Stat 312: Lecture 17
Two sample ttest
Moo K. Chung
November 9, 2004
1. Let X1,· · · , Xnand Y1,···, Ymbe two indepen-
dent samples from normal distributions with
the same population variance, i.e.
Xi∼N(µX, σ2) and Yj∼N(µY, σ2).
The pooled estimator of σis
S2
p=(n−1)S2
X+ (m−1)S2
Y
n+m−2.
The test statistic for testing
H0:µX=µYvs. H1:µX6=µY
T=¯
X−¯
Y−(µX−µY)
Spp1/n + 1/m ∼tn+m−2.
Reject H0if |T|> tα/2,n+m−2.
Example. A study was conducted to compare
the weights of cats and dogs. Weights of cats:
20, 21, 35, 13, 21, 10. Weights of dogs: 31, 10,
20, 40. Assume normality and equal variance
for both cats and dogs. Is there any difference
between the weights of cats and dogs?
> x<-c(20,21,35,13,21,10)
> y<-c(31,10,20,40)
> Sp<-sqrt((5*var(x)+3*var(y))/8)
[1] 10.52824
> t=(mean(x)-mean(y))/(Sp*sqrt(1/5+1/3))
> t
[1] -0.6827026
> qt(0.05,8)
[1] -1.859548
2. Checking the equality of variance. This topic
will be discussed later in detail.
> var(x)
[1] 75.2
> var(y)
[1] 170.25
> var.test(x,y)
F test to compare two variances
... p-value = 0.3975
alternative hypothesis: true ratio of variances
is not equal to 1 95 percent confidence interval:
0.02967475 3.42920369
sample estimates: ratio of variances
0.4417034
The Welch-Satterthwaite ttest is an alternative to
the two sample t-test with equal variance, and is
used when the assumption that the two populations
have equal variances seems unreasonable. The prob-
lem with this method is that it is asymptotically a
t distribution.
>t.test(x,y,alternative="two.sided",
conf.level=0.9)
Welch Two Sample t-test
data:x and y t = -0.7073,df = 4.778
p-value = 0.5124 alternative
hypothesis: true difference in means
is not equal to 0 90 percent
confidence interval:
-20.361647 9.861647
Review problems. Compute a CI for µX−µYin the
previous Example. Example 9.7
