The pooled estimate of the population SD is sqrt{ [10.672 × 9 + 9.302 × 4] / 13 } = 10.27.
The estimated SE of the difference between the sample means is 10.27 × sqrt{ 1/10 + 1/5 } = 5.62
The 97.5 percentile of the t distribution
with 13 degrees of freedom is 2.16. (In R, use
qt(0.975,13).)
Thus the 95% confidence interval is (103-67) ± 2.16 × 5.62 = 36 ± 12 = (24, 48).
In R, if x is the data for
the sample from strain A and y is the data for the
sample from strain B, type
t.test(x,y,var.equal=TRUE) to get the above
confidence interval.
| Last modified: Sun Feb 26 00:14:00 EST 2006 |