Chapter 9: Part 2 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 Suppose we would like to compare the average production for the night shift versus the day shift at a facility that manufactures ceiling tiles. EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

Equation for 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 We can build a confidence interval for μ1 - μ2 (given σ12 and σ22 known) as follows: EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

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) x-barA = 36 mpg x-barB = 42 mpg σA = 6 σB= 8 α = 0.02 Z0.02 = 2.055 6 – 2.055 sqrt(64/75 + 36/50) < μB - μA < 6 + 2.055 sqrt(64/75 + 36/50) 3.4224 < μB - μA < 8.5776 EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

Differences Between Two Means: Variances Unknown Case 1: σ12 and σ22 unknown but “equal” Pages 287 and 288 Where, Note v = n1 + n2 -2 EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

Differences Between Two Means: Variances Unknown (Page 290) Case 2: σ12 and σ22 unknown and not assumed to be equal Where, WOW! EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

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 Note: since n is large, we can estimate σi2 with si2 EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

Solution 95% CI Z.025 = 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 is larger than the other); however, since the CI for μ1 – μ2 is positive, we conclude μ1 is significantly larger than μ2 . X1 – X2 = 12 – 10 = 2 Z.025 = 1.96 2 + 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 is larger than the other); however, since this CI doesn’t contain 0, we conclude μ1 must be larger than μ2 . EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

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 as follows: x1 = 12 minutes x2 = 10 minutes s12 = 12 s22 = 8 n1 = 18 n2 = 14 Construct a 95% C.I. for μ1 – μ2 EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

Solution Governing Equations: Calculations: t0.025,30 = 2.042 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 for the difference between the means contains 0, we conclude the mean downtimes for the two machines are not significantly different. X1 – X2 = 12 – 10 = 2 t.025,30 = 2.042 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 larger than μ2 . EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

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. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

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 Based on the data in Table 9.1 Dbar = -0.87 Sd = 2.9773 n=20 We determine that a (1-0.05)100% CI for μd is: -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. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

C.I. for Proportions The proportion, P, in a binomial experiment may be estimated by where X is the number of successes in n trials. For a sample, the point estimate of the parameter is The mean for the sample proportion is and the sample variance note: variance = npq, but sample variance = npq/n2 = pq/n EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

C.I. for Proportions An approximate (1-α)100% confidence interval for p is: Large-sample C.I. for p1 – p2 is: Interpretation: If the CI contains 0 … interpretation of large-sample CI: if the CI contains 0, there is no reason to believe there is a significant difference between the two population proportions. EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

Interpretation of the Confidence Interval Significance 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? What if the interval were (+0.002, +0.022)? What if the interval were (-0.900, -0.700)? No The interval contains zero, so there is no significant difference in the two proportions (at the alpha level of significance) For p1-p2 CI , both values are positive, therefore conclude p1 > p2 3. For p1-p2 CI, both values are negative, therefore conclude p1 < p2 EGR 252 Ch. 9 Lecture2 9th ed. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition

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. JMB Fall 2018 2019 252Chapter9 JMB F06 8th edition