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STAT 420
Examples for 09/25/
Fall 2008
The (normal) simple linear regression model:
Y
i
= α + β x
i
i
where ε
i
’s are independent Normal ( 0 , σ
) ( iid Normal ( 0 , σ
α , β , and σ
are unknown model parameters.
Suppose x
i
’s are fixed (not random).
Y
i
’s are independent Normal ( α + β x
i
) random variables.
^ (^ )
2
Y
x x
x x
i
i i ~ N
2
2
xi x
ˆ^ = x
Y − ~ N
^ −
2
2 2
,
n x x
x
i
i = N
2
2 2
x x
x
n
i
2 2
S ˆ^
e (^) i i
y x
n
2
i i
y y
n
2
2
n − 2 S e
~ χ
2
( n – 2 )
1. The owner of Momma Leona’s Pizza restaurant chain believes that if a restaurant
is located near a college campus, then there is a linear relationship between sales
and the size of the student population. Suppose data were collected from a
sample of 10 Momma Leona’s Pizza restaurants located near college campuses.
For the i th restaurant in the sample, xi is the size of the student population (in
thousands) and yi is the quarterly sales (in thousands of dollars). The values of xi
and yi for the 10 restaurants in the sample are summarized in the following table:
Restaurant
Student Population
(1000s)
Quarterly Sales
($1000s)
i xi yi
x = 14, y = 130
SXX = 568
SXY = 2,
SYY = 15,
= 5, ˆ^ = 60
y ˆ^ = 60 + 5 ⋅ x
SSE = 1,
R
s e
2 = 191.
Confidence interval for β :
(^ − )
2 2
t
ˆ e
x x
s
i
SXX
s e
2
t
where
2
t
is the appropriate value of t-distribution
with n – 2 degrees of freedom.
Rejection Region:
Reject H
0
if T > t
( 8 df )
t
( 8 df ) = 1.860.
Do NOT Reject H
( 0.05 < p-value < 0.10 )
Confidence interval for α : SXX
x
n
se
2
2
ˆ^ ± t ⋅ +
where 2
t
is the appropriate value of t-distribution
with n – 2 degrees of freedom.
Test statistic for H
: α = α 0
T =
SXX
x
n
se
2
0
ˆ
( n – 2 degrees of freedom )
d) Construct a 90% confidence interval for α.
SXX
x
n
se
2
2
ˆ^ ± t ⋅ +
10 – 2 = 8 degrees of freedom, t
( )
2 −
e) Test H
: α = 75 vs. H 1
: α < 75. Use a 5% level of significance.
Test Statistic:
T =
SXX
x
n
se
2
0
ˆ
( )
568
2 −
Rejection Region:
Reject H
0
if T < – t
( 8 df )
( 8 df ) = – 1.860.
Do NOT Reject H
Confidence interval for σ
2 :
( ) ( )
−
2
2
1
2
2
2
2
n se n s e
−
2
2
1
2
2
2
2
ˆ^ ˆ
n n
where
2
2
2
2
1
and
− α^ α
are the appropriate values of χ
2 distribution
with n – 2 degrees of freedom.
f) Construct a 95% confidence interval for σ
2 .
χ
2
( 8 df ) = 17.54, χ
2
( 8 df ) = 2.180.
( 87.229 , 701.835 )
Mean response ( y ) for a fixed value of x : μ ( x ) = μ
y | x
= α + β x.
To estimate μ ( x ), use y ˆ^ = x
E ( Yˆ^ ) = μ ( x ) = α + β x. Var ( Yˆ^ ) =
( )
SXX
x x
n
2 2 1
Note that α = μ ( x = 0 ), ˆ^ = y ˆ^ if x = 0.
i) Construct a 95% prediction interval for a future value of y corresponding to x = 38.
( Construct 95% limits of prediction if x = 38. )
x = 38 y ˆ^ = 60 + 5 ⋅ x = 60 + 5 ⋅ 38 = 250
( )
SXX
x x
n
y se
2
2
ˆ t 1
10 – 2 = 8 degrees of freedom, t
( )
2 −
j) University of Illinois at Urbana-Champaign has 38 thousand students. The owner
of Momma Leona’s Pizza restaurant chain would agree to open a restaurant near the
UIUC campus, but only if there is enough evidence that the average quarterly sales
would be over $225,000. Test H
: μ ( x = 38 ) = 225 vs. H
: μ ( x = 38 ) > 225.
Use a 5% level of significance.
x = 38. y ˆ^ = 60 + 5 ⋅ x = 60 + 5 ⋅ 38 = 250.
Test Statistic:
T =
( ) (^ )
568
2 2
e SXX
x x
n
s
y
Rejection Region:
Reject H
0
if T > t
( 8 df )
t
( 8 df ) = 1.860.
Accept H
b) Test the assumption that students do not affect the sales. That is, test
H
: β = 0 vs. H
1
: β ≠ 0 ( the significance of regression test ).
Use a 1% level of significance.
Here V
= { a 1 , a ∈ R }, dim ( V 0
) = 1, [ ]
T
ˆ = Y,Y,..., Y
Y ,
V = { a 0
1 + a 1
x , a 0
, a 1
∈ R }, dim ( V ) = 2.
x <- c(2,6,8,8,12,16,20,20,22,26)
y <- c(58,105,88,118,117,137,157,169,149,202)
fit <- lm(y ~ x)
summary(fit)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-21.00 -9.75 -3.00 11.25 18.
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 60.0000 9.2260 6.503 0.000187 ***
x 5.0000 0.5803 8.617 2.55e-05 ***
Signif. codes: 0 ***' 0.001**' 0.01 *' 0.05.' 0.1 ` ' 1
Residual standard error: 13.83 on 8 degrees of freedom
Multiple R-Squared: 0.9027, Adjusted R-squared: 0.
F-statistic: 74.25 on 1 and 8 DF, p-value: 2.549e-
anova(fit)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x 1 14200.0 14200.0 74.248 2.549e-05 ***
Residuals 8 1530.0 191.
Signif. codes: 0 ***' 0.001**' 0.01 *' 0.05.' 0.1 ` ' 1
confint(fit, level=0.90)
5 % 95 %
(Intercept) 42.843745 77.
x 3.920969 6.
> new <- data.frame(x=10)
> predict.lm(fit,new,interval=c("confidence"),level=0.95)
fit lwr upr
[1,] 110 98.583 121.
> new <- data.frame(x=38)
> predict.lm(fit,new,interval=c("confidence"),level=0.95)
fit lwr upr
[1,] 250 216.3396 283.
> predict.lm(fit,new,interval=c("prediction"),level=0.95)
fit lwr upr
[1,] 250 203.6316 296.
