1. 1 If we live with a deep sense of gratitude, our life will be greatly embellished.

2. Chapter 19 Inference about a Population Proportion BPS - 5th Ed.Chapter 192

3. Proportions BPS - 5th Ed.Chapter 193  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: “p-hat”

4. Inference about a Proportion Simple Conditions BPS - 5th Ed.Chapter 194

5. Inference about a Proportion Sampling Distribution BPS - 5th Ed.Chapter 195

6. Case Study BPS - 5th Ed.Chapter 196 Science News, Jan. 27, 1995, p. 451. Comparing Fingerprint Patterns

7. Case Study: Fingerprints BPS - 5th Ed.Chapter 197 u Fingerprints are a “sexually dimorphic trait…which means they are among traits that may be influenced by prenatal hormones.” u 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. u Compare fingerprint patterns of heterosexual and homosexual men.

8. Case Study: Fingerprints Study Results BPS - 5th Ed.Chapter 198  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.

9. Case Study: Fingerprints A Question BPS - 5th Ed.Chapter 199 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%?

10. Case Study: Fingerprints Sampling Distribution BPS - 5th Ed.Chapter 1910

11. Case Study: Fingerprints Answer to Question BPS - 5th Ed.Chapter 1911  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%).

12. Standardized Sample Proportion BPS - 5th Ed.Chapter 1912  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.

13. Building a Confidence Interval Population Proportion BPS - 5th Ed.Chapter 1913

14. Standard Error BPS - 5th Ed.Chapter 1914 Since the population proportion p is unknown, the standard deviation of the sample proportion will need to be estimated by substituting for p.

15. Confidence Interval BPS - 5th Ed.Chapter 1915

16. Case Study: Soft Drinks BPS - 5th Ed.Chapter 1916 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.

17. Case Study: Soft Drinks BPS - 5th Ed.Chapter 1917 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.

18. The Hypotheses for Proportions BPS - 5th Ed.Chapter 1918 u Null: H 0 : p = p 0 u One sided alternatives H a : p > p 0 H a : p < p 0 u Two sided alternative H a : p  p 0

19. Test Statistic for Proportions BPS - 5th Ed.Chapter 1919  Start with the z statistic that results from standardizing : u Assuming that the null hypothesis is true (H 0 : p = p 0 ), we use p 0 in the place of p:

20. P -value for Testing Proportions BPS - 5th Ed.Chapter 1920  H a : p > p 0 v P-value is the probability of getting a value as large or larger than the observed test statistic (z) value.  H a : p < p 0 v P-value is the probability of getting a value as small or smaller than the observed test statistic (z) value.  H a : p ≠ p 0 v P-value is two times the probability of getting a value as large or larger than the absolute value of the observed test statistic (z) value.

21. BPS - 5th Ed.Chapter 1921

22. Case Study BPS - 5th Ed.Chapter 1922 Brown, C. S., (1994) “To spank or not to spank.” USA Weekend, April 22-24, pp. 4-7. Parental Discipline What are parents’ attitudes and practices on discipline?

23. Case Study: Discipline BPS - 5th Ed.Chapter 1923 u Nationwide random telephone survey of 1,250 adults that covered many topics u 474 respondents had children under 18 living at home –results on parental discipline are based on the smaller sample u reported margin of error –5% for this smaller sample Scenario

24. Case Study: Discipline BPS - 5th Ed.Chapter 1924 “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.) Reported Results

25. Case Study: Discipline BPS - 5th Ed.Chapter 1925 u 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. [H 0 : p = 0.50] u Alt: A majority (more than 50%) of parents did not physically discipline their children in 1993. [H a : p > 0.50] The Hypotheses

26. Case Study: Discipline BPS - 5th Ed.Chapter 1926 Based on the sample u n = 474 (large, so proportions follow Normal distribution) u no physical discipline: 51% – – standard error of p-hat: (where.50 is p 0 from the null hypothesis) u standardized score (test statistic) z = (0.51 - 0.50) / 0.023 = 0.43 Test Statistic

27. Case Study: Discipline BPS - 5th Ed.Chapter 1927 P-value = 0.3336 From Table A, z = 0.43 is the 66.64 th percentile. z = 0.43 0.5000.5230.4770.5460.4540.5690.431 012-23-3 z: P-value

28. Case Study: Discipline BPS - 5th Ed.Chapter 1928 1. Hypotheses:H 0 : p = 0.50 H a : p > 0.50 2. Test Statistic: 3. P-value:P-value = P(Z > 0.43) = 1 – 0.6664 = 0.3336 4. Conclusion: Since the P-value is larger than  = 0.10, there is no strong evidence that a majority of parents did not physically discipline their children during 1993.

29. Chapter 20 Comparing Two Proportions BPS - 5th Ed.Chapter 2029

30. Two-Sample Problems BPS - 5th Ed.Chapter 2030  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.

31. Case Study BPS - 5th Ed.Chapter 2031 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. nDefects Machine A1408 Machine B20011 Machine Reliability

32. Inference about the Difference p 1 – p 2 Simple Conditions BPS - 5th Ed.Chapter 2032  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 mean p 1 – p 2 and standard deviation

33. Inference about the Difference p 1 – p 2 Sampling Distribution BPS - 5th Ed.Chapter 2033

34. Standard Error BPS - 5th Ed.Chapter 2034 Since the population proportions p 1 and p 2 are unknown, the standard deviation of the difference in sample proportions will need to be estimated by substituting for p 1 and p 2 :

35. BPS - 5th Ed.Chapter 2035

36. Case Study: Reliability BPS - 5th Ed.Chapter 2036 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. Compute a 90% confidence interval for the difference in reliabilities (as measured by proportion of defectives) for the two machines. Confidence Interval

37. The Hypotheses for Testing Two Proportions BPS - 5th Ed.Chapter 2037 u Null: H 0 : p 1 = p 2 u One sided alternatives H a : p 1 > p 2 H a : p 1 < p 2 u Two sided alternative H a : p 1  p 2

38. Pooled Sample Proportion BPS - 5th Ed.Chapter 2038  If H 0 is true (p 1 =p 2 ), then the two proportions are equal to some common value p.  Instead of estimating p 1 and p 2 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. pooled sample proportion

39. Test Statistic for Two Proportions BPS - 5th Ed.Chapter 2039  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:

40. P -value for Testing Two Proportions BPS - 5th Ed.Chapter 2040  H a : p 1 > p 2 v P-value is the probability of getting a value as large or larger than the observed test statistic (z) value.  H a : p 1 < p 2 v P-value is the probability of getting a value as small or smaller than the observed test statistic (z) value.  H a : p 1 ≠ p 2 v P-value is two times the probability of getting a value as large or larger than the absolute value of the observed test statistic (z) value.

41. BPS - 5th Ed.Chapter 2041

42. Case Study: BPS - 5th Ed.Chapter 2042 u A university financial aid office polled a simple random sample of undergraduate students to study their summer employment. u Not all students were employed the previous summer. Here are the results: u Is there evidence that the proportion of male students who had summer jobs differs from the proportion of female students who had summer jobs. Summer StatusMenWomen Employed718593 Not Employed79139 Total797732 Summer Jobs

43. Case Study: Summer Jobs BPS - 5th Ed.Chapter 2043 u Null: The proportion of male students who had summer jobs is the same as the proportion of female students who had summer jobs. [H 0 : p 1 = p 2 ] u Alt: The proportion of male students who had summer jobs differs from the proportion of female students who had summer jobs. [H a : p 1 ≠ p 2 ] The Hypotheses

44. Case Study: Summer Jobs BPS - 5th Ed.Chapter 2044 u n 1 = 797 and n 2 = 732 (both large, so test statistic follows a Normal distribution) u Pooled sample proportion: u standardized score (test statistic): Test Statistic

45. Case Study: Summer Jobs BPS - 5th Ed.Chapter 2045 1. Hypotheses:H 0 : p 1 = p 2 H a : p 1 ≠ p 2 2. Test Statistic: 3. P-value: P-value = 2P(Z > 5.07) = 0.000000396 (using a computer) P-value = 2P(Z > 5.07) 3.49 (the largest z-value in the table)] 4. Conclusion: Since the P-value is smaller than  = 0.001, there is very strong evidence that the proportion of male students who had summer jobs differs from that of female students.

46. Chapter 21 Inference about Variables: Part III Review

47. BPS - 5th Ed.Chapter 2147