1. IEEM 3201 Two-mean Hypothesis Testing: Two means, Two proportions

2. IEEM 320IEEM151 Notes 23, Page 2 Outline  test concerning two means (variances known)  test concerning two means (variances unknown but equal)  test concerning two means (variances unknown and unequal)  Paired observations

3. IEEM 320IEEM151 Notes 23, Page 3 Test Concerning Two Means (Variance Known)  use the statistic for testing. ? Two-tailed test Hypothesis: H 0 : μ 1 –μ 2 = d 0 H 1 : μ 1 –μ 2 ≠ d 0 Decision: If z z  /2, reject H 0 ; if – z  /2 < z < z  /2, accept H 0. ? For one-tailed tests, we can modify the above accordingly.

4. IEEM 320IEEM151 Notes 23, Page 4 Test Concerning Two Means (Variance Unknown But Equal)  use the statistic for testing, where and v = n 1 +n 2 -2. ? Two-tailed test Hypothesis: H 0 : μ 1 –μ 2 = d 0 H 1 : μ 1 –μ 2 ≠ d 0 Decision: If t t  /2,v, reject H 0 ; if – t  /2,v < t < t  /2,v, accept H 0. ? For one-tailed tests, we can modify the above accordingly.

5. IEEM 320IEEM151 Notes 23, Page 5 Solution: Hypothesis: H 0 : μ 1 –μ 2 = 0 H 1 : μ 1 –μ 2 ≠ 0 Computation: Test Concerning Two Means (Variance Unknown But Equal) Example: Are the two population means the same at a 95% confidence level?

6. IEEM 320IEEM151 Notes 23, Page 6 Decision: v = n 1 +n 2 -2 = 20 t  /2,v = t 0.025,20 = 2.086 t > t  /2,v. So we reject H 0. We have evidence to believe that the two means are different. Test Concerning Two Means (Variance Unknown But Equal)

7. IEEM 320IEEM151 Notes 23, Page 7 Test Concerning Two Means (Variance Unknown And Unequal)  use the statistic for testing, the degrees of freedom ? Two-tailed test Hypothesis: H 0 : μ 1 –μ 2 = d 0 H 1 : μ 1 –μ 2 ≠ d 0 Decision: If t’ t  /2,v, reject H 0 ; if – t  /2,v < t’ < t  /2,v, accept H 0. ? For one-tailed tests, we can modify the above accordingly.

8. IEEM 320IEEM151 Notes 23, Page 8 Paired Observations  D: the difference of a random pair of observations from the two populations.  use the statistic with v = n-1 for testing, ? Two-tailed test Hypothesis: H 0 : μ D = d 0 H 1 : μ D ≠ d 0 Decision: If t t  /2,v, reject H 0 ; if –t  /2,v < t < t  /2,v, accept H 0. ? For one-tailed tests, we can modify the above accordingly.

9. IEEM 320IEEM151 Notes 23, Page 9 Solution: Hypothesis: H 0 : μ 1 –μ 2 = 0 H 1 : μ 1 –μ 2 ≠ 0 Computation: Decision: v = n-1 = 14, t  /2,n-1 = t 0.025,14 = 2.145 - t  /2,n-1 < t < t  /2,n-1. So cannot reject H 0. we do not have significant evidence that the difference is not zero. Example: Is the difference zero at a 95% confidence level? Paired Observations