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UIUC STAT 400 - 400Ex7_1_2ans

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STAT 400 Lecture AL1 Answers for 5.5, 7.1 Spring 2015 Dalpiaz ¼ . Let Z be a N ( 0, 1 ) standard normal random variable. Then X = Z 2 has a chi-square distribution with 1 degree of freedom. M X ( t ) = E ( e t Z 2 ) = zee dzzt 22 2 21π = ze dtz 2 212 21π = 21 211 t, t < 1/2, since  2 21221 221πtzet is the p.d.f. of a N ( 0, t 211) random variable.  X has a  2 ( 1 ) distribution. ½ . Let X and Y be be two independent  2 random variables with m and n degrees of freedom, respectively. Then W = X + Y has a chi-square distribution with m + n degrees of freedom. M X ( t ) = 2 211mt, t < 1/2 , M Y ( t ) = 2 211nt, t < 1/2 . M W ( t ) = M X ( t )  M Y ( t ) = 2 211nmt, t < 1/2 .  W has a  2 ( m + n ) distribution.1. A manufacturer of TV sets wants to find the average selling price of a particular model. A random sample of 25 different stores gives the mean selling price as $342 with a sample standard deviation of $14. Assume the prices are normally distributed. Construct a 95% confidence interval for the mean selling price of the TV model.  is unknown. n = 25 – small. The confidence interval : nst2X . n  1 = 25  1 = 24 degrees of freedom. 95% confidence level,  = 0.05, /2 = 0.025, 2t( 24 ) = 2.064. 2514064.2342  342  5.78 ( 336.22 ; 347.78 )2. The following random sample was obtained from N (  ,  2 ) distribution: 16 12 18 13 21 15 8 17 a) Compute the sample mean and the sample standard deviation. 81208178152113181216nxx = 15. x x 2 x xx 2 xx  16 12 18 13 21 15 8 17 256 144 324 169 441 225 64 289 OR 16 12 18 13 21 15 8 17 1 – 3 3 – 2 6 0 – 7 2 1 9 9 4 36 0 49 4 1912 0 112 78120191212222 nnxxs  = 16. 7112122 nxxs = 16. ss 162 = 4. b) Construct a 95% confidence interval for . Confidence interval: nsx 2αt  . n – 1 = 7 degrees of freedom. 95% confidence level,  = 0.05, /2 = 0.025, t 0.025 = 2.365. 84365.215 15  3.3446 ( 11.6554 , 18.3446 )b ½ ) Construct a 90% confidence interval for . Confidence interval: nsx 2αt  . n – 1 = 7 degrees of freedom. 90% confidence level,  = 0.10, /2 = 0.05, t 0.05 = 1.895. 84895.115 15  2.68 ( 12.32 , 17.68 ) c) Construct a 90% confidence upper bound for . ns α X , t n – 1 = 7 degrees of freedom t 0.10 = 1.415. 841.415 15 , ( – ∞ ; 17 ) d) Construct a 99% confidence lower bound for .  , α X tns n – 1 = 7 degrees of freedom t 0.01 = 2.998.  , 842.998 15 ( 10.76 ; ∞


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UIUC STAT 400 - 400Ex7_1_2ans

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