1 / 21

7.4 Hypothesis Testing for Proportions

7.4 Hypothesis Testing for Proportions. Statistics Mrs. Spitz Spring 2009. Objectives/Assignment. How to use a z-test to test a population proportion, p. Assignment: pp. 344-345 #1-16 all. Hypothesis Test for Proportions.

Download Presentation

7.4 Hypothesis Testing for Proportions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 7.4 Hypothesis Testing for Proportions Statistics Mrs. Spitz Spring 2009

  2. Objectives/Assignment • How to use a z-test to test a population proportion, p. Assignment: pp. 344-345 #1-16 all

  3. Hypothesis Test for Proportions • In this section, you will learn how to test a population proportion, p. If np ≥ 5 and nq ≥ 5 for a binomial distribution, then the sampling distribution for is normal with and

  4. WRITE THIS DOWN!!!

  5. Ex. 1: Hypothesis Test for a Proportion • A medical researcher claims that less than 20% of American adults are allergic to a medication. In a random sample of 100 adults, 15% say they have such an allergy. Test the researcher’s claim at  = 0.01.

  6. SOLUTION • The products np = 100(0.20)= 20 and nq = 100(0.80) = 80 are both greater than 5. So, you can use the z-test. The claim is “less than 20% are allergic to a medication.” So the null and alternative hypothesis are: Ho: p ≥ 0.2 and Ha: p < 0.2 (Claim)

  7. Solution continued . . . • Because the test is a left-tailed test and the level of significance is  = 0.01, the critical value is zo = -2.33 and the rejection region is z < -2.33. Using the z-test, the standardized test statistic is:

  8. SOLUTION Continued . . . • The graph shows the location of the rejection region and the standardized test statistic, z. Because z is not in the rejection region, you should decide not to reject the null hypothesis. In other words, there is not enough evidence to support the claim that less than 20% of Americans are allergic to the medication.

  9. Study Tip: • Remember that when you fail to reject Ho, a type II error is possible. For instance, in Example 1, the null hypothesis, p ≥ 0.20, may be false.

  10. Ex. 2 Hypothesis Test for a Proportion • Harper’s Index claims that 23% of Americans are in favor outlawing cigarettes. You decide to test this claim and ask a random sample of 200 Americas whether they are in favor outlawing cigarettes. Of the 200 Americans, 27% are in favor. At  = 0.05, is there enough evidence to reject the claim?

  11. SOLUTION: • The products np = 200(0.23) = 45 and nq = 200(0.77) = 154 are both greater than 5. So you can use a z-test. The claim is “23% of Americans are in favor of outlawing cigarettes.” So, the null and alternative hypotheses are: Ho: p = 0.23 (Claim)and Ha: p  0.23

  12. SOLUTION continued . . . • Because the test is a two-tailed test, and the level of significance is  = 0.05, the critical values are –zo = -1.96 and zo = 1.96. The rejection regions are z < -1.96 and z> 1.96

  13. SOLUTION continued . . . • Using the z-test, the standardized test statistic is:

  14. SOLUTION Continued . . . • The graph shows the location of the rejection regions and the standardized test statistic, z. • Because z is not in the rejection region, you should fail to reject the null hypothesis. At the 5% level of significance, there is not enough evidence to reject the claim that 23% of Americans are in favor of outlawing cigarettes.

  15. Ex. 3 Hypothesis Test a Proportion • The Pew Research Center claims that more than 55% of American adults regularly watch a network news broadcast. You decide to test this claim and ask a random sample of 425 Americans whether they regularly watch a network news broadcast. Of the 425 Americans, 255 responded yes. At  = 0.05, is there enough evidence to support the claim?

  16. SOLUTION: • The products np = 425(0.55) = 235 and nq = 425(0.45) = 191 are both greater than 5. So you can use a z-test. The claim is “more than 55% of Americans watch a network news broadcast.” So, the null and alternative hypotheses are: Ho: p  0.55 and Ha: p > 0.55 (Claim)

  17. SOLUTION continued . . . • Because the test is a right-tailed test, and the level of significance is  = 0.05, the critical value is zo = 1.645 and the rejection region is z > 1.645.

  18. SOLUTION continued . . . • Using the z-test, the standardized test statistic is: REMEMBER P-hat is equal to x/n.

  19. SOLUTION Continued . . . • The graph shows the location of the rejection region and the standardized test statistic, z. Because z is in the rejection region, you should decide to There is enough evidence at the 5% level of significance, to support the claim that 55% of American adults regularly watch a network news broadcast.

  20. Upcoming dates: • Today and tomorrow – 7.4—Check in 7.3 for all others who did not get it done yesterday! • Thursday/Friday – 7.5 Hypothesis Testing for the Variance and Standard Deviation • Monday – No School – President’s Day • Tuesday/Wednesday—Chapter 7 Review pp. 361-364 #1-46 – Very long so we are taking the extra day for this. • Thursday – Chapter 7 Exam/Binder Check

More Related