Statistical Methods in Medical Research 4ed

160 Analysing variances
Let y ˆ x1=x2, where again x1 and x2 are independent. No general formula
can be given for the variance of y. Indeed, it may be infinite. However, if x2 has a
small coefficient of variation, the distribution of y will be rather similar to a
distribution with a variance given by the following formula:
var…y† ˆ var…x1†
2
‰E…x1†Š
2 ‡
‰E…x2†Š ‰E…x2†Š 4 var…x2†:
Note that if x2 has no variability at all, (5.12) reduces to
var…y† ˆ
…5:12†
var…x1†
x2 ,
2
which is an exact result when x2 is a constant.
Approximate confidence limits for a ratio may be obtained from (5.12), with
p
the usual multiplying factors for SE…y† ‰ˆ var…y† Š based on the normal distribution.
However, if x1 and x2 are normally distributed, an exact expression for
confidence limits is given by Fieller's theorem (Fieller, 1940). This covers a rather
more general situation, in which x1 and x2 are dependent, with a non-zero
covariance. We suppose that x1 and x2 are normally distributed with variances
and a covariance which are known multiples of some unknown parameter s2 ,
and that s2 is estimated by a statistic s2 on f DF. Define E…x1†ˆ m1,
E…x2†ˆ m2, var…x1† ˆv11s2 , var…x2† ˆv22s2 and cov…x1, x2† ˆv12s2 . Denote
the unknown ratio m1=m2 by r, so that m1 ˆ rm2. It then follows that the
quantity z ˆ x1 rx2 is distributed as N‰0, …v11 2rv12 ‡ r2v22†s2 the ratio
Š, and so
T ˆ
x1 rx2
s …v11 2rv12 ‡ r2 p …5:13†
v22†
follows a t distribution on f DF. Hence, the probability is 1 a that
or, equivalently,
tf , a < T < tf,a,
T 2 < t 2 f , a : …5:14†
Substitution of (5.13) in (5.14) gives a quadratic inequality for r, leading to
100…1 a†% confidence limits for r given by
where
r L, r U ˆ
y
gv12
v22
t f , a S
x2
v11 2yv12 ‡ y 2 v22 g v11
1 g
v 2 12
v22
1
2
, …5:15†