Question 2: The following are the final grades of two sections of one statistics course. n1=22students in section 1 have a sample of an average of x_bar1 = 84 with a standard deviation of s1
= 4. n2=18 students in section 2 made a sample average of x_bar2 = 80 with a standard
deviation of s1 = 6. We now want to assess the difference of population means µ1 − µ2 of
grades between two sections.
a) Test for H0 : µ1 − µ2 = 0 vs Ha : µ1 − µ2
0 using a large sample Z-test and report the p
value. Use a significance level of α = 0.025. (2)
b) Compute a 95% confidence interval estimate for µ1 − µ2.
Question 3: Let X1, X2, . . . , Xn be i.i.d. samples from the uniform(θ, θ + 1) distribution. To test
H0 : θ = 0 versus H1 : θ > 0, we use the following test reject H0 if X(1) ≥ k, where k is a constant,
X(1)=min{X1, …, Xn}.
a) Determine k so that the level of the test is α (the probability of Type I error).
b) Find the expression for the probability of Type II error of this test using k and θ. (Hint:
Discuss different values of k.)
Question 4: Consider a simple linear model with normal error:
Y = β0 + β1x + ε, ε ∼ N(0, σ^2 )
Suppose that 10 pairs of observed values from this model given in the following table are
obtained:
a) Calculate the values of the M.L.E’s
.
b) Determine an unbiased estimator of Var( ) and calculate its value.
c) Le
t, where c is a constant. Determine an unbiased estimator ˆθ of θ. For
what value of c will the M.S.E. of be smallest?
Question 5: Suppose n data points {(xi , yi), i = 1, . . . , n} are obtained from the linear model:
Where where E[ε] = 0 and Var(ε) = σ^2 . However, we do not know the above true model and
fit the data points on the simple linear model:
and obtain the least-squares estimator for β0 and β1:
a) Are these estimators unbiased? If yes, prove it. If no, find the bias
b) Derive the variance of
and the covariance between
and
.
Continuous Distributions
Mean
Variance
MomentGenerating
Function
1
; θ ≤ y ≤ θ2
θ2 − θ1 1
θ1 + θ2
2
(θ2 − θ1 )2
12
etθ2 − etθ1
t (θ2 − θ1 )
1
1
2
(y
−
µ)
√ exp −
2σ 2
σ 2π
−∞ < y < +∞
µ
σ2
β
β2
(1 − βt)−1
αβ
αβ 2
(1 − βt)−α
v
2v
(1−2t)−v/2
α
α+β
(α + β) (α + β + 1)
Distribution
Uniform
Normal
Exponential
Probability Function
f (y) =
f (y) =
f (y) =
Gamma
Chi-square
f (y) =
1
α−1 −y/β
e
;
α y
(α)β
0
