ec41f10_hw6_sol

# ec41f10_hw6_sol - Kata Bognar [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ Economics 41...

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Unformatted text preview: Kata Bognar [email protected] Economics 41 Statistics for Economists UCLA Fall 2010 Homework Assignment 6.- suggested solutions by Andrew Zaeske - NOTE: Please show your calculations for Questions 1-5, 7, 9-16. The exercises are from the textbook (Tanis and Hogg: A Brief Course in Mathematical Statistics). t-Interval procedure. 1. Suppose that a sample of size 36 is drawn from a normally distributed population. The sample mean for this sample is 25 and the sample standard deviation is 3. Construct a 90% confidence interval for the population mean. Answer : We are given the sample standard deviation, so we should use the t-distribution critical value rather than one from the normal. The value t . 05 (35) = 1 . 69 . Thus, the 90% confidence interval will be ¯ X- t . 05 (35) · s √ n , ¯ X + t . 05 (35) · s √ n = 25- 1 . 69 · 3 6 , 25 + 1 . 69 · 3 6 = [24 . 155 , 25 . 845] 2. A random sample of 40 families selected from a city showed that the mean amount these families donated to charitable agencies last year was \$600 with a standard deviation of \$160. Construct a 95% confidence interval for the mean amount donated to charitable agencies last year by all families living in this city. Answer : We are given the sample standard deviation, so we should use the t-distribution critical value rather than one from the normal. The value t . 025 (39) = 2 . 023 . Thus the 95% confidence interval will be ¯ X- t . 025 (39) · s √ n , ¯ X + t . 025 (39) · s √ n = 600- 2 . 023 · 160 √ 40 , 600 + 2 . 023 · 160 √ 40 = [548 . 822 , 651 . 178] 3. Solve Exercise 4.2 - 8. Answer : (a) ¯ x = 1 n ∑ x i = 3 . 243 (b) s 2 = 1 n- 1 ∑ ( x i- ¯ x ) 2 = 0 . 2372, s = √ s 2 = 0 . 4870 (c) Using the t-table in the book, we find that t . 05 (11) = 1 . 796. Thus the 90% confidence interval will be, ¯ X- t . 05 (11) · s √ n , ¯ X + t . 05 (11) · s √ n = 3 . 243- 1 . 796 · . 4870 √ 11 , 3 . 243 + 1 . 796 · . 4870 √ 11 = [3 . 2166 , 3 . 2694] 4. A sample of 30 commuters in the Washington, D.C. area yielded the following commute times, in minutes....
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## This note was uploaded on 12/28/2010 for the course ECON 41 taught by Professor Guggenberger during the Fall '07 term at UCLA.

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ec41f10_hw6_sol - Kata Bognar [email protected] Economics 41...

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