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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. JMB Ch10 Lecture 1 9th ed. EGR 252 2016 Slide 1 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 < _____ One-sided JMB Ch10 Lecture 1 9th ed. H0 : μ = _____ H1 : μ ≠ _____ H0 : p = _____ H1 : p > _____ One-sided Two-sided EGR 252 2016 Slide 2 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 < _____ JMB Ch10 Lecture 1 9th ed. EGR 252 2016 Slide 3 Potential errors in decision-making H0 True H0 False Do not reject H0 Correct Decision Type II error Reject H0 Type I error Correct Decision 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) JMB Ch10 Lecture 1 9th ed. 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) EGR 252 2016 Slide 4 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 H0, based on the location of the test statistic. 6. Make an engineering or scientific conclusion. JMB Ch10 Lecture 1 9th ed. EGR 252 2016 Slide 5 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 Pvalue. Draw a graphic representation. 4. Make a decision to reject or fail to reject H0, based on the P-value. 5. Make an engineering or scientific conclusion. Three potential test results (actual test results should include only one arrow per plot): ↓ p = 0.02 P-value 0 JMB Ch10 Lecture 1 9th ed. p = 0.45 0.25 ↓ 0.50 p = 0.85 ↓ 0.75 EGR 252 2016 1.00 P-value Slide 6 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. JMB Ch10 Lecture 1 9th ed. EGR 252 2016 Slide 7 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 JMB Ch10 Lecture 1 9th ed. EGR 252 2016 Slide 8 Single Sample Example (Approach 2) Approach 2 (p-value approach): X T S/ n = -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. Note: the conclusion “matches” the alternate hypothesis - H1: μ < 35 JMB Ch10 Lecture 1 9th ed. EGR 252 2016 Slide 9 Single Sample Example (Approach 1) 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 format is “μ< μ0,” tcrit = -1.729 Step 4: tcalc = -1.11842 Step 5: Decision Fail to reject H0 Step 6: Conclusion: The population mean is not significantly less than 35 mpg. ***Do not conclude that the population mean equals 35 mpg.*** JMB Ch10 Lecture 1 9th ed. EGR 252 2016 Slide 10

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