This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Stat 20: solutions to homework 10 Michael Lugo November 22, 2010 1 Chapter 26, #2 (a) The box model is that the data are like 3800 draws at random with replacement from a box containing 1s and 0s, where the proportion of 1s is p . (b) The null hypothesis is that p = 18 / 38; the alternative hypothesis is that p > 18 / 38. (c) The expected number of reds is 1800. The standard error of the number of reds is √ 3800 × p 18 / 38 × 20 / 38 ≈ 31. Thus we have z = (1890 1800) / 31 ≈ 2 . 9 and P ≈ . 002. (d) Yes, there are too many reds. 2 Chapter 26, # 4 The null hypothesis that the TA assumes here is that the scores in his section are like 30 draws at random from a box containing all 900 scores. The null hypothesis gives an average of 63 and an SD of 20. The standard error of the average of 30 scores is therefore 20 / √ 30 ≈ 3 . 65. So we have z = (55 63) / 3 . 65 ≈  2 . 2, and P ≈ . 01. The answer is no....
View
Full
Document
This note was uploaded on 07/21/2011 for the course STAT 20 taught by Professor Staff during the Spring '08 term at Berkeley.
 Spring '08
 Staff
 Statistics, Probability

Click to edit the document details