1 / 47

If we live with a deep sense of gratitude, our life will be greatly embellished.

If we live with a deep sense of gratitude, our life will be greatly embellished. Chapter 19. Inference about a Population Proportion . “p-hat”. Proportions. The proportion of a population that has some outcome (“ success ”) is p .

wiley
Télécharger la présentation

If we live with a deep sense of gratitude, our life will be greatly embellished.

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. If we live with a deep sense of gratitude, our life will be greatly embellished.

  2. Chapter 19 Inference about a Population Proportion Chapter 19

  3. “p-hat” Proportions • The proportion of a population that has some outcome (“success”) is p. • The proportion of successes in a sample is measured by the sample proportion: Chapter 19

  4. Inference about a ProportionSimple Conditions Chapter 19

  5. Inference about a ProportionSampling Distribution Chapter 19

  6. Case Study Comparing Fingerprint Patterns Science News, Jan. 27, 1995, p. 451. Chapter 19

  7. Case Study: Fingerprints • Fingerprints are a “sexually dimorphic trait…which means they are among traits that may be influenced by prenatal hormones.” • It is known… • Most people have more ridges in the fingerprints of the right hand. (People with more ridges in the left hand have “leftward asymmetry.”) • Women are more likely than men to have leftward asymmetry. • Compare fingerprint patterns of heterosexual and homosexual men. Chapter 19

  8. 66 homosexual men were studied. 20 (30%) of the homosexual men showed leftward asymmetry. 186 heterosexual men were also studied. 26 (14%) of the heterosexual men showed leftward asymmetry. Case Study: FingerprintsStudy Results Chapter 19

  9. Case Study: FingerprintsA Question Assume that the proportion of all men who have leftward asymmetry is 15%. Is it unusual to observe a sample of 66 men with a sample proportion ( ) of 30% if the true population proportion (p) is 15%? Chapter 19

  10. Case Study: FingerprintsSampling Distribution Chapter 19

  11. Case Study: FingerprintsAnswer to Question • Where should about 95% of the sample proportions lie? • mean plus or minus two standard deviations 0.15  2(0.044) = 0.062 0.15 + 2(0.044) = 0.238 • 95% should fall between 0.062 & 0.238 • It would be unusual to see 30% with leftward asymmetry (30% is not between 6.2% & 23.8%). Chapter 19

  12. Standardized Sample Proportion • Inference about a population proportion p is based on the z statistic that results from standardizing : • z has approximately the standard normal distribution as long as the sample is not too small and the sample is not a large part of the entire population. Chapter 19

  13. Building a Confidence IntervalPopulation Proportion Chapter 19

  14. Standard Error Since the population proportion p is unknown, the standard deviation of the sample proportion will need to be estimated by substituting for p. Chapter 19

  15. Confidence Interval Chapter 19

  16. Case Study: Soft Drinks A certain soft drink bottler wants to estimate the proportion of its customers that drink another brand of soft drink on a regular basis. A random sample of 100 customers yielded 18 who did in fact drink another brand of soft drink on a regular basis. Compute a 95% confidence interval (z* = 1.960) to estimate the proportion of interest. Chapter 19

  17. Case Study: Soft Drinks We are 95% confident that between 10.5% and 25.5% of the soft drink bottler’s customers drink another brand of soft drink on a regular basis. Chapter 19

  18. The Hypotheses for Proportions • Null: H0:p = p0 • One sided alternatives Ha:p > p0 Ha:p < p0 • Two sided alternative Ha:p¹ p0 Chapter 19

  19. Test Statistic for Proportions • Start with the z statistic that results from standardizing : • Assuming that the null hypothesis is true(H0: p = p0), we use p0 in the place of p: Chapter 19

  20. P-value for Testing Proportions • Ha: p > p0 • P-value is the probability of getting a value as large or larger than the observed test statistic (z) value. • Ha: p < p0 • P-value is the probability of getting a value as small or smaller than the observed test statistic (z) value. • Ha: p ≠ p0 • P-value is two times theprobability of getting a value as large or larger than the absolute value of the observed test statistic (z) value. Chapter 19

  21. Chapter 19

  22. Case Study Parental Discipline Brown, C. S., (1994) “To spank or not to spank.” USA Weekend, April 22-24, pp. 4-7. What are parents’ attitudes and practices on discipline? Chapter 19

  23. Case Study: Discipline Scenario • Nationwide random telephone survey of 1,250 adults that covered many topics • 474 respondents had children under 18 living at home • results on parental discipline are based on the smaller sample • reported margin of error • 5% for this smaller sample Chapter 19

  24. Case Study: Discipline Reported Results “The 1994 survey marks the first time a majority of parents reported not having physically disciplined their children in the previous year. Figures over the past six years show a steady decline in physical punishment, from a peak of 64 percent in 1988.” • The 1994 sample proportion who did not spank or hit was 51% ! • Is this evidence that a majority of the population did not spank or hit?(Perform a test of significance.) Chapter 19

  25. Case Study: Discipline The Hypotheses • Null: The proportion of parents who physically disciplined their children in 1993 is the same as the proportion [p] of parents who did not physically discipline their children. [H0: p = 0.50] • Alt: A majority (more than 50%) of parents did not physically discipline their children in 1993. [Ha: p > 0.50] Chapter 19

  26. Case Study: Discipline Test Statistic • Based on the sample • n = 474 (large, so proportions follow Normal distribution) • no physical discipline: 51% • standard error of p-hat: • (where .50 is p0 from the null hypothesis) • standardized score (test statistic) • z = (0.51 - 0.50) / 0.023 = 0.43 Chapter 19

  27. 0.431 0.454 0.477 0.500 0.523 0.546 0.569 -3 -2 -1 0 1 2 3 z: z = 0.43 Case Study: Discipline P-value P-value = 0.3336 From Table A, z = 0.43 is the 66.64th percentile. Chapter 19

  28. Hypotheses: H0: p = 0.50 Ha: p > 0.50 • Test Statistic: • P-value: P-value = P(Z > 0.43) = 1 – 0.6664 = 0.3336 • Conclusion: Since the P-value is larger than a = 0.10, there is no strong evidence that a majority of parents did not physically discipline their children during 1993. Case Study: Discipline Chapter 19

  29. Chapter 20 Comparing Two Proportions Chapter 20

  30. Two-Sample Problems • The goal of inference is to compare the responses to two treatments or to compare the characteristics of two populations. • We have a separate sample from each treatment or each population. The units are not matched, and the samples can be of differing sizes. Chapter 20

  31. Case Study Machine Reliability A study is performed to test of the reliability of products produced by two machines. Machine A produced 8 defective parts in a run of 140, while machine B produced 10 defective parts in a run of 200. This is an example of when to use the two-proportion z procedures. Chapter 20

  32. Inference about the Difference p1 – p2Simple Conditions • The difference in the population proportions is estimated by the difference in the sample proportions: • When both of the samples are large, the sampling distribution of this difference is approximately Normal with meanp1 – p2 and standard deviation Chapter 20

  33. Inference about the Difference p1 – p2Sampling Distribution Chapter 20

  34. Standard Error Since the population proportions p1 and p2 are unknown, the standard deviation of the difference in sample proportions will need to be estimated by substituting for p1 and p2: Chapter 20

  35. Chapter 20

  36. Case Study: Reliability Confidence Interval Compute a 90% confidence interval for the difference in reliabilities (as measured by proportion of defectives) for the two machines. We are 90% confident that the difference in proportion of defectives for the two machines is between -3.97% and 4.39%. Since 0 is in this interval, it is unlikely that the two machines differ in reliability. Chapter 20

  37. The Hypotheses for TestingTwo Proportions • Null: H0:p1 = p2 • One sided alternatives Ha:p1 > p2 Ha:p1 < p2 • Two sided alternative Ha:p1¹ p2 Chapter 20

  38. pooled sample proportion Pooled Sample Proportion • If H0 is true (p1=p2), then the two proportions are equal to some common value p. • Instead of estimating p1 and p2 separately, we will combine or pool the sample information to estimate p. • This combined or pooled estimate is called the pooled sample proportion, and we will use it in place of each of the sample proportions in the expression for the standard error SE. Chapter 20

  39. Test Statistic for Two Proportions • Use the pooled sample proportion in place of each of the individual sample proportions in the expression for the standard error SE in the test statistic: Chapter 20

  40. P-value for Testing Two Proportions • Ha: p1> p2 • P-value is the probability of getting a value as large or larger than the observed test statistic (z) value. • Ha: p1< p2 • P-value is the probability of getting a value as small or smaller than the observed test statistic (z) value. • Ha: p1≠ p2 • P-value is two times theprobability of getting a value as large or larger than the absolute value of the observed test statistic (z) value. Chapter 20

  41. Chapter 20

  42. Case Study: Summer Jobs • A university financial aid office polled a simple random sample of undergraduate students to study their summer employment. • Not all students were employed the previous summer. Here are the results: • Is there evidence that the proportion of male students who had summer jobs differs from the proportion of female students who had summer jobs. Chapter 20

  43. Case Study: Summer Jobs The Hypotheses • Null: The proportion of male students who had summer jobs is the same as the proportion of female students who had summer jobs. [H0: p1 = p2] • Alt: The proportion of male students who had summer jobs differs from the proportion of female students who had summer jobs. [Ha: p1 ≠ p2] Chapter 20

  44. Case Study: Summer Jobs Test Statistic • n1 = 797 and n2 = 732 (both large, so test statistic follows a Normal distribution) • Pooled sample proportion: • standardized score (test statistic): Chapter 20

  45. Case Study: Summer Jobs • Hypotheses: H0: p1 = p2 Ha: p1 ≠ p2 • Test Statistic: • P-value:P-value = 2P(Z > 5.07) = 0.000000396 (using a computer)P-value = 2P(Z > 5.07) < 2(1 – 0.9998) = 0.0004 (Table A)[since 5.07 > 3.49 (the largest z-value in the table)] • Conclusion: Since the P-value is smaller than a = 0.001, there is very strong evidence that the proportion of male students who had summer jobs differs from that of female students. Chapter 20

  46. Chapter 21 Inference about Variables:Part III Review

  47. Chapter 21

More Related