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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.
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STA4322 Exam3 (closed book, 9:30 - 10:45, 8/6/2009) Name
Sol: α = P {RR|H 0 } and 1 − β = P {RR|Ha}
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.
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 ≥
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′
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 =
= 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%)