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Unformatted text preview: Answer Key to Homework #4 Econ41 Winter 2008 By Jiyeon Seo 1 : & n 1 = 200 & x 1 = 7 : 9 s 1 = : 65 s 2 & X 1 = s 1 =n 1 = : 0021 n 2 = 300 & x 2 = 5 : 4 s 2 = : 90 s 2 & X 2 = s 2 =n 2 = : 0027 ¡ ! CASE : Independent & & 1 , & 2 are unknown and not equal. s & X 1 & & X 2 = q s 2 & X 1 + s 2 & X 2 = : 0694 ; df = 494 ; ¡ = 0 : 1 ; t ( or z ) = 1 : 645 ) 90% CI = 2 : 3859 to 2 : 6141 : answer is (d) . 2 & 3 : & n 1 = 20 & x 1 = 65 s 1 = 9 n 2 = 25 & x 2 = 58 s 2 = 11 ¡ ! CASE : Independent & & 1 ;& 2 are unknown but equal. 2. s p = s ( n 1 ¡ 1) s 2 1 + ( n 2 ¡ 1) s 2 2 n 1 + n 2 ¡ 2 = 10 : 1649 : answer is (c) . 3. s & X 1 & & X 2 = s p r 1 n 1 + 1 n 2 = 3 : 0495 : answer is (c) . 4 & 5 : & n 1 = 16 & x 1 = 82 s 1 = 14 n 2 = 18 & x 2 = 75 s 2 = 16 ¡ ! CASE : Independent & & 1 ;& 2 are unknown but equal. H : ¢ 1 = ¢ 2 ; H 1 : ¢ 1 > ¢ 2 ( righttailed test ) 4. t ¡ distribution with df = n 1 + n 2 ¡ 2 = 32 ; ¡ = 0 : 01 ! Critical value = 2 : 449 : answer is (b) . 5. s p = 15 : 0955 ;s & X 1 & & X 2 = 5 : 1867 ; observed test stat. = & x 1 ¡ & x 2 s & X 1 & & X 2 = 1 : 3496 : answer is (b) . 6 : Answer is (b) . 10.13. & T999: n 1 = 45 & x 1 = 3300 & 1 = 800 51XPY: n 2 = 51 & x 2 = 3850 & 2 = 1000 ¡ (a) 99% CI for ( ¢ 1 ¡ ¢ 2 ) ! CASE : Independent & & 1 ;& 2 are known. ; hence, z = ¢ & X 1 ¡ & X 2 £ ¡ ( ¢ 1 ¡ ¢ 2 ) & & X 1 & & X 2 & N (0 ; 1) & 2 & X 1 = & 2 1 =n 1 = 14222 : 22 ; & 2 & X 2 = & 2 2 =n 2 = 19607 : 84 ; & & X 1 & & X 2 = q & 2 & X 1 + & 2 & X 2 = 183 : 93 ¡ = : 01 ! z = 2 : 575 ) 99% CI : (& x 1 ¡ & x 2 ) ¢ z £ & & X 1 & & X 2 ! ¡ 1023 : 62 to ¡ 76 : 38 1 (b) Hypothesis Test & = : 01 ; H : ¡ 1 = ¡ 2 ; H 1 : ¡ 1 6 = ¡ 2 (twotailed test)...
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 Winter '07
 Guggenberger
 Critical Point, Null hypothesis, Type I and type II errors, Critical phenomena

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