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µ = µ0 versus H1 : µ = µ0, our second challenge is to find a way to compute
P . We remind the reader that we have assumed that n is large.
Case 1: The population variance is known or specified by the null
hypothesis.
We define two new quantities, the random variable
¯
Xn - µ0
Zn = "
Ã/ n
and the real number
xn - µ0
¯
z = " .
Ã/ n
Under the null hypothesis H0 : µ = µ0, Zn
Ù
Limit Theorem; hence,
¯
P = Pµ0 Xn - µ0 e" |xn - µ0|
¯
¯
= 1 - Pµ0 - |xn - µ0|
¯ ¯
¯
|xn - µ0| Xn - µ0 |xn - µ0|
¯ ¯
= 1 - Pµ0 - "
Ã/ n Ã/ n Ã/ n
= 1 - Pµ0 (-|z|
.
= 1 - [¦(|z|) - ¦(-|z|)]
= 2¦(-|z|),
which can be computed by the S-Plus command
164 CHAPTER 8. INFERENCE
> 2*pnorm(-abs(z))
or by consulting a table. An illustration of the normal probability of interest
is sketched in Figure 8.1.
-4 -3 -2 -1 0 1 2 3 4
z
Figure 8.1: P (|Z| e" |z| = 1.5)
An important example of Case 1 occurs when Xi
case, Ã2 = Var Xi = µ(1 - µ); hence, under the null hypothesis that µ = µ0,
Ã2 = µ0(1 - µ0) and
xn - µ0
¯
z = .
µ0(1 - µ0)/n
Example 1 To test H0 : µ = .5 versus H1 : µ = .5 at significance level
± = .05, we perform n = 2500 trials and observe 1200 successes. Should H0
be rejected?
The observed proportion of successes is xn = 1200/2500 = .48, so the
¯
value of the test statistic is
.48 - .50 -.02
z = = = -2
.5/50
.5(1 - .5)/2500
f(z)
0.0
0.1
0.2
0.3
0.4
8.4. TESTING HYPOTHESES ABOUT A POPULATION MEAN 165
and the significance probability is
. .
P = 2¦(-2) = .0456
Because P d" ±, we reject H0.
Case 2: The population variance is unknown.
Because Ã2 is unknown, we must estimate it from the sample. We will use
the estimator introduced in Section 8.2,
n
1
2
¯
Sn = Xi - Xn 2 ,
n - 1
i=1
and define
¯
Xn - µ0
Tn = " .
Sn/ n
P
2 2
Because Sn is a consistent estimator of Ã2, i.e. Sn ’! Ã2, it follows from
Theorem 6.3 that
lim P (Tn d" z) = ¦(z).
n’! "
Just as we could use a normal approximation to compute probabilities
involving Zn, so can we use a normal approximation to compute probabili-
ties involving Tn. The fact that we must estimate Ã2 slightly degrades the
quality of the approximation; however, because n is large, we should observe
an accurate estimate of Ã2 and the approximation should not suffer much.
Accordingly, we proceed as in Case 1, using
xn - µ0
¯
t = "
sn/ n
instead of z.
Example 2 To test H0 : µ = 1 versus H1 : µ = 1 at significance level
± = .05, we collect n = 2500 observations, observing xn = 1.1 and sn = 2.
¯
Should H0 be rejected?
The value of the test statistic is
1.1 - 1.0
t = = 2.5
2/50
and the significance probability is
. .
P = 2¦(-2.5) = .0124
Because P d" ±, we reject H0.
166 CHAPTER 8. INFERENCE
One-Sided Hypotheses
In Section 8.3 we suggested that, if Robin is not interested in whether or
not penny-spinning is fair but rather in whether or not it favors her brother,
then appropriate hypotheses would be p
and p e" .5 (penny-spinning does not favor Arlen). These are examples of
one-sided (as opposed to two-sided) hypotheses.
More generally, we will consider two canonical cases:
H0 : µ d" µ0 versus H1 : µ > µ0
H0 : µ e" µ0 versus H1 : µ
Notice that the possibility of equality, µ = µ0, belongs to the null hypothesis
in both cases. This is a technical necessity that arises because we compute
significance probabilities using the µ in H0 that is nearest H1. For such a
µ to exist, the boundary between H0 and H1 must belong to H0. We will
return to this necessity later in this section.
Instead of memorizing different formulas for different situations, we will
endeavor to understand which values of our test statistic tend to undermine
the null hypothesis in question. Such reasoning can be used on a case-by-
case basis to determine the relevant significance probability. In so doing,
sketching crude pictures can be quite helpful!
Consider testing each of the following:
(a) H0 : µ = µ0 versus H1 : µ = µ0
(b) H0 : µ d" µ0 versus H1 : µ > µ0
(c) H0 : µ e" µ0 versus H1 : µ
Qualitatively, we will be inclined to reject the null hypothesis if
(a) We observe xn µ0 or xn µ0, i.e. if we observe |xn - µ0| 0.
¯ ¯ ¯
This is equivalent to observing |t| 0, so the significance probability
is
Pa = Pµ0 (|Tn| e" |t|) .
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