02-07-2011 2 Sample Hypothesis Testing

02-07-2011 2 Sample Hypothesis Testing - 2/7/2011 Lecture...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online