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Outline
1. Introduce independent group t confidence intervals
2. Define the pooled variance estimate
3. Derive the distribution for the independent group,
common variance, statistic
4. Cover likelihood methods for the change in the group
means per standard deviation
5. Discuss remedies for unequal variances
Independent group t confidence intervals
• Suppose
that we want to compare the mean blood
pressure between two groups in a randomzied trial;
those who received the treatment to those who received a placebo
• We
cannot use the paired t test because the groups
are independent and may have different sample sizes
• We
now present methods for comparing independent
groups
Notation
• Let X1, . . . , Xnx
• Let Y1, . . . , Yny
be iid N (µx, σ2)
be iid N (µy , σ2)
• Let X̄ , Ȳ , Sx, Sy
be the means and standard deviations
• Using the fact that linear combinations of normals are
again normal, we know that Ȳ − X̄ is also normal with
mean µy − µx and variance σ2( n1x + n1y )
• The
pooled variance estimator
Sp2 = {(nx − 1)Sx2 + (ny − 1)Sy2}/(nx + ny − 2)
is is a good estimator of σ2
Note
• The
pooled estimator is a mixture of the group variances, placing greater weight on whichever has a larger
sample size
• If
the sample sizes are the same the pooled variance
estimate is the average of the group variances
• The
pooled estimator is unbiased
2 ] + (n − 1)E[S 2]
2 + (n − 1)σ 2
(n
−
1)E[S
(n
−
1)σ
x
y
x
y
x
y
2
E[Sp ] =
=
nx + ny − 2
nx + ny − 2
• The
pooled variance estimate is independent of Ȳ − X̄
since Sx is independent of X̄ and Sy is independent of
Ȳ and the groups are independent
Result
• The sum of two independent Chi-squared random vari-
ables is Chi-squared with degrees of freedom equal to
the sum of the degrees of freedom of the sumands
• Therefore
(nx + ny − 2)Sp2/σ 2 = (nx − 1)Sx2 /σ 2 + (ny − 1)Sy2/σ 2
= χ2nx−1 + χ2ny −1
= χ2nx+ny −2
Putting this all together
• The
statistic
Ȳ −X̄−(µy −µx)
1/2
σ n1x + n1y
Ȳ − X̄ − (µy − µx)
r
=
1/2
2
(nx+ny −2)Sp
1 + 1
S
p
2
nx
ny
(nx+ny −2)σ
is a standard normal divided by the square root of
an independent Chi-squared divided by its degrees of
freedom
• Therefore
this statistic follows Gosset’s t distribution
with nx + ny − 2 degrees of freedom
• Notice
the form is (estimator - true value) / SE
Confidence interval
• Therefore a (1 − α) × 100%
confidence interval for µy − µx
is
1
1
+
Ȳ − X̄ ± tnx+ny −2,1−α/2Sp
nx ny
• Remember
1/2
this interval is assuming a constant variance across the two groups
• If
there is some doubt, assume a different variance
per group, which we will discuss later
Likelihood method
• Exactly
as before,
Ȳ − X̄
Sp n1x + n1y
1/2
follows a non-central t distribution with non-centrality
µy −µx
parameter 1/2
σ
• Therefore,
1
1
+
nx ny
we can use this statistic to create a likelihood for (µy − µx)/σ, a standardized measure of the
change in group means
Example
Unequal variances
• Note
that under unequal variances
Ȳ − X̄ ∼ N
• The
statistic
σx2 σy2
+
µy − µx,
nx ny
!
Ȳ − X̄ − (µy − µx)
2
2 1/2
σ
σx1
y
+
nx
ny
approximately follows Gosset’s t distribution with degrees of freedom equal to
2
2
2
Sx/nx + Sy /ny
2 2
S 2 2
Sx
y
/(ny − 1)
/(n
−
1)
+
x
nx
ny