1. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 1 Statistical Hypothesis Testing  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. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 2 Tests of Hypotheses - Graphics I  We can make a decision about our hypotheses based on our understanding of probability.  We can visualize this probability by defining a rejection region on the probability curve.  The general location of the rejection region is determined by the alternate hypothesis. H0 : μ = _____ H1 : μ < _____ H0 : μ = _____ H1 : μ ≠ _____ H0 : p = _____ H1 : p > _____ One-sided Two-sided One-sided

3. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 3 Choosing the Hypotheses  Your turn … Suppose a coffee vending machine claims it dispenses an 8-oz cup of coffee. You have been using the machine for 6 months, but recently it seems the cup isn’t as full as it used to be. You plan to conduct a statistical hypothesis test. What are your hypotheses? H0 : μ = _____ H1 : μ ≠ _____ H0 : μ = _____ H1 : μ < _____

4. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 4 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)

5. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 5 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 or fail to reject H 0, based on the location of the test statistic. 6.Make an engineering or scientific conclusion.

6. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 6 Hypothesis Testing – 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 or fail to reject H 0, based on the P-value. 5.Make an engineering or scientific conclusion. Three potential test results: P-value 01.00 0.250.50 0.75 p = 0.02 ↓ P-value p = 0.45 ↓ p = 0.85 ↓

7. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 7 Hypothesis Testing Tells Us …  Strong conclusion:  If our calculated t-value is “outside” t α,ν (approach 1) or we have a small p-value (approach 2), then we reject H 0 : μ = μ 0 in favor of the alternate hypothesis.  Weak conclusion:  If our calculated t-value is “inside” t α,ν (approach 1) or we have a “large” p-value (approach 2), then we cannot reject H 0 : μ = μ 0.  Failure to reject H 0 does not imply that μ is equal to the stated value (μ 0 ), only that we do not have sufficient evidence to support H 1.

8. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 8 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

9. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 9 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.

10. JMB Ch10 Lecture 1 9th ed. v Mar 2012 EGR 252 Spring 2012 Slide 10 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 format is “μ< μ 0,” t crit = -1.729 Step 4: t calc = -1.11842 Step 5: Decision Fail to reject H 0 Step 6: Conclusion: The population mean is not significantly less than 35 mpg. ******Do not conclude that the population mean is 35 mpg.******