1. TESTS OF HYPOTHESES OF POPULATIONS WITH KNOWN VARIANCE  TWO SAMPLE  Tests on The Difference of Means  Tests on The Ratio of Variances

2. Tests on The Difference of MEANS

3. 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. The standard deviation Of the drying time is 8 minutes. Ten specimens are taken from each formulation. The Two Sample Average drying time are 121 and 112 minutes respectively. What conclusions can be drawn about The effectiveness of the new ingredient. Evaluate the P-Value of the test ____________________________________________________________________________________ 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 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

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

5. 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 Z o ½ P-Value Φ(Z o) Z o P-Value Z 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

6. Step 5 : Perform the hypothesis test by evaluating of the P-Value of the test 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 P-Value Zo = 2.516

7. Example 2 If in Example 1, the measure of Effectiveness of the new ingredient is preset to be that the drying time of new formulation2 is 97% of that of standard formulation1. Do the experimental result support the claim that the new ingredient is effective in reducing the drying time. If by some means the true difference between the two drying times is 1.5 minutes, Evaluate the type II error. What sample size that render the power of test in detecting the true difference More than 90%. ___________________________________________________________________________________ Step 1: Statement of Test Ho : μ1 = (1/0.97) μ2 Ho: μ1 = 1.031 μ2 H1 : μ1 > 1.031 μ2 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 UPPER-SIDED, then the Rejection Zone is defined As shown in the figure to the right H1 :μ1 > 1.031μ2 P-Value 0 REJECT Z o

8. We may Reject Ho with a significance error of 0.061 Step 5 : Perform the hypothesis test We REJECT Ho and then ACCEPT H1 Step 6 : CONCLUSION Since we Reject Ho, H1 (: μ1 > 1.031 μ2 )) is ACCEPTED, then THERE IS ENOUGH EVIDENCE TO STATE THAT THE NEW INGERIEDNT IS EFFECTIVE IN REDUCING DRYING TIME OF THE PAINT

9. δ > 0 δ < 0

10. Type II Error β In order to raise the power of test to be 90%, the Sample size should be: δ < 0

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 From Tables

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 24.524.841.9 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

19. Tests on The RATIOS of VARIANCES

20. Consider the following RANDOM VARIABLE: The numerator and the denominator are CHI-Square Variables with Degrees of Freedom m and n respectively. α α α =0.05 n m251020 2195.7861354.1028213.492828 519.296415.0503293.3258352.71089 1019.39594.7350632.9782372.347878 2019.445774.5581312.7740162.124155

21. 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 Is there any Evidence to support the claim that the variability in Mileage performance is greater for a Volkswagen than for a Mercedes. 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 This is an F- Variable with N1 – 1 and N2 - 1 Degrees of Freedom Therefore, is a Chi-square variable with (n-1) degrees of freedom

22. Step 4: Define The Zone of ACCEPTANCE α =0.05 Step 5 : Perform the hypothesis test by Comparing f o with If f o >, then we are in the REJECT zone, then we REJECT Ho Otherwise, we are not in a position to Reject Ho = 14.035 >> = 3.388, Therefore we REJECT Ho Step 6 : CONCLUSION Since we Reject Ho, is ACCEPTED, then THERE IS STRONG EVIDENCE TO STATE THAT The Variation in Fuel Mileage of Volkswagen is GREATER than that of MERCEDES