## Statistics-STAT 409 Fall 2015 A. Stepanov Homework #9

STAT 409Fall 2015A. StepanovHomework #9( due Friday, November 20, by 4:00 p.m. )Please include your name ( with your last name underlined ), your NetID,and your discussion section number at the top of the first page.No credit will be given without supporting work.1 â 2.Let X have a Binomial distribution with the number of trials n = 15 and withprobability of âsuccessâ p. We wish to test H 0 : p = 0.30 vs. H 1 : p ≠ 0.30.ËRecall that p =1.a)XnFind the values ofis the maximum likelihood estimator of p.Λ(x) =L( p 0 = 0.30 ; x )ËL( p; x )for x = 0, 1, 2, â¦ , 14, 15.If you are using a computer, include the printout.b)2.Suppose we observe X = 7. Find the p-value of this test.c)Likelihood Ratio Test:Reject H 0 if Λ ( x ) ≤ c.Let c = 0.15. Find(i)the significance level,(ii)power when p = 0.20,(iii)power when p = 0.40for the corresponding rejection region.3.Let X have a Binomial distribution with the number of trials n = 15 and withprobability of âsuccessâ p. We wish to test H 0 : p = 0.30 vs. H 1 : p > 0.30.a)Find the best Rejection Rule with the significance level α closest to 0.05.b)Find the power of the test from part (a) at p = 0.40 and at p = 0.50.c)Suppose we observe X = 7. Find the p-value of this test.4 â 6.Let λ > 0 and let X 1 , X 2 , â¦ , X n be independent random variables, each withthe probability density functionf (x; λ ) =λx λ +1x ≥ 1.,nRecallY=∑ ln X iis a sufficient statistic for λ ,∑ ln X ihas a Gamma distribution with α = n and θ = 1 λ ,i =1nY=/i =1nËthe maximum likelihood estimator of λ is λ =n=∑ ln X inY.i =14.We wish to test H 0 : λ = 3 vs. H 1 : λ ≠ 3.a)Likelihood Ratio Test:Reject H 0 if Λ ( y ) ≤ c.Show that Λ ( y ) ≤ c is equivalent to Yb)Let k = eâ3.5.c)n â3Yee≤ k.Then Y = 1 is one ( obvious ) solution of YSuppose n = 7.Yn â3Yn â3Ye=eâ3.Find ( to the fourth decimal place ) the value of d such that≤ eâ 3⇔Y ≤ 1 or Y ≥ dFind(i)the significance level,(ii)power when λ = 2,(iii)power when λ = 4for the rejection region in part (b).6.We wish to test H 0 : λ = 3 vs. H 1 : λ < 3.a)Suppose n = 7. Find the uniformly most powerful rejection region with a 5%level of significance. Round the critical value for Y to the fourth decimal place.b)Find the power of the test in part (a) when λ = 2 and when λ = 1.7 â 8.7.Crosses of mice will produce either gray, brown, or albino offspring. Mendelâsmodel predicts that the probability of a gray offspring is 9/16 , the probabilityof a brown offspring is 3/16 , and the probability of an albino offspring is 4/16 .An experiment to assess the validity of Mendelâs theory produces the following data:36 gray offspring; 21 brown offspring; and 23 albino offspring. TestH 0 : p 1 = 9/16 , p 2 = 3/16 , p 3 = 4/16a)8.at α = 0.05,b)vsH 1 : not H 0at α = 0.10.Suppose the experiment in Problem 7 is repeated, but with twice as many observations.Suppose also that we happened to get the same proportions, namely, 72 gray offspring;42 brown offspring; and 46 albino offspring. Repeat Problem 7 in this case, usingα = 0.01.EXCEL functions you may find helpful:= BINOMDIST( x , n , p , 0 )givesP( X = x )= BINOMDIST( x , n , p , 1 )givesP( X ≤ x )= CHIINV ( α , v )gives2χ α (v ) for χ 2 distribution with v degreesof freedom, x s.t. P ( χ 2 (v ) > x ) = α.= CHIDIST ( x , v )givesχ 2 distributionwith v degrees of freedom, P ( χ 2 (v ) > x ).the upper tail probability for= POISSON( x , λ , 0 )givesP( X = x )= POISSON( x , λ , 1 )givesP( X ≤ x )

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