ENGR 201: Statistics for Engineers

1. ENGR 201: Statistics for Engineers Chapter 10: Statistical Inference for Two Samples • Inference on difference in means of two normal distribution • A non-parametric test on the differences in two means • Paired t-test • Inference on variances, population proportions

2. Large Sample Test for Difference Between Two Population Means

3. Inference on the Difference in Means of Two Normal Distributions, Variances Known Assumptions: • Let be a random sample from population 1. • Let be a random sample from population 2. • The two populations X1and X2are independent. 4. Both X1and X2 are normal.

4. Inference on the Difference in Means of Two Normal Distributions, Variances Known The quantity is normally distributed with N(0, 1).

5. Inference on the Difference in Means of Two Normal Distributions, Variances Known Null hypothesis: H0: 12 = 0 Test statistic:

6. Large Sample Test for Difference Between Two Population Means

7. Large Sample Test for Difference Between Two Population Means Step 2: Test statistic for large sample sizes when n1≥30 and n2≥30 Variance of populations are known Variance of populations are unknown but estimated Step 3: Under H0, the sampling distribution of z is approximately standard normal

8. Large Sample Test for Difference Between Two Population Means Step 3: Find p-value. Compute

9. Example • Is there a difference in the average daily intakes of dairy products for men versus women? Use a = 0.05.

10. Example (Cont.) Decision: since the p-value is greater than a = .05, H0 is not to be rejected. Conclusion: there is insufficient evidence to indicate that men and women have different average daily intakes.

11. Type II Error and Choice of Sample Size Two-sided alternative: • For the two-sided alternative hypothesis with significance level , the sample size n1n2n required to detect a true difference in means of  with power at least 1  is One-sided alternative: • For a one-sided alternative hypothesis with significance level , the sample size n1n2n required to detect a true difference in means of (0) with power at least 1  is

12. Confidence Interval on a Difference in Means, Variances Known If and are the means of independent random samples of sizes n1 and n2 from two independent normal populations with known variance and , respectively, a 100(1 ) confidence interval for 12 is where z2 is the upper 2 percentage point of the standard normal distribution

13. Example Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known. The data obtained are as follows: n1 = 10, , 1 1, n2 = 12, , and 2 1.5. If 1 and 2 denote the true mean tensile strengths for the two grades of spars, we may find a 90% confidence interval on the difference in mean strength 12 as follows:

14. Example Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known. The data obtained are as follows: n1 = 10, , 1 1, n2 = 12, , and 2 1.5. If 1 and 2 denote the true mean tensile strengths for the two grades of spars, we may find a 90% confidence interval on the difference in mean strength 12 as follows:

15. Small Sample Testing the Difference between Two Population Means Note that both population are normally distributed with the same variances

16. Small Sample Testing the Difference between Two Population Means

17. Small Sample Testing the Difference between Two Population Means Step 2: Test statistic for small sample sizes Pool estimator of variance Step 3: Under H0, the sampling distribution of t has a t- distribution with n1+n2-2 degrees of freedom

18. Small Sample Testing the Difference between Two Population Means Step 3: Find p-value. Compute

19. Small Sample Testing the Difference between Two Population Means

20. Example Two training procedures are compared by measuring the time that it takes trainees to assemble a device. A different group of trainees are taught using each method. Is there a difference in the two methods? Use a = 0.01.

21. Example (Cont.)

22. Example Decision: since the p-value is greater than a = .01, H0 is not rejected. Conclusion: there is insufficient evidence to indicate a difference in the population means. df = n1 + n2 – 2 = 10 + 12 – 2 = 20

23. Example Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Specifically, catalyst 1 is currently in use, but catalyst 2 is acceptable. Since catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield. A test is run in the pilot plant and results in the data shown in Table 10-1. Is there any difference between the mean yields? Use  0.05, and assume equal variances. s1 = 2.39 s2 = 2.98

24. Example (cont.) The seven-step hypothesis-testing procedure is as follows: • Parameter of interest: The parameters of interest are 1 and 2, the mean process yield using catalysts 1 and 2, respectively. 2. Null hypothesis: H0: 12 3. Alternative hypothesis: H1: 12 • Test statistic: The test statistic is 5. Reject H0 if: Reject H0 if the P-value is less than 0.05.

25. Example (cont.) • Computations: From Table 10-1 we have , s1 = 2.39, n1=8, , s2 = 2.98, and n2 = 8. Therefore 7. Conclusions: From Appendix Table V we can find t0.40,14 0.258 and t0.25,14 0.692. Since, 0.258  0.35  0.692, we conclude that lower and upper bounds on the P-value are 0.50  P 0.80. Therefore, since the P-value exceeds  0.05, the null hypothesis cannot be rejected. Interpretation:At 5% level of significance, we do not have strong evidence to conclude that catalyst 2 results in a mean yield that differs from the mean yield when catalyst 1 is used.

26. The Paired-Difference Test • We have assumed that samples from two populations are independent. Sometimes the assumption of independent samples is intentionally violated, resulting in a matched-pairs or paired-difference test. • By designing the experiment in this way, we can eliminate unwanted variability in the experiment • Denote data as

27. The Paired-Difference Test

28. The Paired-Difference Test Step 2: Test statistic for small sample sizes Step 3: Under H0, the sampling distribution of t has a Student’s t distribution with n-1 degrees of freedom

29. The Paired-Difference Test Step 3: Find p-value. Compute

30. Paired t Test Example A weight reduction center advertises that participants in its program lose an average of at least 5 pounds during the first week of the participation. Because of numerous complaints, the state’s consumer protection agency doubts this claim. To test the claim at the 0.05 level of significance, 12 participants were randomly selected. Their initial weights and their weights after 1 week in the program appear on the next slide. Set up and perform an appropriate hypothesis test.

31. Paired Sample Example (continued)

32. Paired Sample Example (continued) Each member serves as his/her own pair. weight changes=initial weight–weight after one week

33. Paired Sample Example (continued) Decision: since the p-value is smaller than a = .05, H0 is rejected. Conclusion: there is strong evidence that the mean weight loss is less than 5 pounds for those who took the program for one week.

34. Example 10-11: Shear Strength of Steel Girder An article in the Journal of Strain Analysis [1983, Vol. 18(2)] reports a comparison of several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied to nine specific girders, are shown in the table below. Determine whether there is any difference (on the average) for the two methods.

35. Example 10-11: (Cont.) 1. Parameter of interest: The parameter of interest is the difference in mean shear strength for the two methods. 2. Null hypothesis: H0: µD = 0 3. Alternative hypothesis: H1: µD 0 4. Test statistic: The test statistic is 5. Reject H0if: Reject H0 if the P-value is <0.05. 6. Computations: The sample average and standard deviation of the differences dj are and sd = 0.1350, and so the test statistic is 7. Conclusions: Because t0.0005.8 = 5.041 and the value of the test statistic t0 = 6.15 exceeds this value, the P-value is less than 2(0.0005) = 0.001. Therefore, we conclude that the strength prediction methods yield different results. Interpretation:The data indicate that the Karlsruhe method produces, on the average, higher strength predictions than does the Lehigh method.

36. Key Concepts I. A statistical test of hypothesis consist of five steps • Specify the null hypothesisH0and alternativehypothesisH1 in terms of population parameters • Identify and calculate test statistic • Identify distribution and find p-value • Compare p-value with the given significance level and decide if to reject the null hypothesis • State conclusion

37. Key Concepts II. Errors and Statistical Significance • Type I error: reject the null hypothesis when it is true • Type II error: fail to reject the null hypothesis when it is false • The significance level a = P(type I error) • b =P(type II error) • Thep-valueis the probability of observing a test statistic as extreme as or more than the one observed; also, the smallest value of a for which H0 can be rejected • When the p-valueis less than the significance level a, the null hypothesis is rejected

38. Sample size? H1: µ≠µ0 p-value = 2P(z>|z0|) H1: µ>µ0 p-value= P(z> z0) H1: µ<µ0 p-value= P(z< z0) H1: µ≠µ0 p-value= 2P(t>|t0|) H1: µ>µ0 p-value= 2P(t> t0) H1: µ≠µ0 p-value= P(t< t0) Key Concepts • Test for a population mean H0: µ=µ0 large small

39. Sample size? H1: µ1-µ2≠D0 p-value= 2P(z>|z0|) H1: µ1-µ2>D0 p-value= P(z> z0) H1: µ1-µ2<D0 p-value= P(z< z0) H1:µ1-µ2≠D0 p-value= 2P(t>|t0|) H1: µ1-µ2>D0 p-value= 2P(t> t0) H1: µ1-µ2<D0 p-value= P(t< t0) Key Concepts IV. Test for Difference Between Two Population MeanH0: µ1-µ2=𝚫0 large small

40. Key Concepts V. The Paired-Difference Test P-value: 2P(t>|t0|) • H1: m1-m2 𝚫0(two-sided test) • H1: m1-m2> 𝚫0 (one-sided test) • H1: m1-m2< 𝚫0 (one-sided test) P-value: P(t>t0) P-value: P(t<t0)