STAT 4101 Midterm 1, Spring 2021
First name: Last name:
Hawkid: @uiowa.edu
Please read the following instructions before you start the exam.
• The exam is from Apr 7th 6:30 PM to Apr 8 noon.
• There are 5 problems. The total is 10 × 5 = 50 points.
• Show all work for each problem below. If you only present a correct answer without stating how you get it, no credit will be given.
• Calculator ready solutions are not sufficient to obtain full credits. A numer- ical answer is needed if a problem asks for it.
• It is prohibited to share the exam content or the solution with others. See code of academic honesty: https://clas.uiowa.edu/students/handbook/academic-fraud-honor-code
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1. (10 pts) Multiple choices. Select the most accurate answer.
• For a confidence interval with the level 1 − α, if α becomes larger, than the interval becomes (a) wider.
(b) narrower.
(c) unchanged.
• What is the best plot that we should use when we want to check if data are normally distributed? (a) Boxplot.
(b) q-q plot.
(c) Histogram.
• Which is the following is an unbiased estimator of σ2? (a) 1 ni=1(Xi −X ̄)2.
n−1
(b) 1ni=1(Xi−X ̄)2.
n
(c) 1 ni=1(Xi −X ̄)2.
n+1
• We want to the Chi-square test to test the independence between A and B. What is the degree of freedom?
B1 B2 B3 B4 A1 10 21 15 6 A2 11 27 21 13 A3 6 19 27 24
(a) 6. (b) 9.
(c) 12.
• Let Yn denote the n th order statistic of a random sample of size n from a distribution of the continuous type. Find the smallest value of n for which the inequality P (ξ0.95 < Yn) ≥ 0.55 is true.
(a) 8. (b) 12. (c) 16.
Solution: BBAAC
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2. (10 pts) Let X1, X2, . . . , Xn denote a random sample from the pdf 1 −x/(θ+1)
θ+1e x>0; θ>−1, f(x) =
0 otherwise. (a) Show that X ̄ − 1 is an unbiased estimator for θ.
Solution: We first note that if X ∼ f(x) then X ∼ exp(θ+1). Therefore, μ = E(X) = θ+1, and σ2 =Var(X)=(θ+1)2. HenceE[X ̄−1]=θ+1−1=θforallθ.
(b) Find MSEθ(X ̄ − 1).
Solution: Since X ̄ − 1 is an unbiased estimator for θ,
̄ ̄ ̄ σ2 MSEθ(X−1)=Var(X−1)=Var(X)= n =
(c) Show that X ̄ − 1 is a consistent estimator of θ. Solution: By weak law of large numbers, we see
(θ + 1)2 n .
Hence
̄P
X −→ E(X) = θ + 1.
̄P
X − 1 −→ θ ,
which says X ̄ − 1 is a consistent estimator of θ. (d) Can you find a consistent estimator of |θ|?
Solution: Part c shows X ̄ − 1 is a consistent estimator of θ. Since |θ| is a continuous function in terms of θ, by the continuous mapping theorem, |X ̄ − 1| is a consistent estimator of |θ|.
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3. (10 pts) Suppose that X1, . . . , Xn form a random sample from
x 2 θ 2
f(x;θ) =
where θ > 0. Let Yn = max(X1,…,Xn). (a) Find the pdf of Yn.
Solution:
We first compute the CDF of Yn: FYn (y) =
y2 n 4θ2
y2n y2n = 2θ = 2θ ,
0 < y < 2θ,
0
0 < x < 2θ, otherwise,
thus the PDF of Yn is:
Yn Yn Yn 2θα1/(2n)2n P 2 <θ< 2α1/(2n) =P θ< 2α1/(2n) =1−P(Yn <2θα1/(2n))=1− 2θ
where the first equality is because Yn < 2θ always holds. (c) Show that Yn/2 is a consistent estimator of θ.
fYn(y)=2ny2n−1 1, 0
Yn Yn
=P 2<θ−ε+P 2>θ+ε Yn
=P 2 <θ−ε becauseYn<2θ =P(Yn <2(θ−ε))
2(θ − ε)2n
= 2θ using the CDF of Yn
→0, asn→∞.
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4. (10 pts) Let X1, . . . , Xn be a random sample from exp(1) and X ̄n is the sample mean. (a) Find the limiting distribution of Yn = √n(X ̄n − 1) as n → ∞.
Solution: Using central limit theorem, we directly have √ ̄ D
(1)
Alternatively, you can show
n(Xn − 1) −→ N(0, 1). t/√n √ t/√n−n
MYn(t)=e −(t/n)e
√
,t< n
converges to MZ(t) = exp(t2/2), which is the MGF of N(0,1). (b) Find the limiting distribution of Zn = √n X ̄n2 − 1 as n → ∞.
Solution: Define g(θ) = θ2, and we have g′(θ) = 2θ. Based on (1) and the delta method, we see
which is N(0, 4).
√ ̄D′2 Zn = n(g(Xn) − g(1)) −→ N(0, (g (1)) ),
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5. (10 pts) A poll shows that among 100 students at the University of Iowa, 50 prefer Pepsi over Coca Cola. Another poll shows that among 100 students at the Iowa State University 40 prefer Pepsi. Suppose two samples are independent.
Let p1 be the proportion of UI students who prefer Pepsi, and p2 be the proportion of ISU students who prefer Pepsi. Let pˆ1 = 50/100 and pˆ2 = 40/100 are the sample proportions for UI and ISU, respectively. Let n = 100 be the size of each sample.
(a) Show that
pˆ1 −pˆ2 −(p1 −p2)
D
pˆ1(1−pˆ1) + pˆ2(1−pˆ2) nn
−→ N(0, 1).
Solution: By central limit theorem, we see
which follows that
√D √D n(p1−pˆ1)−→N(0,p1(1−p1))and n(p2−pˆ2)−→N(0,p2(1−p2)).
By independence between pˆ1 and pˆ2, we have
√√ n(p1−pˆ1) D
n(p2−pˆ2) D
p1(1 − p1) −→ N(0, 1) and p2(1 − p2) −→ N(0, 1),
√
n((p1 − pˆ1) + (p2 − pˆ2)) −→ N(0, p1(1 − p1) + p2(1 − p2)),
√n((p1−pˆ1)+(p2−pˆ2)) D
p1(1−p1)+p2(1−p2) −→N(0,1). (2)
pˆ1(1 − pˆ1) −→ p1(1 − p1) and pˆ2(1 − pˆ2) −→ p2(1 − p2). by continuous mapping theorem. We further have
p1(1−p1)+p2(1−p2) P
pˆ1(1 − pˆ1) + pˆ2(1 − pˆ2) −→ 1. (3)
which is the same to
D
By weak law of large numbers, we have pˆ1 −→ p1 and pˆ2 −→ p2, which give PP
PP
Applying Slutsky’s theorem on (2) and (3), we have √n((p1−pˆ1)+(p2−pˆ2)) D
pˆ1(1−pˆ1)+pˆ2(1−pˆ2) −→N(0,1).
(b) Could you provide a large sample 95% confidence interval for p1 − p2, i.e., the difference between the
proportions of students who prefer Pepsi over Coca Cola in Iowa and Georgia? Solution: With z0.025 = 1.96, an approximate 95% confidence interval for p1 − p2 is
pˆ1(1 − pˆ1) pˆ2(1 − pˆ2)
pˆ1 − pˆ2 ± 1.96 n + n = (−0.0372, 0.2372).
We are about 95% confident that difference in the two proportions is between −3.72% and 23.72%. 6
(c) Whatisthep-valuefortestingH0 :p1 =p2 vsH1 :p1 >p2? Solution: Under H0 : p1 = p2, we observe the test statistic
⋆ pˆ 1 − pˆ 2
Z =
The p-value is P(Z > 1.429) = 0.077, where Z ∼ N(0,1). It does not lead to rejection of H0.
pˆ1(1−pˆ1) + pˆ2(1−pˆ2) nn
=1.429.
End of the exam!
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