1. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 1 Estimating the Difference Between Two Means  Given two independent random samples, a point estimate the difference between μ 1 and μ 2 is given by the statistic We can build a confidence interval for μ 1 - μ 2 (given σ 1 2 and σ 2 2 known) as follows: Note: Use this calculation for large n as well (n>30).

2. 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

3. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 3 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 Z 0.02 = 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)

4. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 4 Differences Between Two Means: Variances Unknown  Case 1: σ 1 2 and σ 2 2 unknown but “equal”  Pages 287 and 288 Where,  Note v = n 1 + n 2 - 2

5. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 5 Differences Between Two Means: Variances Unknown (Page 290)  Case 2: σ 1 2 and σ 2 2 unknown and not equal Where,  WOW!

6. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 6 Estimating μ 1 – μ 2  Example ( σ 1 2 and σ 2 2 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: x 1 = 12 minutesx 2 = 10 minutes s 1 2 = 12s 2 2 = 8 n 1 = n 2 = 100  Construct a 95% C.I. for μ 1 – μ 2  Since n is large, we can estimate σ i 2 with s i 2

7. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 7 Solution  95% CI α/2 = ? Z.025, Z.975 = 1.96  (12-10) + 1.96*sqrt(12/100 + 8/100) = 2 + 0.8765  1.1235 < μ 1 – μ 2 < 2.8765  Interpretation: If CI contains 0, then μ 1 – μ 2 may be either positive or negative (can’t say that one population mean is larger than the other); however, since the CI for μ 1 – μ 2 is positive, we conclude μ 1 must be larger than μ 2.

8. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 8 μ 1 – μ 2 : σ i 2 Unknown  Example (σ 1 2 and σ 2 2 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 (  1 2 =  2 2 ). x 1 = 12 minutesx 2 = 10 minutes s 1 2 = 12s 2 2 = 8 n 1 = 18n 2 = 14 Construct a 95% C.I. for μ 1 – μ 2

9. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 9 Solution Governing Equations: Calculations: t 0.025,30 = 2.042 s p 2 = ((17*12)+(13*8))/30 = 10.267 s p = 3.204 2 + 2.042*3.204*sqrt(1/18 + 1/14) = 2 + 2.3314 -0.3314 < μ 1 – μ 2 < 4.3314 Interpretation: Since this CI contains 0, we can’t conclude μ 1 is significantly different from μ 2.

10. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 10 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.

11. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 11 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: μ d = μ 1 – μ 2 Based on the data in Table 9.1 D bar = -0.87 S d = 2.9773 n=20 We determine that a (1-0.05 )100% CI for μ d is: -2.2634 < μ d < 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.

12. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 12 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

13. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 13 C.I. for Proportions  An approximate (1-α)100% confidence interval for p is:  Large-sample C.I. for p 1 – p 2 is:  Interpretation: If the CI contains 0 …

14. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 14 Interpretation of the Confidence Interval Significance 1.If the C.I. for p 1 – p 2 = (-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)?

15. EGR 252 Ch. 9 Lecture2 9th ed. MDH 2015 Slide 15 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