11 Natural cork in wine bottles is subject to deterioration, and as a result wine in such
Example 8.11 Natural cork in wine bottles is subject to deterioration, and as a result wine in such
bottles may experience contamination. The article “Effects of Bottle Closure Type on Consumer Perceptions of Wine Quality” (Amer. J. of Enology and Viticulture, 2007: 182–191) reported that, in a tasting of commercial chardonnays, 16 of 91 bot- tles were considered spoiled to some extent by cork-associated characteristics. Does this data provide strong evidence for concluding that more than 15 of all such bot- tles are contaminated in this way? Let’s carry out a test of hypotheses using a sig- nificance level of .10.
1. p5 the true proportion of all commercial chardonnay bottles considered spoiled to some extent by cork-associated characteristics.
2. The null hypothesis is H 0 : p 5 .15 .
3. The alternative hypothesis is H a : p . .15 , the assertion that the population
percentage exceeds 15.
8.3 Tests Concerning a Population Proportion
4. Since np 0 5 91(.15) 5 13.65 . 10 and nq 0 5 91(.85) 5 77.35 . 10 , the large-
sample z test can be used. The test statistic value is z 5 (pˆ 2 .15) 1(.15)(.85)n .
5. The form of H a implies that an upper-tailed test is appropriate: Reject H 0 if
zz .10 5 1.28 .
6. pˆ 5 16 91 5 .1758 , from which
z 5 (.1758 2 .15)2(.15)(.85)91 5 .0258.0374 5 .69
7. Since .69 < 1.28, z is not in the rejection region. At significance level .10, the null hypothesis cannot be rejected. Although the percentage of contaminated bottles in the sample somewhat exceeds 15, the sample percentage is not large enough to conclude that the population percentage exceeds 15. The difference between the sample proportion .1758 and the null value .15 can adequately be explained by sampling variability.
■
B and Sample Size Determination When H 0 is true, the test statistic Z has approxi- mately a standard normal distribution. Now suppose that H 0 is not true and that p 5 pr.
Then Z still has approximately a normal distribution (because it is a linear function of pˆ ), but its mean value and variance are no longer 0 and 1, respectively. Instead,
pr 2 p
0 pr(1 2 pr)n V(Z) 5
E(Z) 5
2p 0 (1 2 p 0 )n
p 0 (1 2 p 0 )n
The probability of a type II error for an upper-tailed test is b(pr) 5
P(Z . z a when p 5 pr) . This can be computed by using the given mean and vari- ance to standardize and then referring to the standard normal cdf. In addition, if it is desired that the level a test also have b(pr) 5 b for a specified value of b, this equa- tion can be solved for the necessary n as in Section 8.2. General expressions for b(pr) and n are given in the accompanying box.
Alternative Hypothesis
b ( pr)
p 0 2 pr 1 z a 2p 0 (1 2 p 0 )n
H a : p.p 0 c d
2pr(1 2 pr)n p 0 2 pr 2 z
2pr(1 2 pr)n
2pr(1 2 pr)n
p 0 2 pr 2 z a2 2p
2 0 (1 2 p 0 )n c d
2pr(1 2 pr)n
The sample size n for which the level a test also satisfies b(pr) 5 b is
z a 2p 0 (1 2 p 0 )1z b 2pr(1 2 pr) 2
c d pr 2 p one-tailed test n5 0 e
z a2 2p 0 (1 2 p 0 )1z b 2pr(1 2 pr) 2 two-tailed test (an
c pr 2 p d
0 approximate solution)
CHAPTER 8 Tests of Hypotheses Based on a Single Sample
Example 8.12
A package-delivery service advertises that at least 90 of all packages brought to its
office by 9 A . M . for delivery in the same city are delivered by noon that day. Let p
denote the true proportion of such packages that are delivered as advertised and con-
sider the hypotheses H 0 : p 5 .9 versus H a : p , .9 . If only 80 of the packages are
delivered as advertised, how likely is it that a level .01 test based on n 5 225 pack-
ages will detect such a departure from H 0 ? What should the sample size be to ensure that ? b(.8) 5 .01 With a 5 .01, p 0 5 .9, pr 5 .8 , and n 5 225 ,
b(.8) 5 1 2 a b
Thus the probability that H 0 will be rejected using the test when p 5 .8 is .9772—
roughly 98 of all samples will result in correct rejection of H 0 . Using z a 5z b 5 2.33 in the sample size formula yields