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Theory: The Paired t-Test Analysis
Define δ as the true mean of differences between random variables y 1 and y 2 that were observed as
matched pairs under identical experimental conditions. δ will be zero if the means of the populations
from which y 1 and y 2 are drawn are equal. The estimate of δ is the average of differences between n
paired observations:

d = --------- = 1 ∑ ( y 1,i – y 2,i )
∑d i
---
n n
d
Because of measurement error, the value of is not likely to be zero, although it will tend toward zero
if δ is zero.
The sample variance of the differences is:

(
s d = ∑ d i – d) 2
2
-------------------------
n 1
–
The standard error of the average difference is:
d
s = -------
s d
d
n
This is used to establish the 1 – α confidence limits for δ, which are d ± . The correctness of
s t n−1,α/2
d
this confidence interval depends on the data being independent and coming from distributions that are
approximately normal with the same variance.

Case Study Solution
The differences were calculated by subtracting the electrode measurements from the Winkler measure-
ments. The average of the paired differences is:
– ( 0.4) ++ – ( 0.5) + – ( 1.0) + … + – ( 0.4)
0
d = ------------------------------------------------------------------------------------------------------ = – 0.329 mg/L
14
and the variance of the paired differences is:

– [ 0.4 – – ( 0.329)] + [ 0 ( – 0.329)] + … + – [ 0.4 – – ( 0.329)] 2
2
2
–
2
s d = ----------------------------------------------------------------------------------------------------------------------------------------------------------- = 0.244 mg/L) 2
(
14 1
–
giving s d = 0.494 mg/L. The standard error of the average of the paired differences is:
------------- =
s = 0.494 0.132 mg/L
d
14
The (1 – α)100% confidence interval is computed using the t distribution with ν = 13 degrees of freedom
at the α/2 probability point. For (1 − α) = 0.95, t 13,0.025 = 2.160, and the 95% confidence interval of the
true difference δ is:
ds t 13,0.025 < δ < d +
–
d s d t 13,0.025
© 2002 By CRC Press LLC