1. 1/49 EF 507 QUANTITATIVE METHODS FOR ECONOMICS AND FINANCE FALL 2008 Chapter 9 Estimation: Additional Topics

2. 2/49 Estimation: Additional Topics Chapter Topics Population Means, Independent Samples Population Means, Dependent Samples Population Variance Group 1 vs. independent Group 2 Same group before vs. after treatment Variance of a normal distribution Examples: Population Proportions Proportion 1 vs. Proportion 2

3. 3/49 1. Dependent Samples Tests Means of 2 Related Populations  Paired or matched samples  Repeated measures (before/after)  Use difference between paired values:  Eliminates Variation Among Subjects  Assumptions:  Both Populations Are Normally Distributed Dependent samples d i = x i - y i

4. 4/49 Mean Difference The i th paired difference is d i, where d i = x i - y i The point estimate for the population mean paired difference is d : n is the number of matched pairs in the sample The sample standard deviation is: Dependent samples

5. 5/49 Confidence Interval for Mean Difference The confidence interval for difference between population means, μ d, is Where n = the sample size (number of matched pairs in the paired sample) Dependent samples

6. 6/49  The margin of error is  t n-1,  /2 is the value from the Student’s t distribution with (n – 1) degrees of freedom for which Confidence Interval for Mean Difference (continued) Dependent samples

7. 7/49  Six people sign up for a weight loss program. You collect the following data: Paired Samples Example Weight: Person Before (x) After (y) Difference, d i 1 136 125 11 2 205 195 10 3 157 150 7 4 138 140 - 2 5 175 165 10 6 166 160 6 42 d =  didi n = 7.0

8. 8/49  For a 95% confidence level, the appropriate t value is t n-1,  /2 = t 5,0.025 = 2.571  The 95% confidence interval for the difference between means, μ d, is Paired Samples Example (continued) Since this interval contains zero, we can not be 95% confident, given this limited data, that the weight loss program helps people lose weight

9. 9/49 2. Difference Between Two Means Population means, independent samples Goal: Form a confidence interval for the difference between two population means, μ x – μ y x – y  Different data sources  Unrelated  Independent  Sample selected from one population has no effect on the sample selected from the other population  The point estimate is the difference between the two sample means:

10. 10/49 Difference Between Two Means Population means, independent samples Confidence interval uses z  /2 Confidence interval uses a value from the Student’s t distribution σ x 2 and σ y 2 assumed equal σ x 2 and σ y 2 known σ x 2 and σ y 2 unknown σ x 2 and σ y 2 assumed unequal (continued)

11. 11/49 Population means, independent samples 2. σ x 2 and σ y 2 Known Assumptions:  Samples are randomly and independently drawn  both population distributions are normal  Population variances are known * σ x 2 and σ y 2 known σ x 2 and σ y 2 unknown

12. 12/49 Population means, independent samples …and the random variable has a standard normal distribution When σ x and σ y are known and both populations are normal, the variance of X – Y is (continued) * σ x 2 and σ y 2 known σ x 2 and σ y 2 unknown σ x 2 and σ y 2 Known

13. 13/49 Population means, independent samples The confidence interval for μ x – μ y is: Confidence Interval, σ x 2 and σ y 2 Known * σ x 2 and σ y 2 known σ x 2 and σ y 2 unknown

14. 14/49 Population means, independent samples 3.a) σ x 2 and σ y 2 Unknown, Assumed Equal Assumptions:  Samples are randomly and independently drawn  Populations are normally distributed  Population variances are unknown but assumed equal * σ x 2 and σ y 2 assumed equal σ x 2 and σ y 2 known σ x 2 and σ y 2 unknown σ x 2 and σ y 2 assumed unequal

15. 15/49 Population means, independent samples (continued) Forming interval estimates:  The population variances are assumed equal, so use the two sample standard deviations and pool them to estimate σ  use a t value with (n x + n y – 2) degrees of freedom * σ x 2 and σ y 2 assumed equal σ x 2 and σ y 2 known σ x 2 and σ y 2 unknown σ x 2 and σ y 2 assumed unequal σ x 2 and σ y 2 Unknown, Assumed Equal

16. 16/49 Population means, independent samples The pooled variance is (continued) * σ x 2 and σ y 2 assumed equal σ x 2 and σ y 2 known σ x 2 and σ y 2 unknown σ x 2 and σ y 2 assumed unequal σ x 2 and σ y 2 Unknown, Assumed Equal

17. 17/49 The confidence interval for μ 1 – μ 2 is: Where * Confidence Interval, σ x 2 and σ y 2 Unknown, Equal σ x 2 and σ y 2 assumed equal σ x 2 and σ y 2 unknown σ x 2 and σ y 2 assumed unequal

18. 18/49 Pooled Variance Example You are testing two computer processors for speed. Form a confidence interval for the difference in CPU speed. You collect the following speed data (in Mhz): CPU x CPU y Number Tested 17 14 Sample mean 3,004 2,538 Sample std dev 74 56 Assume both populations are normal with equal variances, and use 95% confidence

19. 19/49 Calculating the Pooled Variance The pooled variance is: The t value for a 95% confidence interval is:

20. 20/49 Calculating the Confidence Limits  The 95% confidence interval is We are 95% confident that the mean difference in CPU speed is between 416.69 and 515.31 Mhz.

21. 21/49 Population means, independent samples 3.b) σ x 2 and σ y 2 Unknown, Assumed Unequal Assumptions:  Samples are randomly and independently drawn  Populations are normally distributed  Population variances are unknown and assumed unequal * σ x 2 and σ y 2 assumed equal σ x 2 and σ y 2 known σ x 2 and σ y 2 unknown σ x 2 and σ y 2 assumed unequal

22. 22/49 Population means, independent samples σ x 2 and σ y 2 Unknown, Assumed Unequal (continued) Forming interval estimates:  The population variances are assumed unequal, so a pooled variance is not appropriate  use a t value with degrees of freedom, where σ x 2 and σ y 2 known σ x 2 and σ y 2 unknown * σ x 2 and σ y 2 assumed equal σ x 2 and σ y 2 assumed unequal

23. 23/49 The confidence interval for μ 1 – μ 2 is: * Confidence Interval, σ x 2 and σ y 2 Unknown, Unequal σ x 2 and σ y 2 assumed equal σ x 2 and σ y 2 unknown σ x 2 and σ y 2 assumed unequal Where

24. 24/49 Two Population Proportions Goal: Form a confidence interval for the difference between two population proportions, P x – P y The point estimate for the difference is Population proportions Assumptions: Both sample sizes are large (generally at least 40 observations in each sample)

25. 25/49 Two Population Proportions Population proportions (continued)  The random variable is approximately normally distributed

26. 26/49 Confidence Interval for Two Population Proportions Population proportions The confidence limits for P x – P y are:

27. 27/49 Example: Two Population Proportions Form a 90% confidence interval for the difference between the proportion of men and the proportion of women who have college degrees.  In a random sample, 26 of 50 men and 28 of 40 women had an earned college degree

28. 28/49 Example: Two Population Proportions Men: Women: (continued) For 90% confidence, Z  /2 = 1.645

29. 29/49 Example: Two Population Proportions The confidence limits are: so the confidence interval is -0.3465 < P x – P y < -0.0135 Since this interval does not contain zero we are 90% confident that the two proportions are not equal (continued)

30. 30/49 Confidence Intervals for the Population Variance Population Variance  Goal: Form a confidence interval for the population variance, σ 2  The confidence interval is based on the sample variance, s 2  Assumed: the population is normally distributed

31. 31/49 Confidence Intervals for the Population Variance Population Variance The random variable follows a chi-square distribution with (n – 1) degrees of freedom (continued) The chi-square value denotes the number for which

32. 32/49 Confidence Intervals for the Population Variance Population Variance The (1 -  )% confidence interval for the population variance is (continued)

33. 33/49 Example You are testing the speed of a computer processor. You collect the following data (in Mhz): CPU x Sample size 17 Sample mean 3,004 Sample std dev 74 Assume the population is normal. Determine the 95% confidence interval for σ x 2

34. 34/49 Finding the Chi-square Values  n = 17 so the chi-square distribution has (n – 1) = 16 degrees of freedom   = 0.05, so use the the chi-square values with area 0.025 in each tail: probability α/2 = 0.025  2 16 = 28.85  2 16 = 6.91 probability α/2 =0.025

35. 35/49 Calculating the Confidence Limits  The 95% confidence interval is Converting to standard deviation, we are 95% confident that the population standard deviation of CPU speed is between 55.1 and 112.6 Mhz

36. 36/49 Sample PHStat Output

37. 37/49 Sample PHStat Output Input Output (continued)

38. 38/49 Sample Size Determination For the Mean Determining Sample Size For the Proportion

39. 39/49 Margin of Error  The required sample size can be found to reach a desired margin of error (ME) with a specified level of confidence (1 -  )  The margin of error is also called sampling error  the amount of imprecision in the estimate of the population parameter  the amount added and subtracted to the point estimate to form the confidence interval

40. 40/49 For the Mean Determining Sample Size Margin of Error (sampling error) Sample Size Determination

41. 41/49 For the Mean Determining Sample Size (continued) Now solve for n to get Sample Size Determination

42. 42/49 Sample Size Determination  To determine the required sample size for the mean, you must know:  The desired level of confidence (1 -  ), which determines the z  /2 value  The acceptable margin of error (sampling error), ME  The standard deviation, σ (continued)

43. 43/49 Required Sample Size Example If  = 45, what sample size is needed to estimate the mean within ± 5 with 90% confidence? (Always round up) So the required sample size is n = 220

44. 44/49 Determining Sample Size For the Proportion Margin of Error (sampling error) Sample Size Determination

45. 45/49 Determining Sample Size For the Proportion Substitute 0.25 for and solve for n to get (continued) Sample Size Determination cannot be larger than 0.25, when = 0.5

46. 46/49  The sample and population proportions, and P, are generally not known (since no sample has been taken yet)  P(1 – P) = 0.25 generates the largest possible margin of error (so guarantees that the resulting sample size will meet the desired level of confidence)  To determine the required sample size for the proportion, you must know:  The desired level of confidence (1 -  ), which determines the critical z  /2 value  The acceptable sampling error (margin of error), ME  Estimate P(1 – P) = 0.25 (continued) Sample Size Determination

47. 47/49 Required Sample Size Example How large a sample would be necessary to estimate the true proportion defective in a large population within ±3%, with 95% confidence?

48. 48/49 Required Sample Size Example Solution: For 95% confidence, use z 0.025 = 1.96 ME = 0.03 Estimate P(1 – P) = 0.25 So use n = 1,068 (continued)

49. 49/49 PHStat Sample Size Options