STA4322 Exam 3: Hypothesis Testing, Min. Variance Unbiased Estimator, and Power, Exams of Statistics

Solutions to exam 3 of the statistics course sta4322. It covers topics such as hypothesis testing, minimum variance unbiased estimators, and the power of tests. Students are expected to understand concepts related to the rejection region, types of errors, minimum variance unbiased estimators, uniformly most powerful tests, and confidence intervals.

Typology: Exams

2019/2020

Uploaded on 06/08/2020

stagist
stagist 🇺🇸

4.1

(27)

265 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
STA4322 Exam3 (closed book, 9:30 - 10:45, 8/6/2009) Name
1. In hypothesis testing, we need two hypotheses H0and Ha, a rejection region RR, and
two types of errors αand β. State the two main relations between αand βand the
other terms. (10%)
Sol: α=P{RR|H0}and 1 β=P{RR|Ha}
2. (i) When we say that ˆ
θis a MVUE (minimum variance unbiased estimator) for an
unknown parameter θ, what do we means. Explain it using commonly used statistical
notation. (10%)
Sol: (i) E[ˆ
θ] = θand for and other unbiased estimator ˜
θ,V ar(˜
θ)V ar(ˆ
θ)
(ii) What we say that a test is the uniformly most powerful test in hypothesis testing,
what do we mean? Use an example to explain this. (10%)
Sol: Let H0:p=p0and Ha:p > p0. Then the uniformly most powerful test is the
most powerful for test for all the simple hypothesis Ha:p=pawith pa> p0.
3. The minimum requirement for the hardness index of an alloy is 65. Fifty specimen of
the alloy was collected and the sample mean is 68 with sample standard deviation 8.
Can the company claim that their alloy meets the hardness requirement? Make you
conclusion by p-value. (15%)
Sol: To guarantee the product meets the requirement, we need to put Haas µ > 65.
Thus H0is µ65.
p=value =P( ¯y68)
=P(Z68 65
8/50 )
=P(Z2.65) = 0.004
4. Let y1,y2,..., ynbe a random sample from a Poisson distribution with unknown mean
λ. Find the most powerful test for H0:λ= 1 against Ha:λ= 2. You are allowed to
have one threshold to be determined by the type I error probability. (10%)
–1–
pf3

Partial preview of the text

Download STA4322 Exam 3: Hypothesis Testing, Min. Variance Unbiased Estimator, and Power and more Exams Statistics in PDF only on Docsity!

STA4322 Exam3 (closed book, 9:30 - 10:45, 8/6/2009) Name

  1. In hypothesis testing, we need two hypotheses H 0 and Ha, a rejection region RR, and two types of errors α and β. State the two main relations between α and β and the other terms. (10%)

Sol: α = P {RR|H 0 } and 1 − β = P {RR|Ha}

  1. (i) When we say that θˆ is a MVUE (minimum variance unbiased estimator) for an unknown parameter θ, what do we means. Explain it using commonly used statistical notation. (10%)

Sol: (i) E[θˆ] = θ and for and other unbiased estimator θ˜, V ar(θ˜) ≥ V ar(θˆ)

(ii) What we say that a test is the uniformly most powerful test in hypothesis testing, what do we mean? Use an example to explain this. (10%)

Sol: Let H 0 : p = p 0 and Ha : p > p 0. Then the uniformly most powerful test is the most powerful for test for all the simple hypothesis Ha : p = pa with pa > p 0.

  1. The minimum requirement for the hardness index of an alloy is 65. Fifty specimen of the alloy was collected and the sample mean is 68 with sample standard deviation 8. Can the company claim that their alloy meets the hardness requirement? Make you conclusion by p-value. (15%)

Sol: To guarantee the product meets the requirement, we need to put Ha as μ > 65. Thus H 0 is μ ≤ 65.

p = value = P (¯y ≥ 68)

= P (Z ≥

= P (Z ≥ 2 .65) = 0. 004

  1. Let y 1 , y 2 ,..., yn be a random sample from a Poisson distribution with unknown mean λ. Find the most powerful test for H 0 : λ = 1 against Ha : λ = 2. You are allowed to have one threshold to be determined by the type I error probability. (10%)

Sol: l(y 1 , y 2 , ..., yn) =

λy^1 +y^2 +...+yn^ e−nλ y 1 !y 2 !...yn!

Thus, most the powerful test is

l(y 1 , y 2 , ..., yn|λ = 1) l(y 1 , y 2 , ..., yn|λ = 2)

< k

⇒ 2 −^

∑ yi (^) e−n (^) < k

∑ yi > k′

  1. A study was conducted to the DDT level in the brain tissue of brown pelicans. The purpose to test the difference between the juveniles and nestings. The following data summary was obtained. (We can assume that the data are normally distributed.)

Juveniles Nestings Sample size n 1 = 8 n 2 = 6 sample mean (in ppm) y¯ 1 = 0. 041 y¯ 2 = 0. 022 sample standard deviation s 1 = 0. 017 s 2 = 0. 034

(i) Can we claim that juveniles have a higher mean concentration? Draw your conclu- sion with an approximate p-value. (15%)

Sol: We are testing H 0 : μ 1 = μ 2 against Ha : μ 1 6 = μ 2. We use the two sample t-test.

S p^2 =

7 × 0. 0172 + 5 × 0. 0342

= 6. 5 × 10 −^4 or Sp = 0. 0255.

By formula T =

√ 1 /8 + 1/ 6

= 1. 38 ∼ t 12

p-value ∼ 0.2.

(ii) Suppose the two variances for the two populations are σ^21 and σ 22 respectively and they are unknown. We also know that

s^21 /σ^21 s^22 /σ^22

∼ F (^) nn 21 −− 21.

How do we use an F-table to construct a 95% confidence interval for σ^22 /σ 12. Note: F (^) νν 21 , 1 −a = 1/F (^) νν 12 ,a. (10%)