234 0.014797 2.00172
W2 = 2F(.95; 2, 58) = 2(3.15593) = 6.31186, W = 2.512342, 84.9468 ± 2.512342(1.05515), 82.296 ≤β0 +
β1Xh ≤87.598, yes, yes
SAS CODE
2.6/2.25
data new;
input ampule transfer;
datalines;
(HERE IS THE DATASET)
;
run;
proc reg data=new;
model ampule=transfer/clb p r;
output out=new1 p=pred r=resid;
run;
(CALCULATE THE POWER)
data new2;
n=10; sigb0=0.75; beta0=11; df=n-2;alpha=0.025;
delta=(beta0-9)/sigb0; t_c=tinv(1-alpha/2,df);
power=1-probt(t_c, df,delta)+probt(-t_c,df,delta);
output;
proc print data=new2;
run;
2.13
data new;
input GPA ACT;
datalines;
(here is the dataset)
;
//add a new observation
data new1;
ACT=28;output;
data new; set new new1;
run;
//find the 95% mean inverval
proc reg data=new;
model GPA=ACT/clm;
id ACT;
run;
//find the 95% predict interval
proc reg data=new;
model GPA=ACT/cli;
id ACT;
run;
//find the bound
data new; n=120; alpha=0.05;
dfn=2; dfd=n-2;
t=tinv(1-alpha/2,dfd);
w1=2*finv(1-alpha,dfn,dfd);
w=sqrt(w1);
alphat=2*(1-probt(w,dfd));
t_c=tinv(1-alphat/2,dfd);
output;
proc print data=new;
run;
2.28
data new;
input mass age;
datalines;
(HERE IS THE DATASET)
;
data new1;
age=60; output;
data new; set new new1;
proc reg data=new;
model mass=age/clm;
id age;
run;
proc reg data=new;
model mass=age/cli;
id age;
run;
data new;
n=60; alpha=0.05;
dfn=2;dfd=n-2;
t=tinv(1-alpha/2,dfd);
w1=2*finv(1-alpha,dfn,dfd);
w=sqrt(w1);
alphat=2*(1-probt(w,dfd));
t_c=tinv(1-alpha/2,dfd);
output;
proc print data=new;
run;