how many beer drinkers in the sample have tried at least one brand of “cold-filtered beer”?
The Cheese Spread Case
Recall that the soft cheese spread producer has decided that replacing the current spout with the new spout is profitable only if p, the true proportion of all current purchasers who would stop buying the cheese spread if the new spout were used, is less than .10. The producer feels that it is unwise to change the spout unless it has very strong evidence that p is less than .10. Therefore, the spout will be changed if and only if the null hypothesis H0: p = .10 can be rejected in favor of the alternative hypothesis Ha: p < .10 at the .01 level of significance.
In order to see how to test this kind of hypothesis, remember that when n is large, the sampling distribution of
is approximately a standard normal distribution. Let p0 denote a specified value between 0 and 1 (its exact value will depend on the problem), and consider testing the null hypothesis H0: p = p0. We then have the following result:
A Large Sample Test about a Population Proportion
Define the test statistic
If the sample size n is large, we can test H0: p = p0 versus a particular alternative hypothesis at level of significance α by using the appropriate critical value rule, or, equivalently, the corresponding p-value.
Here n should be considered large if both np0 and n(1 − p0) are at least 5.3
EXAMPLE 9.6: The Cheese Spread Case
We have seen that the cheese spread producer wishes to test H0: p = .10 versus Ha: p < .10, where p is the proportion of all current purchasers who would stop buying the cheese spread if the new spout were used. The producer will use the new spout if H0 can be rejected in favor of Ha at the .01 level of significance. To perform the hypothesis test, we will randomly select n = 1,000 current purchasers of the cheese spread, find the proportion of these purchasers who would stop buying the cheese spread if the new spout were used, and calculate the value of the test statistic z in the summary box. Then, since the alternative hypothesis Ha: p < .10 is of the form Ha: p < p0, we will reject H0: p = .10 if the value of z is less than − z α = − z.01= −2.33. (Note that using this procedure is valid because np0 = 1,000(.10) = 100 and n(1 − p0) = 1,000(1 − .10) = 900 are both at least 5.) Suppose that when the sample is randomly selected, we find that 63 of the 1,000 current purchasers say they would stop buying the cheese spread if the new spout were used. Since
, the value of the test statistic is
Because z = −3.90 is less than −z.01 = −2.33, we reject H0: p = .10 in favor of Ha: p < .10. That is, we conclude (at an α of .01) that the proportion of current purchasers who would stop buying the cheese spread if the new spout were used is less than .10. It follows that the company will use the new spout. Furthermore, the point estimate says we estimate that 6.3 percent of all current customers would stop buying the cheese spread if the new spout were used.
Although the cheese spread producer has made its decision by setting α equal to a single, prechosen value (.01), it would probably also wish to know the weight of evidence against H0 and in favor of Ha. The p-value is the area under the standard normal curve to the left of z = − 3.90. Table A.3 (page 862) tells us that this area is .00005. Because this p-value is less than .001, we have extremely strong evidence that Ha: p < .10 is true. That is, we have extremely strong evidence that fewer than 10 percent of current purchasers would stop buying the cheese spread if the new spout were used.
EXAMPLE 9.7
Recent medical research has sought to develop drugs that lessen the severity and duration of viral infections. Virol, a relatively new drug, has been shown to provide relief for 70 percent of all patients suffering from viral upper respiratory infections. A major drug company is developing a competing drug called Phantol. The drug company wishes to investigate whether Phantol is more effective than Virol. To do this, the drug company will test a hypothesis about the true proportion, p, of all patients whose symptoms would be relieved by Phantol. The null hypothesis to be tested is H0: p = .70, and the alternative hypothesis is Ha: p > .70. If H0 can be rejected in favor of Ha at the .05 level of significance, the drug company will conclude that Phantol helps more than the 70 percent of patients helped by Virol. To perform the hypothesis test, we will randomly select n = 300 patients having viral upper respiratory infections, find the proportion of these patients whose symptoms are relieved by Phantol and calculate the value of the test statistic z in the summary box. Then, since the alternative hypothesis Ha: p > .70 is of the form Ha: p> p0, we will reject H0: p = .70 if the value of z is greater than zα = z.05 = 1.645. (Note that using this procedure is valid because np0 = 300(.70) = 210 and n(1 − p0) = 300(1 − .70) = 90 are both at least 5.) Suppose that when the sample is randomly selected, we find that Phantol provides relief for 231 of the 300 patients. Since
the value of the test statistic is
Because z = 2.65 is greater than z.05 = 1.645, we reject H0: p = .70 in favor of Ha: p > .70. That is, we conclude (at an α of .05) that Phantol will provide relief for more than 70 percent of all patients suffering from viral upper respiratory infections. More specifically, the point estimate of p says that we estimate that Phantol will provide relief for 77 percent of all such patients. Comparing this estimate to the 70 percent of patients whose symptoms are relieved by Virol, we conclude that Phantol is somewhat more effective.
The p-value for testing H0: p = .70 versus Ha: p > .70 is the area under the standard normal curve to the right of z = 2.65. This p-value is (1.0 − .9960) = .004 (see Table A.3, page 863), and it provides very strong evidence against H0: p = .70 and in favor of Ha: p > .70. That is, we have very strong evidence that Phantol will provide relief for more than 70 percent of all patients suffering from viral upper respiratory infections.
EXAMPLE 9.8: The Electronic Article Surveillance Case
Suppose that a company selling electronic article surveillance devices claims that the proportion, p, of all consumers who would never shop in a store again if the store subjected them to a false alarm is no more than .05. A store considering installing such a device is concerned that p is greater than .05 and wishes to test H0: p = .05 versus Ha: p > .05. To perform the hypothesis test, the store will calculate a p-value and use it to measure the weight of evidence against H0 and in favor of Ha. In an actual systematic sample, 40 out of 250 consumers said they would never shop in a store again if the store subjected them to a false alarm. Therefore, the sample proportion of lost consumers is Since np0 = 250(.05) = 12.5 and n(1 − p0) = 250(1 − .05) = 237.5 are both at least 5, we can use the test statistic z in the summary box. The value of the test statistic is
Noting that Ha: p > .05 is of the form Ha: p>p0, the p-value is the area under the standard normal curve to the right of z = 7.98. The normal table tells us that the area under the standard normal curve to the right of 3.99 is (1.0 − .99997) = .00003. Therefore, the p-value is less than .00003 and provides extremely strong evidence against H0: p = .05 and in favor of Ha: p > .05. That is, we have extremely strong evidence that the proportion of all consumers who say they would never shop in a store again if the store subjected them to a false alarm is greater than .05. Furthermore, the point estimate says we estimate that the percentage of such consumers is 11 percent more than the 5 percent maximum claimed by the company selling the electronic article surveillance devices. A 95 percent confidence interval for p is
This interval says we are 95 percent confident that the percentage of consumers who would never shop in a store again if the store subjected them to a false alarm is between 6.46 percent and 15.54 percent more than the 5 percent maximum claimed by the company selling the electronic article surveillance devices. The rather large increases over the claimed 5 percent maximum implied by the point estimate and the confidence interval would mean substantially more lost customers and thus are practically important. Figure 9.11 gives the MegaStat output for testing H0: p = .05 versus Ha: p > .05. Note that this output includes a 95 percent confidence interval for p. Also notice that MegaStat expresses the p-value for this test in scientific notation. In general, when a p-value is less than .0001, MegaStat (and also Excel) express the p-value in scientific notation. Here the p-value of 7.77 E-16 says that we must move the decimal point 16 places to the left to obtain the decimal equivalent. That is, the p-value is .000000000000000777.
Figure 9.11: The MegaStat Output for Testing H0: p = .05 versus Ha: p > .05
Exercises for Section 9.4
CONCEPTS
9.63 If we test a hypothesis to provide evidence supporting the claim that a majority of voters prefer a political candidate, explain the difference between p and .
9.64 If we test a hypothesis to provide evidence supporting the claim that more than 30 percent of all consumers prefer a particular brand of beer, explain the difference between p and .
9.65 If we test a hypothesis to provide evidence supporting the claim that fewer than 5 percent of the units produced by a process are defective, explain the difference between p and .
9.66 What condition must be satisfied in order to appropriately use the methods of this section?
METHODS AND APPLICATIONS
9.67 For each of the following sample sizes and hypothesized values of the population proportion p, determine whether the sample size is large enough to use the large sample test about p given in this section:
a n = 400 and p0 = .5.
b n = 100 and p0 = .01.
c n = 10,000 and p0 = .01.
d n = 100 and p0 = .2.
e n = 256 and p0 = .7.
f n = 200 and p0 = .98.
g n = 1,000 and p0 = .98.
h n = 25 and p0 = .4.
9.68 Suppose we wish to test H0: p ≤ .8 versus Ha: p > .8 and that a random sample of n = 400 gives a sample proportion .
a Test H0 versus Ha at the .05 level of significance by using a critical value. What do you conclude?
b Find the p-value for this test.
c Use the p-value to test H0 versus Ha by setting α equal to .10, .05, .01, and .001. What do you conclude at each value of α?
9.69 Suppose we test H0: p = .3 versus Ha: p ≠ .3 and that a random sample of n = 100 gives a sample proportion .
a Test H0 versus Ha at the .01 level of significance by using a critical value. What do you conclude?
b Find the p-value for this test.
c Use the p-value to test H0 versus Ha by setting α equal to .10, .05, .01, and .001. What do you conclude at each value of α?
9.70 Suppose we are testing H0: p ≤ .5 versus Ha: p > .5, where p is the proportion of all beer drinkers who have tried at least one brand of “cold-filtered beer.” If a random sample of 500 beer drinkers has been taken and if equals .57, how many beer drinkers in the sample have tried at least one brand of “cold-filtered beer”?