10-3 INFERENCE FOR THE DIFFERENCE IN MEANS OF TWO NORMAL DISTRIBUTIONS, VARIANCES UNKNOWN 339Given the assumptions of this section, the quantity(10-13)has a tdistribution with n 1 n 2 2 degrees of freedom.TX 1 X 2 1 1 22Sp
B1
n 11
n 2The use of this information to test the hypotheses in Equation 10-11 is now straightfor-
ward: simply replace 1 2 by and the resulting test statistichas a tdistribution with
n 1 n 2 2 degrees of freedom under H 0 : 1 2 . Therefore, the reference distribu-
tion for the test statistic is the tdistribution with n 1 n 2 2 degrees of freedom. The location
of the critical region for both two- and one-sided alternatives parallels those in the one-sample
case. Because a pooled estimate of variance is used, the procedure is often called the pooled
t-test. 0 0 ,Null hypothesis: H 0 : 1 2 Test statistic: (10-14)Alternative Hypothesis Rejection CriterionH 1 : 1 2 0 t 0 t ,n 1 n 2 2H 1 : 1 2
0 t 0
t ,n 1 n 2 2t 0 t 2,n 1 n 2 2H 1 : 1 2 0 t 0
t 2,n 1 n 2 2 orT 0 X 1 X 2 0Sp
B1
n 11
n 2 0Definition:
The Two-Sample
or Pooled t-Test**While we have given the development of this procedure for the case where the sample sizes could be different, there
is an advantage to using equal sample sizes n 1 n 2 n. When the sample sizes are the same from both populations,
the t-test is more robust to the assumption of equal variances. Please see Section 10-3.2 on the CD.EXAMPLE 10-5 Two catalysts are being analyzed to determine how they affect the mean yield of a chemical
process. Specifically, catalyst 1 is currently in use, but catalyst 2 is acceptable. Since catalyst
2 is cheaper, it should be adopted, providing it does not change the process yield. A test is run
in the pilot plant and results in the data shown in Table 10-1. Is there any difference between
the mean yields? Use 0.05, and assume equal variances.
The solution using the eight-step hypothesis-testing procedure is as follows:- The parameters of interest are 1 and 2 , the mean process yield using catalysts
1 and 2, respectively, and we want to know if 1 2 0. - H 0 : 1 2 0, or H 0 : 1 2
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