32
4. Randomized block designs
Condence intervals. The following treatment
seems general enough to cover the cases of prac-
tical interestAny -ratio can be written in the
form
=
()
where is a quantity to be estimated (e.1.
2), is an estimate of (e.g. ¯1 ¯2) and
() is an estimate of the standard deviation of
r 1 1
(e.g. + ). Then if ± 2are the points
1 2
on the horizontal axis which each have 2 of
the probability of the -distribution lying beyond
them, we have
³ ´
1 =
Ã 2 2 !
= 2 2
³ () ´
= · () + · ()
2 2 33
after a rearrangement. Thus, before we take the
sample, we know that the random interval
h i
= 2· () + 2 · ()
will, with probability 1, contain the true value
of . After the sample is taken, and () cal-
culated numerically, we call a 100(1 )%
condence interval.
In the mortar example, the di erence in averages
was = 16 764 17 922 = 1 158; then from
the computer output,
1 158 1 158
9 1094 = = () () = 9 1094 = 1271
There were 18 d.f., and so for a 95% interval we
nd 18 2on R:
( 975 18)
[1] 2 100922
Thus = 1 158 ± 2 1009 · 1271 = 1 158 ±
2670 = [ 1 4250 8910], in agreement with
the computer output. 34
Sample size calculations. The power of a test is
the probability of rejecting the null hypothesis when
it is false.Suppose that we would like a power of
at least .99 when the mortar means di er by =
= 5. Furthermore, suppose that
we will reject 0 if the p-value is 05. (We use
= 05.) We need an estimate of , here Ill use
= 4, which is somewhat larger than was.
> power.t.test(delta = .5, sd = .4,
sig.level = .05,power = .99,type = "two.sample",
alternative = "two.sided")
Two-sample t test power calculation
n = 24.52528
delta = 0.5
sd = 0.4
sig.level = 0.05
power = 0.99
alternative = two.sided
NOTE: n is number in *each* group
Thus in a future study, if = 4 is accurate, we will
need two equal samples of at least 25 each in order to
get the required power. Use help(power.t.test)
to get more details on this function. 35
Paired comparisons. Su

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