Applied Statistics and Probability for Engineers

352 CHAPTER 10 STATISTICAL INFERENCE FOR TWO SAMPLES

assuming that both populations X 1 and X 2 have identical variances ^2. Furthermore, SD^2 n

estimates the variance of. Whenever there is positive correlation within the pairs, the de-

nominator for the paired t-test will be smaller than the denominator of the two-sample t-test.

This can cause the two-sample t-test to considerably understate the significance of the data if

it is incorrectly applied to paired samples.

Although pairing will often lead to a smaller value of the variance of , it does

have a disadvantage—namely, the paired t-test leads to a loss of n1 degrees of freedom in

comparison to the two-sample t-test. Generally, we know that increasing the degrees of free-

dom of a test increases the power against any fixed alternative values of the parameter.

So how do we decide to conduct the experiment? Should we pair the observations or not?

Although there is no general answer to this question, we can give some guidelines based on

the above discussion.

1. If the experimental units are relatively homogeneous (small ) and the correlation
   within pairs is small, the gain in precision attributable to pairing will be offset by the
   loss of degrees of freedom, so an independent-sample experiment should be used.


2. If the experimental units are relatively heterogeneous (large ) and there is large pos-
   itive correlation within pairs, the paired experiment should be used. Typically, this
   case occurs when the experimental units are the samefor both treatments; as in
   Example 10-9, the same girders were used to test the two methods.
   Implementing the rules still requires judgment, because and are never known precisely.
   Furthermore, if the number of degrees of freedom is large (say, 40 or 50), the loss of n1 of
   them for pairing may not be serious. However, if the number of degrees of freedom is small
   (say, 10 or 20), losing half of them is potentially serious if not compensated for by increased
   precision from pairing.

A Confidence Interval for D

To construct the confidence interval for D 1  2 , note that

follows a tdistribution with n1 degrees of freedom. Then, since P(t 2,n 1 T

t 2,n 1 ) 1 , we can substitute for Tin the above expression and perform the necessary

steps to isolate D 1  2 between the inequalities. This leads to the following 100(1)%

confidence interval on  1  2.

T

DD

SD 

1 n

X 1 X 2

D

If and sDare the sample mean and standard deviation of the difference of nrandom

pairs of normally distributed measurements, a 100(1 )% confidence interval on

the difference in means D 1  2 is

(10-23)

where t /2,n 1 is the upper 2% point of the t-distribution with n1 degrees of

freedom.

dt 2, n 1 sD
1 nDd t 2, n 1 sD
1 n

d

Definition

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