1. TESTS OF HYPOTHESES OF POPULATIONS WITH UNKNOWN VARIANCE T-TESTS  SINGLE SAMPLE Prof. Dr. Ahmed Farouk Abdul Moneim

2. A Car Tires Producer is claiming that the Mean Life time of a tire is more than 60 000 km A Sample of 16 tires are RANDOMLY selected and tested for life. The results of the experiments are given below. Example 1 60613 59836 59154 60252 59784 60221 60311 50040 60545 60257 60000 59997 69947 60135 60221 60523 Do these results support the claim of the producer or not ? Take α = 0.06 Step 1: Statement of Test Ho : μ = 60 000 km H1 : μ > 60 000 km Step 2: Experimental Data Processing Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho Since, is a Normal variable and population variance is NOT known the relevant Test Statistic will take the following form: Prof. Dr. Ahmed Farouk Abdul Moneim

3. Step 4: Define The Zone of ACCEPTANCE The Alternative Hypothesis H1 is UPPER-SIDED, then the Rejection Zone is defined As shown in the figure. H1 : μ > 60 000 km α T α,N-1 ACCEPT 0 REJECT α = 0.06, then from tables of T distribution T 0.06, 15 = 1.6 Step 5 : Perform the hypothesis test by Comparing To with T α, N-1 If To > T α, N-1, then we are in the REJECT zone, then we REJECT Ho Otherwise, we are not in a position to Reject Ho To = 0.0234 < < T 0.06, 15 = 2.034, Therefore we ACCEPT Ho Step 6 : CONCLUSION Since we accept Ho, H1 is STRONGLY RREJECTED, then THERE IS NO ENOUGH EVIDENCE TO STATE THAT THE TIRE LIFE IS LONGER THAN 60 000 km Prof. Dr. Ahmed Farouk Abdul Moneim

4. The yield of a chemical process is being studied.. The past FIVE days of plant operation have resulted in the following yields: 91.6%, 88.75%, 90.8%, 89.95%, 91.3%. (use α = 0.05) a) Is there evidence that the yield is not 90%? b) Is there evidence that the yield is less than 90%? c) Evaluate the P-Value of the test. d) What sample size would be required to detect a true mean yield of 85% with probability 0.95. e) What is the type II error probability, if the true mean yield is 92%? ___________________________________________________ Example 2 Step1: Statement of Test Ho : μ = 90 H1 : μ ≠ 90 Step 2: Experimental Data Processing Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho Step 4: Define The Zone of ACCEPTANCE The Alternative Hypothesis H1 is TWO-SIDED, then the Rejection Zones are Symmetrically placed as shown in the figure H1 : μ ≠ 90 ACCEPT Reject

5. α = 0.05, then (the area in the center part)= 0.95 And the Rejection areas on both sides, each = 0.025 Therefore, T α/2,N-1 = T 0.025, 4 = 3.495 (from tables or Excel) 0.025 T α/2,N-1 = 3.495 Step 5 : Perform the hypothesis test by Comparing To with T α, N-1 To = 0.934 < < T 0.025, 4 = 3.485, Therefore we ACCEPT Ho Step 6 : CONCLUSION Since we accept Ho, H1 is RREJECTED, then THERE IS NO ENOUGH EVIDENCE TO STATE THAT THE MEAN YIELD of the chemical process is NOT 90% Prof. Dr. Ahmed Farouk Abdul Moneim

6. The P-Value of a Test Prof. Dr. Ahmed Farouk Abdul Moneim

7. In order to decide whether ACCEPT or REJECT Ho WITHOUT HAVING α The P-Value should be calculated as follows depending on the Statement of the Test H1 : μ ≠ μo H1 : μ > μo H1 : μ < μo T o ½ P-Value Φ(Z o) T o P-Value T o P-Value RULE of DECISION by use of P-Value If P-Value ≤ 0.01 REJECT Ho absolutely If P-Value ≥ 0.1 ACCEPT Ho absolutely Otherwise it is up to concerned parties Prof. Dr. Ahmed Farouk Abdul Moneim

8. In the last exercise Example 2 H1 : μ ≠ 90To = 0.934 Therefore, we ACCEPT Ho NO Evidence that the Yield is NOT 90 FINDING The SAMPLE SIZE that enables Analysts To DETECT The TRUE MEAN different from Hypothetical MEAN with Given Power (1 - β) Prof. Dr. Ahmed Farouk Abdul Moneim

9. Example 2 continued μo = 90, μ True = 85, β = 1-0.95 = 0.05, δ = 85 – 90 = -5 Prof. Dr. Ahmed Farouk Abdul Moneim

11. TESTS OF HYPOTHESES OF POPULATIONS WITH UNKNOWN VARIANCE  TWO SAMPLE  Tests on The Difference of Means Prof. Dr. Ahmed Farouk Abdul Moneim

12. Tests on The Difference of MEANS The Two Populations are of NEARLY Equal Variances The Two Populations are of Non Equal Variances Prof. Dr. Ahmed Farouk Abdul Moneim

13. The Two Populations are of Nearly Equal Variances Prof. Dr. Ahmed Farouk Abdul Moneim

14. Example 1 Two formulations of paints are tested for Drying Time. Formulation1 is the standard chemistry, and formulation2 has a new drying ingredient that reduce the drying time.. Ten specimens are taken from each formulation. The Two Sample Average drying time are 121 and 112 minutes respectively. The Two Samples standard deviations are 8 and 6 minutes respectively. What conclusions can be drawn about The effectiveness of the new ingredient. If we assume that variances of populations are nearly equal (α=0.05) _ ___________________________________________________________________________________ Step 1: Statement of Test Ho : μ1 = μ2 H1 : μ1 > μ2 Step 2: Experimental Data Processing Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho Since the Two populations are of Nearly equal variances then, evaluating the Pooled Variance and Pooled degrees of Freedom of the two samples as follows: The Test Statistic in this case will take the form: Prof. Dr. Ahmed Farouk Abdul Moneim

15. Step 4: Define The Zone of ACCEPTANCE The Alternative Hypothesis H1 is UPPER-SIDED, then the Rejection Zone is defined As shown in the figure to the right H1 :μ1 > μ2 α ACCEPT 0 REJECT Tα, ν Step 5 : Perform the hypothesis test by Comparing To with Tα, ν To = 2.846 > T 0.05, 18 =1.734 Therefore the Test Result is SIGNIFICANT we REJECT Ho Step 6 : CONCLUSION Since we Reject Ho, H1 (: μ1 > μ2 )) is ACCEPTED, then THERE IS ENOUGH EVIDENCE TO STATE THAT THE NEW INGERIEDNT IS EFFECTIVE IN REDUCING DRYING TIME OF THE PAINT Prof. Dr. Ahmed Farouk Abdul Moneim

16. The Two Populations are of NONEQUAL Variances Prof. Dr. Ahmed Farouk Abdul Moneim

17. Example 1 A fuel economy study was conducted on two German automobiles, Mercedes and Volkswagen One vehicle of each brand was selected, and the mileage performance was observed for 10 tanks of fuel in each car. The data are as follows: MercedesVolkswagen 24.724.941.742.8 24.824.642.342.4 24.923.941.639.9 24.724.939.540.9 24.524.841.929.6 Do the data support the claim that the mean mileage for the Volkswagen is at least 15 Km pg higher than that for the Mercedes (α = 0.04) Step 1: Statement of Test Step 2: Experimental Data Processing Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho Since the Two populations are of NONEQUAL variances then the combined Degrees of freedom will be:

18. The Test Statistic in this case will take the form: Step 4: Define The Zone of ACCEPTANCE The Alternative Hypothesis H1 is UPPER-SIDED, then the Rejection Zone is defined As shown in the figure to the right α ACCEPT 0 REJECT Tα, ν Step 5 : Perform the hypothesis test by Comparing To with Tα, ν To = 4.56 > > T 0.025, 10 = 2.228 Therefore the Test Result is SIGNIFICANT we REJECT Ho Step 6 : CONCLUSION Since we Reject Ho, H1 (: μ1 > μ2 )) is ACCEPTED, then THERE IS ENOUGH EVIDENCE TO STATE THAT THE VOLKSWAGEN HAS GREATER MILEAGE THAN MERCEDES BY 15 MPG