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: H0 : null hypothesis H1 : alternate hypothesis Example H0 : μ = 17 H1 : μ ≠ 17 We sometimes refer to the null hypothesis as the “equals” hypothesis. Draw critical region, critical value. JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition

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 > _____ Draw critical region, critical value. JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition

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 = 8 oz. H1 : μ < 8 oz. or μ ≠ 8 oz Draw pictures ? H0 : μ = _____ H1 : μ < _____ JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition

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 H0 | H0 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 H0 | H1 is true) Power of the test = 1- β JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition

Hypothesis Testing – Approach 1 Approach 1 - Fixed probability of Type 1 error. State the null and alternative hypotheses. Choose a fixed significance level α. Specify the appropriate test statistic and establish the critical region based on α. Draw a graphic representation. Calculate the value of the test statistic based on the sample data. Make a decision to reject or fail to reject H0, based on the location of the test statistic. Make an engineering or scientific conclusion. recall our question about the amount of coffee in the cup … 1. H0 : μ = 8 oz. H1 : μ < 8 oz. 2. α = 0.05 3. zα = -1.645 5. if zcalc < -1.645, reject H0 if zcalc > -1.645, fail to reject H0 6. e.g., coffee in the cup is significantly less than 8 oz. or coffee in the cup is not significantly less than 8 oz. JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition

Hypothesis Testing – Approach 2 Approach 2 - Significance testing based on the calculated P-value State the null and alternative hypotheses. Choose an appropriate test statistic. Calculate value of test statistic and determine P-value. Draw a graphic representation. Make a decision to reject or fail to reject H0, based on the P-value. Make an engineering or scientific conclusion. recall our question about the amount of coffee in the cup … 1. H0 : μ = 8 oz. H1 : μ < 8 oz. 2. if variance known or n large, z-test (assume z in this case) 3. from zcalc, determine p-value from table A.3 or as given by statistical software packages. 4. P = 0, H0 rejected / not plausible (e.g., coffee in the cup is significantly less than 8 oz.) P = 1, H0 is not rejected (coffee in the cup is not significantly less than 8 oz.) Note: Approach 1 is the classical method. Approach 2 is gaining acceptance, partly because of the increasing availability of statistical software packages. The conclusion based on a P-value requires judgment. The smaller the P-value, the less plausible is the null hypothesis. p = 0.05 ↓ P-value 0 0.25 0.50 0.75 1.00 P-value JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition

Hypothesis Testing Tells Us … Strong conclusion: If our calculated t-value is “outside” tα,ν (approach 1) or we have a small p-value (approach 2), then we reject H0: μ = μ0 in favor of the alternate hypothesis. Weak conclusion: If our calculated t-value is “inside” tα,ν (approach 1) or we have a “large” p-value (approach 2), then we cannot reject H0: μ = μ0. Failure to reject H0 does not imply that μ is equal to the stated value (μ0), only that we do not have sufficient evidence to support H1. END of 10.1 – 10.4 JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition

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. H0: μ = 35 H1: μ < 35 Step 2: Determine the appropriate test statistic. σ unknown, n = 20 Therefore, use t distribution H0 : μ = 35 H1 : μ < 35 n = 20 use t-distribution JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition

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 does not differ significantly from 35 mpg. t = -1.11842 =(34.271-35)/(2.915/(SQRT(20))) P = 0.138665 =TDIST(1.118,19,1) draw the graphs. NOTE: if we look at table A.4, pg. 672, the α value associated to 1.118 (by symmetry) falls between 0.15 and 0.10. 13.86% of the area under the curve lies to the left of t = -1.118. Judge that H0 is plausible (fail to reject) and conclude that μ does not differ significantly from 35. JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition

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 H0 region from the do not reject H0 region. t, n-1 = t0.05,19 = 1.729 Since H1 specifies “μ< μ0,” tcrit = -1.729 Step 4: tcalc = -1.11842 Step 5: Decision Fail to reject H0 Step 6: Conclusion: Conclusion: The mean does not differ significantly from 35 mpg. (one-sided or one-tailed test) t0.05,19 = 1.729, tcrit = -1.729 t = -1.11842 draw the picture … JMB Ch 10 Lecture 1 8th edition EGR 252 Spring 2008 EGR 252 F06 Ch. 10 8th edition