02-07-2011 2 Sample Hypothesis Testing

# 02-07-2011 2 Sample Hypothesis Testing - Lecture 5 February...

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Unformatted text preview: 2/7/2011 Lecture 5: February 7, 2011 • Today: – Review the classical approach and the p‐value approach to hypothesis testing: one sample z for Ford trucks – The t‐distribution – Two‐Sample Hypothesis Tests • Homework 4 opens at 10:00 p.m. tonight: – Page 318: Problems 8.18 and 8.20 – Confidence intervals using t – Pages 363‐364: Problem 9.23, 9.28, and 9.33 – Hypothesis tests of one mean using t – Homework 4 closes Wednesday, February 9, at 6:00 p.m. – Unlimited tries until it closes! Feedback! • Non‐Connect Practice problems: – Page 318: Problems 8.19 and 8.21 – Confidence intervals using t – Page 364: Problem: 9.35 – Hypothesis tests of one mean using t – Answers are in the back of the book. – If you need help with these, see a TA during office hours. • Worked‐Out Problems – On my web site – Page 318: Problem 8.17 – Confidence interval using t – Page 364: Problem 9.31 – Hypothesis test of one mean using t – These will be ready some time on Tuesday afternoon. Lecture 5: February 7, 2011 Readings • Student’s t distribution: – Confidence intervals – Chapter 8, pages 310‐318 – One‐sample hypothesis tests – Chapter 9, pages 359‐ 364 • Two‐Sample Hypothesis Tests: – End of today and next two classes – Chapter 10, pages 390‐401 – Put on your seat belt! – Type of problems to look at: • pages 400‐401: 10.1 and 10.2 What we did last class: • We did a one‐sample hypothesis test of the population mean. • We were given the set‐up on the following slide. 1 2/7/2011 A One‐Sample Hypothesis Test of the Population Mean • Test the null hypothesis that the true mean is 20 mpg versus the alternative that the true mean is greater than 20 mpg. • Assume that the standard deviation of the sample mean (or standard error) for each sampling distribution is: X / n 1.818 • You gather 25 observations and calculate X=24 • Test at the 0.025 level of significance. • Assume the underlying variable of interest is normally distributed. What we did last class (continued): • Next, we followed the steps in the handout titled Revised Steps In Hypothesis Testing. • The approach that we took with these five steps is sometimes called the classical approach to hypothesis testing. • Again, here are the dynamics. Dynamics Of The Classical Approach To Hypothesis Testing H0: µ0=20 Test statistic: X = 24 1‐α X = 1.818 α=0.025 Is far from μ0 = 20? X X µ0=20 Standardized test statistic: z = 2.20 X = 24 Reject H0 Fail to Reject H0 z.025=1.96 z=2.20 HA: µ1>26 β z Reject H0 … because z>z.025 1‐β µ1>20 X 2 2/7/2011 The P‐Value Approach To Hypothesis Testing H0 True α=0.025 P‐value=0.0139 X µ0=20 X = 24 Reject H0 Fail to Reject H0 Z.025=1.96 • z z=2.20 Compare α and p‐value • Reject H0 • Because p‐value < α Usefulness of the p‐value approach • • • • • • In our hypothesis test problem What is the meaning of P(z>2.20) = 0.0139? Here’s what it means and how it is used. 0 It gets at the issue: Is this value of far away from ? X If the p‐value is small, this means is far away from . X 0 This is evidence against the null hypothesis. Interpretation Of A P‐Value H0 True α=0.025 P‐value=0.0139 X µ0=20 X = 24 A p‐value is the observed level of significance for our X For this example, the p‐value = 0.0139 Interpretation: It is the probability of observing a value ofX=24 or greater if the null hypothesis is true. It is calculated asP(X>24) with μ=20 and X =1.818. Meaning: If H0 is true, we would observe a value of or greater only about 1 time in 100. X=24 Translation: We have extremely strong evidence against the null hypothesis. Decisions reached with both the p‐value approach and the classical approach are the same. 3 2/7/2011 New Topic Moving beyond the z‐distribution (Chapter 8, pages 310‐312) (1) Strict definition of z (Chapter 8): z X- X / n (a) z has one shape: standard normal (b) z is centered at zero. (c) z has standard deviation one. (d) As long as is given, i.e., known, X can be converted to z. (In fact, this is what we just did.) Moving beyond the z‐distribution (continued) (2) Q: In what situations did we use z? X A: Making probability statements about . Confidence intervals. Hypothesis tests. (3) (a) In practical situations, we do not have . (b) We must use its estimate, which is the sample standard deviation s. (c) If s replaces in the z formula, then the formula is really no longer z! 4 2/7/2011 Moving beyond the z‐distribution (continued) This image cannot currently be display ed. (4) Q: What is: X - s n A: t (5) Q: Is there a relationship between t and z? A: Yes. The relationship depends on the magnitude of n. Notice the t and z formulas. As n increases, s begins to resemble better. t becomes z. Moving beyond the z‐distribution (continued) This image cannot currently be display ed. Here’s what we have so far: z X- X / n t = X - s n As n increases, s begins to resemble better, t becomes z. (6) Let’s apply Points (1)‐(4) from the previous two slides and develop a set of pictures for t. Moving beyond the z‐distribution (continued) (6) Pictures of t, continued ... (a) Shape: a number of shapes, depending on something called degrees of freedom (df): df=n‐1 (See p. 311.) (b) Centered at zero; symmetrical about zero. (c) t is flatter and more spread out than the z curve for small values of n; t becomes more like z as n → . This image cannot currently be display ed. 5 2/7/2011 Moving beyond the z‐distribution (continued) (7) Each curve reports tail area probabilities () and corresponding t‐values (t) (in Appendix D on page 766 of the text) for a variety of different sample sizes, i.e., for different n. This image cannot currently be display ed. (8) If I am given n (or df) and , I can find t. df t0.10 t0.05 t0.025 t0.01 t0.005 1 6.314 12.706 19 1.729 2.093 1000 1.282 1.646 1.962 2.330 2.581 z 1.282 1.645 1.960 2.326 2.576 Moving beyond the z‐distribution (continued) This is the same as the bottom of the previous slide. t0.05 t0.025 t0.01 t0.005 df t0.10 1 6.314 12.706 19 1.729 2.093 1000 1.282 1.646 1.962 2.330 2.581 z 1.282 1.645 1.960 2.326 2.576 This image cannot currently be display ed. e.g., Given: n 2 .05 df t 20 .05 1001 .05 z .05 6 ...
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