lect15_16_17

# lect15_16_17 - Chapter 9 HYPOTHESIS TESTS Ref Devore 6e...

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Chapter 9 HYPOTHESIS TESTS Ref: Devore 6e, Chaps 8 and 9. 9.1 Sampling distributions and P-values Null and alternate hypothesis, one sample Z-test and T-test, P-value, significance level, critical value, statistical significance versus practical significance. 9.2 Relation between 2-sided hypothesis tests and confidence intervals. 9.3 Decision theory and error probabilities Type I and type II errors, α and β , operating characteristic (OC) curve, size ( α ) and power (1 - β ). Sample size determination. 9.4 Comparing two independent samples. Two sample T-test. Pooled and unequal variances versions, F-test for equality of variances. Two-sided P-values for asymmetric sampling distributions. 9.5 Paired samples. 9.6 Testing a hypothesis about a proportion 9.7 Testing the difference of two proportions 1

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9.1 SAMPLING DISTRIBUTIONS AND P-VALUES Ref: Devore 6e, Sections 8,1, 8.2, 8.4 MINITAB Handbook 5e Sections 8.1, 8.3. Motivating Example: Suppose we have been manufacturing an alloy with tensile strength of 2000 lbs. A new design is proposed which it is claimed leads to greater strength. We take a sample of n = 25 specimens from the new design and measure their tensile strengths: X 1 , X 2 , ..., X n . Model: X 1 , X 2 , ..., X n indep N ( μ, σ 2 ) We set up two hypotheses: Null hypothesis: H 0 : μ = μ 0 = 2000 Alternative hypothesis H 1 : μ > 2000 The null hypothesis can be thought of as the “skeptic’s” or “straw man” hypothesis. The alternative hypothesis can be thought of as a “research” hypothesis. Suppose the sample mean turned out to be X obs = 2060. Is the 60 lb excess simply due to chance or does it reflect a genuine design improvement? We answer this by supposing that H 0 were indeed true and using the sampling distribution of X to compute the probability of observing a sample mean as large as 2060 or more extreme (ie. greater). Case 1: Variance σ 2 known. Then, under H 0 , Z = X - μ 0 σ/ n is distributed as N (0 , 1) . Thus P 0 [ X > 2060] = P 0 [ Z > 60 n/σ ] = 1 - Φ(60 n/σ ) Since μ 0 = 2000. Here we use the subscript 0 (or H 0 ) on P to remind ourselves we are computing this probability assuming H 0 is true. Here n = 25 and suppose σ = 125. Then P 0 [ X > 2060] = P [ Z > 2 . 4] = 0 . 0082 . This is rather small, and we might conclude that such a large X is unlikely to have occurred by chance under H 0 , and thus conclude that H 1 is true. This quantity (here 0.0082) is called the P-value. (See Devore 6e Sec 8.4.) Cutoff (or “significance”) values of 5% (or sometimes 1%) are often used for deciding that H 0 is false. This cutoff value for the P-value is often denoted by the symbol α , and α -values of 0.05 or 0.01 are typical. 2
Using a significance value for the P-value of 5% means here that we would reject H 0 if Z 1 . 645, whereas using a significance value of 1% means we would reject H 0 if Z 2 . 33. The constant 1.645 (or respectively 2.33) is called the “critical value” for the test.

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