代做Introduction to Mathematical Statistics II (4202) Midterm 1 - Practice Exan代做Python程序

2024-09-28 代做Introduction to Mathematical Statistics II (4202) Midterm 1 - Practice Exan代做Python程序

Introduction to Mathematical Statistics II (4202)

Midterm 1 - Practice Exan

1. Let X1, X2, . . . , Xn be independent and identically distributed (i.i.d.) as Binomial(k, p) with the probability mass function (p.m.f.) given by:

where k is known and 0 ≤ p ≤ 1.

It is known that E(X1) = kp and Var(X1) = kp(1 − p).

(a) Find a sufficient statistic for p.

(b) Find the Minimum Variance Unbiased Estimator (MVUE) of p.

(c) The moment generating function of X1 is given by:

Find the Maximum Likelihood Estimator (MLE) of MX1 (t) for any fixed t ∈ R.

2. Let Y1, Y2, . . . , Yn be independent and identically distributed (i.i.d.) as Uniform(a, θ+ a), with the probability density function (p.d.f.) given by:

where θ > 0 and a is a known constant. It is known that E(Y1) = 2/θ+2a and Var(Y1) = 12/θ2.

We are interested in the properties of the Maximum Likelihood Estimator (MLE) and the Method of Moments Estimator (MoME) for θ.

(a) Is Y(n) = max(Y1, . . . , Yn) a consistent estimator for θ? Justify your answer.

HINT: Use the fact that if Z ∼ Uniform(a, θ + a), then Y = θ/Z−a ∼ Uniform(0, 1), and for Y(n) = max(Y1, . . . , Yn), E(Y(n)) = n+1/n , V ar(Y(n)) = (n+1)2(n+2)/n ..

(b) Compare the efficiency of the MoME with the MLE. Which estimator is more efficient? Assume n > 2.

3. A study has been made to compare the nicotine contents of two brands of cigarettes. Ten cigarettes of Brand A had an average nicotine content of 3.1 mil-ligrams with a standard deviation of 0.5 milligram, while eight cigarettes of Brand B had an average nicotine content of 2.7 milligrams with a standard deviation of 0.7 milligram. Assuming that the two sets of data are independent random sam-ples from normal populations with equal variances, construct a 95% confidence interval for the difference between the mean nicotine contents of the two brands of cigarettes.

[Useful quantiles: t0.025,16 = 2.12.]