SECTION 7.4 Hypothesis Testing for Proportions 389 Hypothesis Test for a Proportion A researcher claims that less than 69% of U.S. adults who use live television streaming platforms have upgraded to advertisement-free service tiers. In a random sample of 100 U.S. adults who use live television streaming platforms, 65% say they have upgraded to advertisement-free service tiers. At a = 0.01, is there enough evidence to support the researcher’s claim? (Adapted from The Harris Poll) SOLUTION The products np = 10010.692 = 69 and nq = 10010.312 = 31 are both greater than 5. So, you can use a z@test. The claim is “less than 69% of U.S. adults who use live television streaming platforms have upgraded to advertisement-free service tiers.” So, the null and alternative hypotheses are H0: p Ú 0.69 and Ha: p 6 0.69. (Claim) Because the test is a left-tailed test and the level of significance is a = 0.01, the critical value is z0 = -2.33 and the rejection region is z 6 -2.33. The standardized test statistic is z = np - p 2 pq n Because np Ú 5 and nq Ú 5, you can use the z-test. = 0.65 - 0.69 2 10.69210.312 100 Assume p = 0.69. ≈ -0.86. Round to two decimal places. The figure shows the location of the rejection region and the standardized test statistic z. Because z is not in the rejection region, you fail to reject the null hypothesis. Interpretation There is not enough evidence at the 1% level of significance to support the claim that less than 69% of U.S. adults who use live television streaming platforms have upgraded to advertisement-free service tiers. TRY IT YOURSELF 1 A researcher claims that more than 36% of U.S. adults would consider buying a product or service directly from an advertisement that is run during streaming programming. In a random sample of 150 adults, 33% say they would consider buying a product or service directly from an advertisement that is run during streaming programming. At a = 0.01, is there enough evidence to support the researcher’s claim? (Adapted from The Harris Poll) Answer: Page A41 To use a P-value to perform the hypothesis test in Example 1, you can use technology, as shown at the left, or you can use Table 4. Using Table 4, the area corresponding to z = -0.86 is 0.1949. Because this is a left-tailed test, the P-value is equal to the area to the left of z = -0.86. So, P = 0.1949. (This value differs from the one found using technology due to rounding.) Because the P-value is greater than a = 0.01, you fail to reject the null hypothesis. Note that this is the same result obtained in Example 1. 1% Level of Significance z −3 −4 −2 −1 0 1 2 3 4 z0 = −2.33 z ≈ −0.86 To explore this topic further, see Activity 7.4 on page 393. 7.4 Study Tip Remember that when you fail to reject H0, a type II error is possible. For instance, in Example 1 the null hypothesis, p Ú 0.69, may be false. TI-84 PLUS prop<0.69 z=-0.8648775001 p=0.1935529676 p=0.65 n=100 1PropZTest See TI-84 Plus steps on page 415. EXAMPLE 1
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