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Probability and statistics Lecturer: Prof. Dr. Coordinator: Yilin not due Mete SONER WANG Solution Series 13 Q1. We toss 100 times a coin and we get 60 head. We want to do a test to know whether the coin is fair. (a) Test the hypothesis with a 0.01 level of signicance. Should this test be one or twotailed?. (b) What is the biggest amount of head should we have in 100 tossings so we cannot discard H0 := (c) The coin biased towards tail. Calculate all p0 so that the null hypothesis H0 (p0 ) := Probability of head is p0 , would not be rejected in a test with 0.05 level of signicance. Hint: It will be useful to use the central limit theorem in all of this question. Solution: Let (a) (Xi )ni=1 an i.i.d sequence of Bernoulli(p). Let X= P100 i=1 Xi , then X ∼ Bin(100, p). We want to know whether the coin is fair or not, so our hypothesis are H0 : The H1 : The coin is fair, coin is not fair. That is to say 1 H0 : p = p0 = , 2 H1 : p 6= p0 . Then we should use a two-tailed test. Under H0 and V ar0 (X) = np0 (1 − p0 ) = 25. By central X−50 can be approximated by the standard normal 5 be the c.d.f. of the standard normal distribution. Now, take c1 and we have that E0 (X) = 50 limit theorem, the distribution of distribution. Let c2 φ such that 0.01 ≥ P0 (X ∈ / (c1 , c2 )) c1 − 50 X − 50 c2 − 50 = 1 − P0 ≤ ≤ 5 5 5 c2 − 50 c1 − 50 ∼1−φ +φ . 5 5 1 Probability and statistics not due We would like to make the rejection zone as biggest as we can, given that φ is symmetric, we just have to make c1 − 50 c2 − 50 =− , 5 5 ⇒c1 = 100 − c2 . Finally c2 − 50 5 0.01 ≥ 1 − φ c2 − 50 ≥ 0.995 ⇒φ 5 c2 − 50 ⇒ ≥ 2.5758 5 ⇒c2 ≥ 62.9, +1−φ c2 − 50 5 , then K1% = [0; 37] ∪ [63; 100]. Given that (b) 60 ∈ / K1% we cannot reject H0 with 0.01 level of signicance. We have to take our hypothesis H0 : p = p0 = 1 2 H1 : p > p0 , now we have to nd c such that 0.01 ≥ P0 (X ≥ c) ∼ 1 − φ Then we have to choose H0 (c) c − 50 5 . c ≥ 61.6, from what we have that K1% = [62; 100]. So we reject if we have 62 or more heads. Take p0 ∈ (0, 1) and H0 : p = p0 , H1 : p 6= p0 . Note that E0 (X) = 100p0 and V ar0 (X) = 100p0 (1 − p0 ). Then, take c1 and that 0.01 ≥ P0 (X ∈ / (c1 , c2 )) = 1 − P0 ∼1−φ c1 − 100p0 X − 100p0 c2 − 100p0 p ≤ p ≤ p 10 p0 (1 − p0 ) 10 p0 (1 − p0 ) 10 p0 (1 − p0 ) ! ! c1 − 100p0 c2 − 100p0 p p +φ . 10 p0 (1 − p0 ) 10 p0 (1 − p0 ) 2 ! c2 such Probability and statistics not due p a we have that c1 (p0 ) = 100p0 − 19.6 p0 (1 − p0 ) p c2 (p0 ) = 100p0 + 19.6 p0 (1 − p0 ). Then we are interested in the set n o p p p : 100p − 19.6 p(1 − p) ≤ 60 ≤ 100p + 19.6 p(1 − p) = p : (60 − 100p)2 ≤ 19.62 p(1 − p) =[0.502; 0.691]. Doing the same that in question Let δ be a test procedure and S1 the critical region of δ , Θ the parameter space. π(θ|δ), called the power function of the test δ , is determined by the relation π(θ|δ) = P(X ∈ S1 |θ), Suppose that mean µ X1 , · · · , Xn for and The function θ ∈ Θ. form a random sample from the normal distribution with unknown and known variance 1, and it is desired to test the following hypotheses: H0 : 0.1 ≤ µ ≤ 0.2 H1 : µ < 0.1 or µ > 0.2. H0 is rejected if either Xn ≤ c1 or Xn ≥ c2 , and let π(µ|δ) denote the power function of δ . Suppose that the sample size is n = 25. Determine the values of the constants c1 and c2 such that Consider a test procedure δ such that the hypothesis π(0.1|δ) = π(0.2|δ) = 0.07. Solution: Given n = 25, the mean of Xi is µ and variance is 1, Xn ∼ N(µ, σ 2 /n) = N(µ, 1/25). Thus 5(Xn − µ) ∼ N(0, 1). Let Y be a standard normal distributed random variable and Φ its c.d.f.: π(0.1|δ) = P(Xn ≤ c1 |µ = 0.1) + P(Xn ≥ c2 |µ = 0.1) = P(5Xn − 0.5 ≤ 5c1 − 0.5|µ = 0.1) + P(5Xn − 0.5 ≥ 5c2 − 0.5|µ = 0.1) = P(Y ≤ 5c1 − 0.5) + P(Y ≥ 5c2 − 0.5) = Φ(5c1 − 0.5) + (1 − Φ(5c2 − 0.5)). Similarly, π(0.2|δ) = Φ(5c1 − 1) + (1 − Φ(5c2 − 1)). We have to nd c1 and c2 such that Φ(5c2 − 0.5) − Φ(5c1 − 0.5) = 0.93 = Φ(y + 0.5) − Φ(x + 0.5) 3 (1) Probability and statistics not due and Φ(5c2 − 1) − Φ(5c1 − 1) = 0.93 = Φ(y) − Φ(x) where we set x = 5c1 − 1 It is not hard to see that and (2) y = 5c2 − 1. x + 0.5 = −y . In fact, taking the dierence of (1) and (2), one gets: Φ(y + 0.5) − Φ(y) = Φ(x + 0.5) − Φ(x). t 7→ Φ(t + 0.5) − Φ(t), any value can have at most two preimages, and the fact that 0.93 = Φ(y)−Φ(x) implies x 6= y . Hence x+0.5 = −y . By studying the monotonicity of the function Substitute it in (2): 0.93 = 1 − Φ(−y) − Φ(x) = 1 − Φ(x + 0.5) − Φ(x) 0.07 = Φ(x + 0.5) + Φ(x). It has a unique solution x = x0 (by monotonicity of −x0 − 0.5. Hence c1 = x0 /5 + 1/5 and c2 = −x0 /5 + 1/10. 4 x 7→ Φ(x + 0.5) + Φ(x)). And y =