For the specific interaction model (B) we have
been considering, the variance of^lis given by
the formula shown here.
In computing this variance, there is an issue
concerning round-off error. Computer packages
typically maintain 16 decimal places for calcu-
lations, with final answers rounded to a set
number of decimals (e.g., 4) on the printout.
The variance results we show here were
obtained with such a program (see Computer
Appendix) rather than using the rounded values
from the variance–covariance matrix presented
at left.
For this model,W 1 is CHL andW 2 is HPT.
As an example of a confidence interval calcula-
tion, we consider the values CHL equal to 220
and HPT equal to 1. Substituting^b;^d 1 , and^d 2
into the formula for^l, we obtain the estimatel^
equals 0.1960.
The corresponding estimated variance is
obtained by substituting into the above vari-
ance formula the estimated variances and co-
variances from the variance–covariance matrix
Vb. The resulting estimate of the variance of^lis
equal to 0.2279. The numerical values used in
this calculation are shown at left.
We can combine the estimates of^land its vari-
ance to obtain the 95% confidence interval.
This is given by exponentiating the quantity
0.1960 plus or minus 1.96 times the square
root of 0.2279. The resulting confidence limits
are 0.48 for the lower limit and 3.10 for the
upper limit.
The 95% confidence intervals obtained for
other combinations of CHL and HPT are
shown here. For example, when CHL equals
200 and HPT equals 1, the confidence limits
are 0.10 and 0.91. When CHL equals 240 and
HPT equals 1, the limits are 1.62 and 14.52.
EXAMPLE
dvarl^
¼dvar ^b
þðÞW 12 vard^d 1
þðÞW 22 vard^d 2
þ 2 W 1 Covd ^b;^d 1
þ 2 W 2 Covd ^b;^d 2
þ 2 W 1 W 2 Covd ^d 1 ;^d 2
W 1 ¼CHL,W 2 ¼HPT
CHL¼220, HPT¼1:
^l¼^bþ^d 1 ðÞþ 220 ^d 2 ðÞ 1
¼ 0 : 1960
dvarl^
¼ 0 : 2279
Vb¼
dvar^b
dcov^b;^d 1
dvar^d 1
dcov^b;^d 2
dcov^d 1 ;^d 2
dvar^d 2
2
(^66)
4
3
(^77)
5
¼
9 : 6389
0 :0437 0: 0002
0 : 0049 0 :0016 0: 5516
2
4
3
5
95% CI for adjusted OR:
CI limits: (0.48, 3.10)
exp[0.1960±1.96 0.2279]
HPT¼ 0 HPT¼ 1
CHL¼ 200
OR : 3.16d
CI : (0.89, 11.03)
dOR : 0.31
CI : (0.10, 0.91)
CHL¼ 220
dOR : 12.61
CI : (3.65, 42.94)
dOR : 1.22
CI : (0.48, 3.10)
CHL¼ 240
dOR : 50.33
CI : (11.79, 212.23)
dOR : 4.89
CI : (1.62, 14.52)
152 5. Statistical Inferences Using Maximum Likelihood Techniques