1. EGR252 F11 Ch 10 9th edition rev2 Slide 1 Statistical Hypothesis Testing Review  A statistical hypothesis is an assertion concerning one or more populations.  In statistics, a hypothesis test is conducted on a set of two mutually exclusive statements: H 0 : null hypothesis H 1 : alternate hypothesis  Example H 0 : μ = 17 H 1 : μ ≠ 17  We sometimes refer to the null hypothesis as the “equals” hypothesis.

2. EGR252 F11 Ch 10 9th edition rev2 Slide 2 Potential errors in decision-making  α  Probability of committing a Type I error  Probability of rejecting the null hypothesis given that the null hypothesis is true  P (reject H 0 | H 0 is true)  β  Probability of committing a Type II error  Power of the test = 1 - β (probability of rejecting the null hypothesis given that the alternate is true.)  Power = P (reject H 0 | H 1 is true)

3. EGR252 F11 Ch 10 9th edition rev2 Slide 3 Hypothesis Testing – Approach 1  Approach 1 - Fixed probability of Type 1 error. 1.State the null and alternative hypotheses. 2.Choose a fixed significance level α. 3.Specify the appropriate test statistic and establish the critical region based on α. Draw a graphic representation. 4.Calculate the value of the test statistic based on the sample data. 5.Make a decision to reject H 0 or fail to reject H 0, based on the location of the test statistic. 6.Make an engineering or scientific conclusion.

4. EGR252 F11 Ch 10 9th edition rev2 Slide 4 Hypothesis Testing – Approach 2  Approach 2 - Significance testing based on the calculated P-value 1.State the null and alternative hypotheses. 2.Choose an appropriate test statistic. 3.Calculate value of test statistic and determine P-value. Draw a graphic representation. 4.Make a decision to reject H 0 or fail to reject H 0, based on the P-value. 5.Make an engineering or scientific conclusion. P-value 01.00 0.250.50 0.75 p = 0.05 ↓ P-value

5. EGR252 F11 Ch 10 9th edition rev2 Slide 5 Example: Single Sample Test of the Mean P-value Approach A sample of 20 cars driven under varying highway conditions achieved fuel efficiencies as follows: Sample mean x = 34.271 mpg Sample std dev s = 2.915 mpg Test the hypothesis that the population mean equals 35.0 mpg vs. μ < 35. Step 1: State the hypotheses. H 0 : μ = 35 H 1 : μ < 35 Step 2: Determine the appropriate test statistic. σ unknown, n = 20 Therefore, use t distribution

6. EGR252 F11 Ch 10 9th edition rev2 Slide 6 Single Sample Example (cont.) Approach 2: = -1.11842 Find probability from chart or use Excel’s tdist function. P(x ≤ -1.118) = TDIST (1.118, 19, 1) = 0.139665 p = 0.14 0______________1 Decision: Fail to reject null hypothesis Conclusion: The mean is not significantly less than 35 mpg.

7. EGR252 F11 Ch 10 9th edition rev2 Slide 7 Example (concl.) Approach 1: Predetermined significance level (alpha) Step 1: Use same hypotheses. Step 2: Let’s set alpha at 0.05. Step 3: Determine the critical value of t that separates the “reject H 0 region” from the “do not reject H 0 region”. t , n-1 = t 0.05,19 = 1.729 Since H 1 specifies “< ” we declare t crit = -1.729 Step 4: Using the equation, we calculate t calc = -1.11842 Step 5: Decision Fail to reject H 0 Step 6: Conclusion: The mean is not significantly less than 35 mpg.

8. EGR252 F11 Ch 10 9th edition rev2 Slide 8 Your turn … same data, different hypotheses A sample of 20 cars driven under varying highway conditions achieved fuel efficiencies as follows: Sample mean x = 34.271 mpg Sample std dev s = 2.915 mpg Test the hypothesis that the population mean equals 35.0 mpg vs. μ ≠ 35 at an α level of 0.05. Be sure to draw the picture. Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 (Conclusion will be different.)

9. EGR252 F11 Ch 10 9th edition rev2 Slide 9 Two-Sample Hypothesis Testing A professor has designed an experiment to test the effect of reading the textbook before attempting to complete a homework assignment. Four students who read the textbook before attempting the homework recorded the following times (in hours) to complete the assignment: 3.1, 2.8, 0.5, 1.9 hours Five students who did not read the textbook before attempting the homework recorded the following times to complete the assignment: 0.9, 1.4, 2.1, 5.3, 4.6 hours

10. EGR252 F11 Ch 10 9th edition rev2 Slide 10 Two-Sample Hypothesis Testing  Define the difference in the two means as: μ 1 - μ 2 = d 0 where d 0 is the actual value of the hypothesized difference  What are the Hypotheses? H 0 : _______________ H 1 : _______________ or H 1 : _______________ or H 1 : _______________

11. EGR252 F11 Ch 10 9th edition rev2 Slide 11 Our Example Using Excel Reading:n 1 = 4mean x 1 = 2.075s 1 2 = 1.363 No reading:n 2 = 5mean x 2 = 2.860s 2 2 = 3.883 If we have reason to believe the population variances are “equal”, we can conduct a t- test assuming equal variances in Minitab or Excel. t-Test: Two-Sample Assuming Equal Variances ReadDoNotRead Mean2.0752.860 Variance1.36253.883 Observations45 Pooled Variance2.8027857 Hypothesized Mean Difference0 df7 t Stat-0.698986 P(T<=t) one-tail0.2535567 t Critical one-tail1.8945775 P(T<=t) two-tail0.5071134 t Critical two-tail2.3646226

12. EGR252 F11 Ch 10 9th edition rev2 Slide 12 Your turn …  Lower-tail test (μ 1 - μ 2 < 0)  “Fixed α” approach (“Approach 1”) at α = 0.05 level.  “p-value” approach (“Approach 2”)  Upper-tail test (μ 2 – μ 1 > 0)  “Fixed α” approach at α = 0.05 level.  “p-value” approach  Two-tailed test (μ 1 - μ 2 ≠ 0)  “Fixed α” approach at α = 0.05 level.  “p-value” approach Recall 

13. EGR252 F11 Ch 10 9th edition rev2 Slide 13 Our Example – Hand Calculation Reading: n 1 = 4 mean x 1 = 2.075s 1 2 = 1.363 No reading: n 2 = 5 mean x 2 = 2.860s 2 2 = 3.883 To conduct the test by hand, we must calculate s p 2. = 2.803 s = 1.674 and = ????

14. EGR252 F11 Ch 10 9th edition rev2 Slide 14 Lower-tail test (μ 1 - μ 2 < 0) Why?  Draw the picture:  Approach 1: df = 7, t 0.5,7 = 1.895  t crit = -1.895  Calculation:  t calc = ((2.075-2.860)-0)/(1.674*sqrt(1/4 + 1/5)) = -0.70  Graphic:  Decision:  Conclusion:

15. EGR252 F11 Ch 10 9th edition rev2 Slide 15 Upper-tail test (μ 2 – μ 1 > 0) Conclusions  The data do not support the hypothesis that the mean time to complete homework is less for students who read the textbook. or  There is no statistically significant difference in the time required to complete the homework for the people who read the text ahead of time vs those who did not. or  The data do not support the hypothesis that the mean completion time is less for readers than for non-readers.

16. EGR252 F11 Ch 10 9th edition rev2 Slide 16 Our Example Using Excel Reading:n 1 = 4mean x 1 = 2.075s 1 2 = 1.363 No reading:n 2 = 5mean x 2 = 2.860s 2 2 = 3.883 What if we do not have reason to believe the population variances are “equal”? We can conduct a t- test assuming unequal variances in Minitab or Excel. t-Test: Two-Sample Assuming Equal Variances ReadDoNotRead Mean2.0752.860 Variance1.36253.883 Observations45 Pooled Variance2.8027857 Hypothesized Mean Difference0 df7 t Stat-0.698986 P(T<=t) one-tail0.2535567 t Critical one-tail1.8945775 P(T<=t) two-tail0.5071134 t Critical two-tail2.3646226 t-Test: Two-Sample Assuming Unequal Variances ReadDoNotRead Mean2.0752.86 Variance1.36253.883 Observations45 Hypothesized Mean Difference0 df7 t Stat-0.7426759 P(T<=t) one-tail0.2409258 t Critical one-tail1.8945775 P(T<=t) two-tail0.4818516 t Critical two-tail2.3646226

17. EGR252 F11 Ch 10 9th edition rev2 Slide 17 Another Example: Low Carb Meals Suppose we want to test the difference in carbohydrate content between two “low-carb” meals. Random samples of the two meals are tested in the lab and the carbohydrate content per serving (in grams) is recorded, with the following results: n 1 = 15x 1 = 27.2s 1 2 = 11 n 2 = 10x 2 = 23.9s 2 2 = 23 t calc = ______________________ ν = ______________ (using equation in table 10.3)

18. EGR252 F11 Ch 10 9th edition rev2 Slide 18 Example (cont.)  What are our options for hypotheses? H 0 : μ 1 - μ 2 = 0 orH 0 : μ 1 - μ 2 = 0 H 1 : μ 1 - μ 2 > 0 H 1 : μ 1 - μ 2 ≠ 0  At an α level of 0.05,  One-tailed test, t 0.05, 15 = 1.753  Two-tailed test, t 0.025, 15 = 2.131  How are our conclusions affected?  Our data don’t support a conclusion that the carb content of the two meals are different at an alpha level of.05 (What is H 1 ?)  Our data do support a conclusion that meal 1 has more carbs than meal 2 at an alpha level of.05 (What is H 1 ?)

19. EGR252 F11 Ch 10 9th edition rev2 Slide 19 Special Case: Paired Sample T-Test Which designs are paired-sample? A.CarRadialBelted 1 ** **Radial, Belted tires 2 ** ** placed on each car. 3 ** ** 4 ** ** B.Person Pre Post 1 ** **Pre- and post-test 2 ** **administered to each 3 ** **person. 4 ** ** C.Student Test1 Test2 1 ** **4 scores from test 1, 2 ** **4 scores from test 2. 3 ** ** 4 ** **

20. EGR252 F11 Ch 10 9th edition rev2 Slide 20 Sheer Strength Example* An article in the Journal of Strain Analysis compares several methods for predicting the shear strength of steel plate girders. Data for two of these methods, when applied to nine specific girders, are shown in the table on the next slide. We would like to determine if there is any difference, on average, between the two methods. Procedure: We will conduct a paired-sample t-test at the 0.05 significance level to determine if there is a difference between the two methods. * adapted from Montgomery & Runger, Applied Statistics and Probability for Engineers.

21. EGR252 F11 Ch 10 9th edition rev2 Slide 21 Sheer Strength Example Data Girder Karlsruhe Method Lehigh Method Difference (d) 11.1861.0610.125 21.1510.9920.159 31.3221.0630.259 41.3391.0620.277 51.2001.0650.135 61.4021.1780.224 71.3651.0370.328 81.5371.0860.451 91.5591.0520.507

22. EGR252 F11 Ch 10 9th edition rev2 Slide 22 Sheer Strength Example Calculations Hypotheses: H 0 : μ D = 0 H 1 : μ D ≠ 0 t 0.025,8 = 2.306 Why 8? Calculation of difference scores (d), mean and standard deviation, and t calc … d = 0.2739 s d = 0.1351 t calc = ( d – d 0 ) = (0.2739 - 0) = 6.082 s d / sqrt(n) (1.1351 / 3)

23. EGR252 F11 Ch 10 9th edition rev2 Slide 23 What does this mean?  Draw the picture:  Decision:  Conclusion:

24. EGR252 F11 Ch 10 9th edition rev2 Slide 24 Goodness-of-Fit Tests  Procedures for confirming or refuting hypotheses about the distributions of random variables.  Hypotheses: H 0 : The population follows a particular distribution. H 1 : The population does not follow the distribution. Examples: H 0 : The data come from a normal distribution. H 1 : The data do not come from a normal distribution.

25. EGR252 F11 Ch 10 9th edition rev2 Slide 25 Goodness of Fit Tests: Basic Method  Test statistic is χ 2  Draw the picture  Determine the critical value χ 2 with parameters α, ν = k – 1  Calculate χ 2 from the sample  Compare χ 2 calc to χ 2 crit  Make a decision about H 0  State your conclusion

26. EGR252 F11 Ch 10 9th edition rev2 Slide 26 Tests of Independence  Example:500 employees were surveyed with respect to pension plan preferences.  Hypotheses H 0 : Worker Type and Pension Plan are independent. H 1 : Worker Type and Pension Plan are not independent.  Develop a Contingency Table showing the observed values for the 500 people surveyed. Worker Type Pension Plan Total #1#2#3 Salaried16014040340 Hourly4060 160 Total200 100500

27. EGR252 F11 Ch 10 9th edition rev2 Slide 27 Calculation of Expected Values 2. Calculate expected probabilities P(#1 ∩ S) = P(#1)*P(S) = (200/500)*(340/500)=0.272 E(#1 ∩ S) = 0.272 * 500 = 136 Worker Type Pension Plan Total #1#2#3 Salaried16014040340 Hourly4060 160 Total200 100500 #1#2#3 S (exp.)136?68 H (exp.) 64?32

28. EGR252 F11 Ch 10 9th edition rev2 Slide 28 Calculate the Sample-based Statistic Calculation of the sample-based statistic = (160-136)^2/(136) + (140-136)^2/(136) + … (60-32)^2/(32) = 49.63

29. EGR252 F11 Ch 10 9th edition rev2 Slide 29 The Chi-Squared Test of Independence 5. Compare to the critical statistic, χ 2 α, r where r = (a – 1)(b – 1) For our example, suppose α = 0.01 χ 2 0.01,2 = ___________ χ 2 calc = ___________ Decision: Conclusion:

30. EGR252 F11 Ch 10 9th edition rev2 Slide 30 The Chi-Squared Test in Minitab 15 Chi-Square Test: pp1, pp2, pp3 Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts pp1 pp2 pp3 Total 1 160 140 40 340 136.00 136.00 68.00 4.235 0.118 11.529 2 40 60 60 160 64.00 64.00 32.00 9.000 0.250 24.500 Total 200 200 100 500  Chi-Sq = 49.632, DF = 2, P-Value = 0.000