Let - *»0 ¼. Test 0 : = 1 against 1 : = 2 using a sample ¹-1 -2º of size 2, by rejecting 0 if either - 7 075 or at least one of -1 and -2 is greater than 1. Compute ¹1º and ¹2º.
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- Null hypothesis (H0): μ = 1 - Alternative hypothesis (H1): μ = 2 Show more…
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In Example $8.1 .2$ of this section, let the simple hypotheses read $H_{0}: \theta=$ $\theta^{\prime}=0$ and $H_{1}: \theta=\theta^{\prime \prime}=-1 .$ Show that the best test of $H_{0}$ against $H_{1}$ may be carried out by use of the statistic $\bar{X}$, and that if $n=25$ and $\alpha=0.05$, the power of the test is $0.9996$ when $H_{1}$ is true.
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For x ∈ ℝ and ̸ ∈ (0, 1), define f̸(x) = {̸² if -1 ≤ x < 0; 1 - ̸² if 0 ≤ x ≤ 1; 0 otherwise.} Let X₁, ..., Xₙ be i.i.d. random variables with density f̸, for some unknown ̸ ∈ (0, 1). Consider the following hypotheses: H₀ : X ∼ Unif (-1, 1) H₁ : X not distributed as Unif (-1, 1). Write down the test statistic Tₙᵀᵃᵌᵀ for Wald's test for the above hypothesis. Use the value of ̸ that defines H₀ as the argument of the asymptotic variance V(̸). Hint: Rewrite the hypothesis in terms of the parameter ̸. (Enter hattheta for ̸̂ᴹᴸᴱ.) (To avoid double jeopardy, you may use V for the asymptotic variance V(̸) under H₀.) Find C such that Wald's test has asymptotic level 5%. (Enter a numerical value accurate to at least 2 decimal places.) C = You obtain a sample of size n = 100, of which 40 of the Xᵢ are negative (Xᵢ < 0) and 60 of the Xᵢ are non-negative (Xᵢ ≥ 0). Do we reject H₀ at asymptotic level 5%? We reject H₀. We fail to reject H₀. Cannot be determined without more information. What is the p-value for this test? (Again, use the value of ̸ that defines H₀ as the argument of the asymptotic variance V(̸).) (Enter a numerical value accurate to at least 2 decimal place) p-value:
Adi S.
(c) The (b) The p Test the Ho Test whether Pa] select 2 H critical values test statistic Z0 #P2 J # #P2 null are 5 answverts hypothesis Sample hypothesis the #Pz Idata are 30 , correct conclusion below decimal decimal places 254, places 38 and nz
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