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The final exam solutions Part I, #1, Central limit theorem Let X1,X2, …, Xn be a sequence of i.i.d. random variables each having mean μ and variance σ2 The sum of a large number of independent random variables has a distribution that is approximately normal X 1 X 2 ... X n is approximat ely normal with mean n and variance n 2 X 1 X 2 ... X n n is approximat ely a standard normal random variable n Part II #2 If the successful probability of trial is very small, then the accumulatively many successful trials distribute as a Poisson. The Poisson parameter λ is the mean frequency of interest within a large bundle of experiment The exponential distribution is often used to describe the distribution of the amount of time until the specific event first occurs. The exponential distribution seems to partition the Poisson distribution with a parameter λ by every two occurrences into several small time intervals and focuses on a specific time interval. The same distributional parameter λ, The exponential distribution emphasizes the cycle time, 1/λ, while the Poisson on the frequency λ. Part I, #3 If Z and Xn2 are independent random variables, with Z having a std. normal dist. And Xn2 having a chi-square dist. with n degrees of freedom, Tn Z12 Z 22 ...Z n2 X 2n , 2 n Xn / n n Z •For large n, the t distribution approximates to the standard normal distribution Part I, #4 When these two population variances are unknown but equal, we calculate the pooled variance estimator, Sp2, by means of weighted average of individual sample 2 2 variance, S1 and S2 Part I, #5 It is not correct. He had better say that the statistic or random variables used to obtain this confidence interval is such that 95% of the time that it is employed it will result in an interval covered the μ. He can assert with probability 0.95 the interval will cover the μ only before the sample data are observed. Whereas after the sample data are observed and computed the interval, he can only assert that the resultant interval indeed contains μ with confidence 0.95. Part I, #6 Suppose the i.i.d. random variables X1 , X 2 , X n , whose joint distributi on is assumed given except for an unknown parameter , are to be observed and constitute d a random sample. The value of likelihood function f (x 1 , x 2 , , x n / ) will be determined by the observed sample (x 1 , x 2 , , x n ) if the true value of could also be found. ^ The maximum likelihood estimator of , denoted by , would maximize the probabilit y of likelihood function of observed values. Max f (x 1 , x 2 , , x n / ) means df d log f 0, (or 0) dθ dθ Part I, #7 (a) if d(X) is the estimator of θ, and E[d(X)]=θ, then the d(x) is unbiased. (x X ) (b) the MLE estimator is not n unbiased. (x X ) (c) the sample variance S n 1 is an 2 unbiased estimator of σ n ^ i i 1 2 n 2 2 i 1 i 2 Part II #1 Mean=np=90, Var=np(1-p)=150(.6)(.4)=36, ∴standard deviation=6 P{X≦80}=p{X<80.5}=P{(X90)/6<(80.5-90)/6}=P{Z<-1.583} =1-P{Z≦1.583} (=1-0.943=0.057) Part II #2,3 _ if σ is known, ( X ) ~ Z (0,1), / n _ _ ( X ) P{ z } 1 , P{ X z / n }, / n (90450 1.645(9400 / 4), ) (86584.25, ) with 95% confidence _ if σ is unknown, ( X ) ~ t n 1 , S/ n _ _ ( X ) P{t ,n 1 } 1 , P{ X t ,n 1S / n }, S/ n we say (90450 1.735(9400 / 4), ) (86330.45, ) with 95% confidence The lower bound of 95% confidence upper interval is smaller wh en the is unknown Part II #4 Suppose the deffective rate is p. So the 100 product items seem to obey a binomial distributi on with mean 17 and variance 100(.17)(. 83), ^ X-n p ^ ^ ~ N (0,1), (we assume that n is large enough for normality approximat ion) n p ( 1- p ) ^ ^ ^ ^ ^ ^ P p z / 2 p ( 1- p ) / n p p z / 2 p ( 1- p ) / n 1 P .17 1.96 (0.17)(. 83) / 100 p .17 1.96 (.17)(. 83) / 100 0.95 p (0.096, 0.244) with 95% confidence Part II #5,6 (0.5)(0.5) 1.96 2 (0.5)(0.5) z0.25 0.03, n 2 n (0.03) n 1067.11,1068 is the necessary sample size if the servey result is 0.3 the margin of error 1.96 (0.3)(0.7) 0.0274 2.75%, lightly decreasing 1068