9.3 Hypothesis Tests for Population Proportions LEARNING GOAL Understand and interpret hypothesis tests for claims made about population proportions Page 11 Copyright © 2009 Pearson Education, Inc.

Should the candidate be confident of winning? The sample proportion is Suppose a political candidate commissions a poll in advance of a close election. Using a random sample of n 400 likely voters, the poll finds that 204 people support the candidate. Should the candidate be confident of winning? The sample proportion is p = = 0.510 204 400 We can now formulate the null and alternative hypotheses. As usual, we set up our null hypothesis as an equality: H0: p = 0.5 (50% of voters favor the candidate) Page 394 The candidate wants to know if she has majority support, so the alternative hypothesis is right-tailed: Ha: p > 0.5 (more than 50% of voters favor the candidate) Copyright © 2009 Pearson Education, Inc. Slide 9.3- 2

Calculations for Hypothesis Tests with Proportions How do we determine whether there is enough evidence in the sample to reject the null hypothesis? We will use 4-steps of the hypothesis test process on page 376. Step 1 is to formulate the hypotheses H0: p = 0.5 (more than 50% of voters favor the candidate) Ha: p > 0.5 (more than 50% of voters favor the candidate) Step 2 is to collect the sample data. Page 394 p = = 0.510 204 400 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 3

Copyright © 2009 Pearson Education, Inc. Step 3 is to determine the likelihood that the sample result could have arisen by chance if the null hypothesis is true. Under the starting assumption that the null hypothesis is true (that the proportion of people in the population who favor the candidate is 0.5), the peak of this distribution will be the population proportion claimed by the null hypothesis, p = 0.5. The standard deviation of distribution of sample proportions is Page 394 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 4

Copyright © 2009 Pearson Education, Inc. • If the sample result is close to the peak of the sampling distribution, then we have no reason to think the null hypothesis is wrong and we do not reject the null hypothesis. • If the sample result is far from the peak of the sampling distribution, then the more likely explanation is that the sampling distribution does not really peak where the null hypothesis claims, in which case we reject the null hypothesis. Page 395 Figure 9.7 The distribution of sample proportions for an election poll with a mean of p = 0.5 and a standard deviation of 0.025. The sample proportion has a standard score of 0.4 and a P-value of 0.34. Copyright © 2009 Pearson Education, Inc. Slide 9.3- 5

sample proportion - population proportion The formula for the standard score (z) for the distribution of sample proportions, is sample proportion - population proportion standard deviation of sampling distribution The sample size is the n = 400 people interviewed, the sample proportion is the 51% of the sample that supports the candidate and the population proportion is the p = 0.5 claimed by the null hypothesis. Page 395 In other words, the sample used in the poll has a proportion that is 0.4 standard deviation above the peak of the distribution of sample proportions. Copyright © 2009 Pearson Education, Inc. Slide 9.3- 6

Copyright © 2009 Pearson Education, Inc. Standard Score for the Sample Proportion in a Hypothesis Test Given the sample size (n), the sample proportion ( ), and the claimed population proportion (p), the standard score for the sample proportion is p ˆ Page 395 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 7

Copyright © 2009 Pearson Education, Inc. Significance Levels and P-Values Because the election poll example uses a right-tailed test, we find the critical values for significance in Table 9.1 (page 383). Step 4 is to decide whether to reject or not reject the null hypothesis. Using statistical significance--the standard score of z = 0.4 is not greater than the critical value of z = 1.645 for significance at the 0.05 level, confirming that the result should not cause us to reject the null hypothesis. Page 396 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 8

Copyright © 2009 Pearson Education, Inc. To find the P-value, we use the tables in Appendix A. The tables show that the area to the left of a standard score of z = 0.4 is 0.6554. Because this is a right-tailed test, we subtract this value from 1 to find the area to the right, which is the P-value for this hypothesis test. That is, the P-value is 1 – 0.6554 = 0.3446, telling us that if the null hypothesis is true, there is a more than 0.34 chance of randomly selecting a sample as extreme as the one found in this poll. Step 4 using the P-value--with such a high probability of drawing such a sample by chance, we have no reason to reject the null hypothesis, so the candidate cannot assume that she has the support of more than 50% of voters. Page 396 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 9

Copyright © 2009 Pearson Education, Inc. Summary of Hypothesis Tests with Proportions Because we are dealing with population proportions, the null hypothesis has the form p = claimed value. To decide whether to reject or not reject the null hypothesis, we must determine whether a sample as extreme as the one found in the hypothesis test is likely or unlikely to occur if the null hypothesis is true. We determine this likelihood from the standard score (z) of the sample proportion, which we compute from the formula Page 396 ˆ p where n is the sample size, is the sample proportion, and p is the population proportion claimed by the null hypothesis. Copyright © 2009 Pearson Education, Inc. Slide 9.3- 10

Copyright © 2009 Pearson Education, Inc. Summary of Hypothesis Tests with Proportions (cont.) We then use the standard score in the same two ways it was used for hypothesis tests with means: 1. We can assess its level of statistical significance by comparing the standard score to the critical values given in Table 9.1 ( page 383) for one-tailed tests and in the box on page 386 for two-tailed tests. 2. We can determine its P-value with standard score tables like those in Appendix A. For a left-tailed test, the P-value is the area under the normal curve to the left of the standard score; for a right-tailed test, it is the area under the normal curve to the right of the standard score; and for a two-tailed test, it is double the value we would find if we were calculating the P-value for a one-tailed test. Page 396 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 11

Copyright © 2009 Pearson Education, Inc. Summary of Hypothesis Tests with Proportions (cont.) If the result is statistically significant at the chosen level (usually either the 0.05 or the 0.01 significance level), we reject the null hypothesis. If it is not statistically significant, we do not reject the null hypothesis. Page 396 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 12

Copyright © 2009 Pearson Education, Inc. EXAMPLE 2 Left-Handed Population A random sample of n = 750 people is selected, of whom 92 are left-handed. Use these sample data to test the claim that 10% of the population is left-handed. Solution: We again follow the four-step process. Step 1. The claim concerns the proportion of the population that is left-handed, so this is a test with a population proportion. The null hypothesis is the claim that 10% of the population is left-handed, or H0: p = 0.1. To test this claim, we need to account for the possibility that the actual population proportion is either less than or greater than 10%. Therefore, the alternative hypothesis is Ha: p ≠ 0.1, which calls for a two-tailed test. Pages 397-398 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 13

Copyright © 2009 Pearson Education, Inc. EXAMPLE 2 Left-Handed Population Solution: (cont.) Step 2. The sample statistics are the sample size, n = 750, and the proportion of left-handed people in the sample: Step 3. The standard score for this sample proportion is Page 398 From the box on page 386, the critical values for significance at the 0.05 level in a two-tailed test are standard scores less than -1.96 or greater than 1.96. Copyright © 2009 Pearson Education, Inc. Slide 9.3- 14

Copyright © 2009 Pearson Education, Inc. EXAMPLE 2 Left-Handed Population Solution: Step 3. (cont.) The standard score of 2.09 for this test is greater than 1.96, so we conclude that the test is significant at the 0.05 level. From Appendix A, the area to the right of a standard score of 2.09 is 1 – 0.9817 = 0.0183. This would be the P-value if we were conducting a one-tailed test. Because we have a two-tailed test, we double it to find that the P-value is 2 × 0.0183 = 0.0366. Figure 9.9 (next slide) shows the meaning of this P-value on the sampling distribution. Page 398 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 15

Copyright © 2009 Pearson Education, Inc. Page 398 Figure 9.9 This graph shows the position of the sample proportion on the distribution of sample proportions for Example 2. Because this is a two-tailed test, the P-value corresponds to the area under the curve more than 2.09 standard deviations from the peak in either direction. Copyright © 2009 Pearson Education, Inc. Slide 9.3- 16

Copyright © 2009 Pearson Education, Inc. EXAMPLE 2 Left-Handed Population Solution: (cont.) Step 4. Because the test is significant at the 0.05 level, we reject the null hypothesis and conclude that the proportion of the population that is left-handed is not equal to 10%. Remembering that the significance level is the probability of a type I error, we recognize that there is a 0.05 probability that we made a type I error of rejecting a hypothesis that is actually true. Page 398 Copyright © 2009 Pearson Education, Inc. Slide 9.3- 17

Copyright © 2009 Pearson Education, Inc. The End Copyright © 2009 Pearson Education, Inc. Slide 9.3- 18