Page 580 - Probability and Statistical Inference

12. Large-Sample Inference 557

See Johnson and Kotz (1969, Chapter 3, Section 8.1) for a variety of other
related approximations.
For large n, one may use either (12.4.4) or (12.4.5) to derive an approxi-
mate 100(1 − α)% confidence interval for p. Also, in order to test the null
hypothesis H : p = p , for large n, one may use the test statistic
0 0
.
to come up with an approximate level α test against an appropriate alternative
hypothesis. The details are left out for brevity.
Example 12.4.1 (Example 12.3.5 Continued) Let p denote the proportion
of voters in the town who favored the proposed computerization. We wish to
test a null hypothesis H : p = .60 against an alternative hypothesis H : p < .60.
1
0
In a random sample of 300 voters, 175 indicated that they favored the pro-
posed computerization. Now, = 175/300 = .58333 and we have
≈ −.588 in view of
(12.4.6), since n = 300 is large. Now, one has z = 2.33 and thus at 1% level,
.01
we fail to reject H . In other words, we conclude that at an approximate 1%
0
level, we do not have enough evidence to validate the claim that less than sixty
percent of the voters are in favor of computerizing the library’s present cata-
loging system. !
Example 12.4.2 (Example 12.3.6 Continued) The manager of a local chain
of fitness centers claimed that 65% of their present members would renew
memberships for the next year. A random sample of 100 present members
showed that 55 have renewed memberships for the next year. At 1% level, is
there sufficient evidence to claim that the manager is off the mark? Let p
denote the proportion of present members who would renew membership.
The null hypothesis is H : p = .65 and the alternative hypothesis is H : p <
1
0

.65. We have = 55/100 = .55 and z = 2.33. Since n = 100 is large, in view
.01
of (12.4.6), we obtain ≈ −
2.0453. At an approximate 1% level, we do not reject H since z > −z . So,
0
.01
calc
we do not have sufficient evidence to claim that the manager’s belief is en-
tirely wrong. !
Example 12.4.3 (Examples 12.3.5 and 12.4.1 Continued) In a random
sample of 300 voters, 175 indicated that they favored the proposed com-
puterization. Let p denote the proportion of voters in the whole town who
favor the proposed computerization. Then, = .58333. Since n =
300 is large, in view of (12.4.4), an approximate 90% confidence interval
for which reduces to
.86912 ± .04749. We conclude that lies between .82163 and