280 Some Elementary Statistical InferencesConsider, first, the one-sided hypothesesH 0 :μ=μ 0 versusH 1 :μ>μ 0 ,whereμ 0
is specified. Write the rejection rule asRejectH 0 in favor ofH 1 ,ifX≥k, (4.6.11)whereXis the sample mean. Previously we have specified a level and then solved
fork. In practice, though, the level is not specified. Instead, once the sample
is observed, the realized valuexofXis computed and we ask the question: Isx
sufficiently large to rejectH 0 in favor ofH 1? To answer this we calculate thep-value
which is the probability,
p-value =PH 0 (X≥x). (4.6.12)
Note that this is a data-based “significance level” and we call it the observed
significance levelor thep-value. The hypothesisH 0 is rejected at all levels greater
than or equal to thep-value. For example, if thep-value is 0.048, and the nominal
αlevel is 0.05 thenH 0 would be rejected; however, if the nominalαlevel is 0.01,
thenH 0 would not be rejected. In summary, the experimenter sets the hypotheses;
the statistician selects the test statistic and rejection rule; the data are observed
and the statistician reports thep-value to the experimenter; and the experimenter
decides whether thep-value is sufficiently small to warrant rejection ofH 0 in favor
ofH 1. The following example provides a numerical illustration.
Example 4.6.5.Recall the Darwin data discussed in Example 4.5.5. It was a
paired design on the heights of cross and self-fertilizedZea maysplants. In each of
15 pots, one cross-fertilized and one self-fertilized were grown. The data of interest
are the 15 paired differences, (cross−self). As in Example 4.5.5, letXidenote the
paired difference for theith pot. Letμbe the true mean difference. The hypotheses
of interest areH 0 :μ=0versusH 1 :μ>0. The standardized rejection rule is
RejectH 0 in favor ofH 1 ifT≥k,whereT =X/(S/√
15), whereXandSare respectively the sample mean and
standard deviation of the differences. The alternative hypothesis states that on the
average cross-fertilized plants are taller than self-fertilized plants. From Example
4.5.5 thet-test statistic has the value 2.15. Lettingt(14) denote a random variable
with thet-distribution with 14 degrees of freedom, and using R thep-value for the
experiment is
P[t(14)> 2 .15] = 1−pt(2.15,14) = 1− 0 .9752 = 0. 0248. (4.6.13)In practice, with thisp-value,H 0 would be rejected at all levels greater than or
equal to 0.0248. This observed significance level is also part of the output from the
Rcallt.test(cross-self,mu=0,alt="greater").
Returning to the discussion above, suppose the hypotheses areH 0 :μ=μ 0
versusH 1 : μ<μ 0. Obviously, the observed significance level in this case is
p-value =PH 0 (X≤x).For the two-sided hypothesesH 0 :μ=μ 0 versusH 1 :μ =
μ 0 , our “unspecified” rejection rule is
RejectH 0 in favor ofH 1 ,ifX≤lorX≥k. (4.6.14)