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Estimating the Difference Between Two Means Given two independent random samples, a point estimate the difference between μ1 and μ2 is given by the statistic x1 x 2 We can build a confidence interval for μ1 - μ2 (given σ12 and σ22 known) as follows: ( x 1 x 2 ) z / 2 12 22 12 22 1 2 ( x 1 x 2 ) z / 2 n1 n2 n1 n2 Note: Use this calculation for large n as well (n>30). EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 1 Example: Estimating Difference Btwn Two Means A study was conducted and two types of engines, A and B, were compared. Fifty experiments were performed using engine A and 75 using B. The average gas mileage for A was 36 mpg, and 42 mpg for B. Find a 96% confidence interval on B-A where A and B are population mean gas mileages for engines A and B. Assume population standard deviations are A = 6 and B = 8. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 2 Example 9.10 Page 286 Find a 96% Confidence Interval xbarA = 36 mpg σA = 6 nA = 50 xbarB = 42 mpg σB = 8 nB = 75 α=0.04 α/2 =0.02 Z0.02 = 2.055 Calculations: 6 – 2.055 sqrt(64/75 + 36/50) < (μB - μA) < 6 + 2.055 sqrt(64/75 + 36/50) Results: 3.4224 < (μB - μA) < 8.5776 96% CI is (3.4224, 8.5776) EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 3 Differences Between Two Means: Variances Unknown Case 1: σ12 and σ22 unknown but “equal” Pages 287 and 288 ( x1 x 2 ) t ( / 2,n1 n2 2 )S p 1 1 1 1 1 2 ( x1 x 2 ) t ( / 2,n1 n2 2 )S p n1 n2 n1 n2 2 2 ( n 1 ) S ( n 1 ) S 1 2 2 Where, Sp2 1 n1 n2 2 Note v = n1 + n2 - 2 EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 4 Differences Between Two Means: Variances Unknown (Page 290) Case 2: σ12 and σ22 unknown and not equal ( x 1 x 2 ) t / 2, Where, s12 s22 s12 s22 1 2 ( x 1 x 2 ) t / 2, n1 n2 n1 n2 (S12 / n1 S22 / n2 )2 S2 / n 2 S2 / n 2 1 1 2 2 n1 1 n2 1 WOW! EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 5 Estimating μ1 – μ2 Example (σ12 and σ22 known) : A farm equipment manufacturer wants to compare the average daily downtime of two sheet-metal stamping machines located in two different factories. Investigation of company records for 100 randomly selected days on each of the two machines gave the following results: x1 = 12 minutes x2 = 10 minutes s12 = 12 s22 = 8 n1 = n2 = 100 Construct a 95% C.I. for μ1 – μ2 Since n is large, we can estimate σi2 with si2 EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 6 Solution ( x 1 x 2 ) z / 2 95% CI 12 22 12 22 1 2 ( x 1 x 2 ) z / 2 n1 n2 n1 n2 α/2 = ? Z.025 , Z.975 = 1.96 (12-10) + 1.96*sqrt(12/100 + 8/100) = 2 + 0.8765 1.1235 < μ1 – μ2 < 2.8765 Interpretation: If CI contains 0, then μ1 – μ2 may be either positive or negative (can’t say that one population mean is larger than the other); however, since the CI for μ1 – μ2 is positive, we conclude μ1 must be larger than μ2 . EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 7 μ1 – μ2 : σi2 Unknown Example (σ12 and σ22 unknown but “equal”): Suppose the farm equipment manufacturer was unable to gather data for 100 days. Using the data they were able to gather, they would still like to compare the downtime for the two machines. The data they gathered is shown below. Assume population standard deviations are equal (12 = 22). x1 = 12 minutes s12 = 12 n1 = 18 x2 = 10 minutes s22 = 8 n2 = 14 Construct a 95% C.I. for μ1 – μ2 EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 8 Solution Governing Equations: ( x 1 x 2 ) t / 2,n1 n2 2Sp S Calculations: t0.025,30 = 2.042 2 p 1 1 1 1 1 2 ( x 1 x 2 ) t / 2,n1 n2 2Sp n1 n2 n1 n2 ( n1 1) S12 (n2 1) S 22 n1 n2 2 sp2 = ((17*12)+(13*8))/30 = 10.267 sp = 3.204 2 + 2.042*3.204*sqrt(1/18 + 1/14) = 2 + 2.3314 -0.3314 < μ1 – μ2 < 4.3314 Interpretation: Since this CI contains 0, we can’t conclude μ1 is significantly different from μ2 . EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 9 Paired Observations Suppose we are evaluating observations that are not independent … For example, suppose a teacher wants to compare results of a pretest and posttest administered to the same group of students. Paired-observation or Paired-sample test … Example: murder rates in two consecutive years for several US cities. Construct a 90% confidence interval around the difference in consecutive years. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 10 Calculation of CI for Paired Data Example 9.13 We have 20 pairs of values. We calculate the difference for each pair. We calculate the sample standard deviation for the difference values. The appropriate equations are: μd = μ1 – μ2 sd 2 ( d d ) i n 1 Based on the data in Table 9.1 Dbar = -0.87 We determine that a (1-0.05)100% CI for μd is: sd d t / 2,n 1( ) n Sd = 2.9773 n=20 -2.2634 < μd < 0.5234 Interpretation: If CI contains 0, then μ1 – μ2 may be either positive or negative (can’t say that one is larger than the other). Since this CI contains 0, we conclude there is no significant difference between the mean TCDD levels in the fat tissue. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 11 C.I. for Proportions The proportion, P, in a binomial experiment may be estimated by X P n where X is the number of successes in n trials. For a sample, the point estimate of the parameter is x p n The mean for the sample proportion is p p and the sample variance 2 pq n p EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 12 C.I. for Proportions An approximate (1-α)100% confidence interval for p is: p z / 2 pq n Large-sample C.I. for p1 – p2 is: ( p1 p2 ) z / 2 p1 q1 p2 q2 n1 n2 Interpretation: If the CI contains 0 … EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 13 Interpretation of the Confidence Interval Significance 1. If the C.I. for p1 – p2 = (-0.0017, 0.0217), is there reason to believe there is a significant decrease in the proportion defectives using the new process? 2. What if the interval were (+0.002, +0.022)? 3. What if the interval were (-0.900, -0.700)? EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 14 Determining Sample Sizes for Developing Confidence Intervals Requires specification of an error amount е Requires specification of a confidence level Examples in text Example 9.3 Page 273 • Single sample estimate of mean Example 9.15 Page 299 • Single sample estimate of proportion EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 15

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