Dr. Neal, WKU
MATH 183
Confidence Interval for
Difference of Means of Normally
Distributed Measurements
Here we discuss an improvement for a confidence interval for the difference of means
of two normally distributed measurements, both of which have unknown standard
deviations. The results are based on independent random samples, of sizes n1 and n2
respectively, taken on each measurement. For calculations on the TI-83/84, we can use
the 2–SampTInt screen from the STAT TESTS menu.
This feature require that we specify whether or not we wish to use the pooled
deviation S p . We should specify “Yes” only under the assumption that the two
measurements have the same standard deviation. When the normal measurements are
assumed to have the same (unknown) deviation, then the pooled variance and standard
deviation estimates are
S 2p =
(n1 ! 1)S12 + (n2 ! 1)S22
n1 + n2 ! 2
and
Sp =
2
Sp =
(n1 ! 1)S12 + (n2 ! 1)S22
n1 + n2 ! 2
The confidence interval for the difference in means is now given by
µ1 ! µ2 " (x1 ! x 2 ) ± t# / 2
S 2p
n1
+
S2p
n2
= (x1 ! x 2 ) ± t# / 2 Sp
1
1
+
n1 n2
where t! / 2 is the t -score from the t(n1 + n2 ! 2) distribution.
(The 2–SampTInt computes this type of confidence interval when set to Pooled.)
When we do not specify a pooled variance, then the confidence interval is
µ1 ! µ2 ≈ (x1 ! x2 ) ± t" / 2
S12 S22
+
n1 n2
where the degrees of freedom for the t -score are given by the greatest integer k
satisfying
" S2 S 2 % 2
$$ 1 + 2 ''
# n1 n2 &
k!
2
2.
1 "$ S12 %'
1 "$ S22 %'
$ ' +
$ '
n1 ( 1 # n1 &
n2 ( 1 # n2 &
Dr. Neal, WKU
Example. IQ scores are generally found to be normally distributed. The IQ's of a group
of 29 second-grade children who did not attend pre-school yielded an average score of
97 with a standard deviation of 13. Another group of 68 second-grade children that did
attend pre-school yielded an average score of 109 with a standard deviation of 11.
Let µ1 be the true average IQ score among all similar second-grade children who
did not attend pre-school, and let µ 2 be the true average IQ score among all similar
second-grade children who did attend pre-school. Find a 98% confidence interval for
µ1 ! µ2 .
Solution. Assuming a common standard deviation among both groups, then the pooled
deviation is
Sp =
(n1 ! 1)S12 + (n2 ! 1)S22
=
n1 + n2 ! 2
28 " 132 + 67 " 112
# 11.6253 .
29 + 68 ! 2
Then using the 98% t -score from the t(29 + 68 ! 2) = t(95) distribution, we have
1 1
= (97 – 109) ± 2.366 !11.6253 !
µ1 ! µ2 " (x1 ! x2 ) ± t# / 2 Sp
+
n1 n2
So, –18.1 ≤ µ1 – µ 2 ≤ –5.9
or
1
1
= –12 ± 6.1
+
29 68
5.9 ≤ µ 2 – µ1 ≤ 18.1.
It is easier to use the 2–SampTInt screen which also displays the pooled deviation Sxp:
In words: The average IQ score among pre-schooled second graders is from 5.9 points
higher to 18.1 points higher than the average IQ score among non-pre-schooled second
graders.
When we do not pool the deviation, then
the margin of error is a little larger. Now
we have 5.352 ≤ µ 2 – µ1 ≤ 18.65