STAT 4101 Quiz 3, Spring 2021
First name: Last name:
Hawkid: @uiowa.edu
Please read the following instructions before you start the exam.
• There are three pages including the cover page.
• The quiz is available from Mar 12th, 2021 noon and due on Mar 15th, 2021 noon.
• There are 2 problems. The total is 10 ∗ 2 = 20 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.
• You must stop writing immediately after the invigilator signals the end of the exam.
• Any distribution of this material is prohibited.
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1. (10 pts) Let X1, X2 be a random sample of size n = 2 from the Poisson distribution with mean θ. We reject H0 : θ = 2 and accept H1 : θ = 1 if the observed values of X1,X2, say x1,x2, are such that
f (x1;2)f (x2;2) ≤ 1 f (x1;1)f (x2;1) 2
Here Ω = {θ : θ = 1, 2}. Find the significance level of the test and the power of the test when H0 is false.
Solution: Reject H0 if
e−22×1 e−22×2
x1! x2! = e−22×1+x2 ≤ 1.
e−11×1 e−11×2 x1! x2!
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Equivalently, reject H0 if x1 + x2 ≤ 2 + log(0.5) = 1.307.
Since both x1 and x2 are integers, the above decision becomes reject H0 if x1 + x2 ≤ 1.
The significant test is the probability of rejecting H0 when H0 is true, say θ = 2. In such case, T = X1 + X2 ∼ Poisson(4). Thus the significant test equals
e−4 40 e−4 41
P(T ≤1)=P(T =0)+P(T =1)= 0! + 1! =0.092.
The power of the test is the probability of rejecting H0 when H0 is false, say θ = 1. In such case, T = X1 + X2 ∼ Poisson(2). Thus the power equals
e−2 20 e−2 21
P(T ≤1)=P(T =0)+P(T =1)= 0! + 1! =0.406.
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2. (10 pts) Let X1, X2, . . . , Xn1 and Y1, Y2, . . . , Yn2 represent two independent random samples from the re-
spective normal distributions N μ1, σ12 and N μ2, σ2. It is given that σ12 = 3σ2, but σ2 is unknown.
PerformatestH0 :μ1 =μ2 againstH1 :μ1 ̸=μ2 withthesizeα=0.05. Supposewearegiventhatn1 =9,
n2=12,X ̄=0.5,Y ̄=1.0,thesamplevariance 1 (Xi−X ̄)2=0.42,and 1 (Yi−Y ̄)2=0.11.
What is the p-value and what is your decision? Solution: Let
n1−1 i n2−1 i
S 12 = S 2 2 =
1n1 2 X i − X ̄ ,
3(n1 −1) i=1 1n2 2
Y i − Y ̄ . n2 −1 i=1
We have
The pooled estimator of σ2 is and
n1 i=1
Xi −X ̄2 (n1 −1)S2 = 1
n2 i=1
Yi −Y ̄2 σ2
=
(n2 −1)S2 σ2
σ12 σ2
∼χ2n1−1, 2 ∼χ2n2−1.
(n1 −1)S12 +(n2 −1)S2 n1 + n2 − 2 ,
2
̄ ̄ σ12 σ2 23 1 X−Y∼N 0,n +n ∼N 0,σ2 n +n ,
12 12
X ̄ − Y ̄
Sp =
(n1 + n2 − 2)Sp2
σ2 ∼ χ2n1+n2−2.
Under H0,
and thus by Student’s theorem,
σ3+1 ̄ ̄ ⋆2n1n2 X−Y
T= = ∼tn1+n2−2. (n1+n2−2)Sp2/σ2 Sp 3 +1
2 n1 n2
n1 +n2 −2
With the size α, we reject H0 if |T⋆| ≥ tα/2,n1+n2−2. The p-value is 2 × P(T > |T⋆|), where T ∼ tn1+n2−2.
By plugging in the values, we see
S12 = 0.42/3 = 0.14, S2 = 0.11, which gives Sp2 = 0.123 and T ⋆ = −2.212.
Because |T⋆| > tα/2,n1+n2−2 = 2.093, we reject H0.
The p-value is 2 × P (T > 2.212) = 0.039 < 0.05, which also leads to the rejection of H0.
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End of the exam!
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