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(a) Find the probability mass function of (X₁ + X₂). (b) Using the result of (a), guess what the probability mass function of X, will be and i=1 substantiate your guess by using the method of induction. In (c), (d), (e), and (f), consider the problem of testing at a = 0.038406. Ho p= 0.5 versus Ha p = 0.2 : n (c) Write down the likelihood function L(p; x), where x = (x1,x2,. .., n) is the data sample realized from the random sample X = (X1, X2,..., Xn). points. (d) Write down the expression for the Neyman-Pearson test statistic T(x) and the corresponding rejection region T as outlined in the posted notes on Neyman-Pearson Lemma. (e) When n = 5, obtain a description of the rejection region T that does not depend on the value 0.2 of the parameter p specified under Ha. (f) When n = 5, find the value of 3 up to six places of decimals.

(a) Find the probability mass function of (X₁ + X₂). (b) Using the result of (a), guess what the probability mass function of X, will be and i=1 substantiate your guess by using the method of induction. In (c), (d), (e), and (f), consider the problem of testing at a = 0.038406. Ho p= 0.5 versus Ha p = 0.2 : n (c) Write down the likelihood function L(p; x), where x = (x1,x2,. .., n) is the data sample realized from the random sample X = (X1, X2,..., Xn). points. (d) Write down the expression for the Neyman-Pearson test statistic T(x) and the corresponding rejection region T as outlined in the posted notes on Neyman-Pearson Lemma. (e) When n = 5, obtain a description of the rejection region T that does not depend on the value 0.2 of the parameter p specified under Ha. (f) When n = 5, find the value of 3 up to six places of decimals.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.CR: Chapter 13 Review
Problem 6CR
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(a) Find the probability mass function of (X₁ + X₂). n (b) Using the result of (a), guess what the probability mass function of EX, will be and i=1 substantiate your guess by using the method of induction. In (c), (d), (e), and (f), consider the problem of testing at a 0.038406. Ho p= 0.5 versus Ha: p = 0.2 (c) Write down the likelihood function L(p; x), where x = (x1, x2, sample realized from the random sample X = (X₁, X2,..., Xn). points. (d) Write down the expression for the Neyman-Pearson test statistic T(x) and the corresponding rejection region T as outlined in the posted notes on Neyman-Pearson Lemma. (g) When n (e) When n = 5, obtain a description of the rejection region T that does not depend on the value 0.2 of the parameter p specified under Ha. (f) When n = 5, find the value of 3 up to six places of decimals. ,xn) is the data = = 5, use the decision rule (T, T) you formulated in (e) to test Ho: p= 0.5 versus Ho: P < 0.5 at a = 0.038406. Calculate your Type II Error probability at p = 0.16 up to six places of decimals. (h) If I use an arbitrary decision rule (S, S) to test Ho p = 0.5 versus Ho: p < 0.5 : at a ≤ 0.038406 with n = 5, how will your Type II Error probability at p = 0.16 calculated in (g) compare with that of mine? Explain.
Transcribed Image Text:(a) Find the probability mass function of (X₁ + X₂). n (b) Using the result of (a), guess what the probability mass function of EX, will be and i=1 substantiate your guess by using the method of induction. In (c), (d), (e), and (f), consider the problem of testing at a 0.038406. Ho p= 0.5 versus Ha: p = 0.2 (c) Write down the likelihood function L(p; x), where x = (x1, x2, sample realized from the random sample X = (X₁, X2,..., Xn). points. (d) Write down the expression for the Neyman-Pearson test statistic T(x) and the corresponding rejection region T as outlined in the posted notes on Neyman-Pearson Lemma. (g) When n (e) When n = 5, obtain a description of the rejection region T that does not depend on the value 0.2 of the parameter p specified under Ha. (f) When n = 5, find the value of 3 up to six places of decimals. ,xn) is the data = = 5, use the decision rule (T, T) you formulated in (e) to test Ho: p= 0.5 versus Ho: P < 0.5 at a = 0.038406. Calculate your Type II Error probability at p = 0.16 up to six places of decimals. (h) If I use an arbitrary decision rule (S, S) to test Ho p = 0.5 versus Ho: p < 0.5 : at a ≤ 0.038406 with n = 5, how will your Type II Error probability at p = 0.16 calculated in (g) compare with that of mine? Explain.
This problem explores the question of testing Ho p= 0.5 versus Ho p < 0.5, . based on a random sample X₁, X2, Xn from the Geometric (p) distribution. Recall that there is an ambiguity surrounding the definition of the Geometric (p) distribution, so let us unequivocally specify which distribution we are going to work with. In the experiment of repeatedly tossing a coin with the probability of getting a Head in a single toss of the coin equal to p = (0, 1) till a Head is obtained, let X denote the number of Tails obtained before the experiment is terminated. Then X has the Geometric (p) distribution. Further recall that X₁, X2,, Xn is a random sample of size n from the Geometric (p) distribution means X (X₁, X2, Xn) is a collection of mutually independent random variables, with each X; having the Geometric (p) distribution. Before proceeding further, let us recall the following definition. Definition A collection of random variables {Y₁, Y2,..., Yr} defined on the same sample space, with S, denoting the support of Y; for 1≤ i ≤r, is said to be mutually independent if, V (S1, S2, , Sr) EX S, the Cartesian product of (S₁, S2, ..., Sr), 1 = 1 are independent. r P(‚ñ‚ (Y₁ = $i)) = [[P(Y₁ = 81). 1=1 Remember that if {Y₁, Y2,, Yr} is a collection of mutually independent random variables, then for any 1 ≤ m ≤r, the random variables U = g(Y₁, Y2₂, •, Ym) and V = h(Ym+1, Ym+2,, Yr)
Transcribed Image Text:This problem explores the question of testing Ho p= 0.5 versus Ho p < 0.5, . based on a random sample X₁, X2, Xn from the Geometric (p) distribution. Recall that there is an ambiguity surrounding the definition of the Geometric (p) distribution, so let us unequivocally specify which distribution we are going to work with. In the experiment of repeatedly tossing a coin with the probability of getting a Head in a single toss of the coin equal to p = (0, 1) till a Head is obtained, let X denote the number of Tails obtained before the experiment is terminated. Then X has the Geometric (p) distribution. Further recall that X₁, X2,, Xn is a random sample of size n from the Geometric (p) distribution means X (X₁, X2, Xn) is a collection of mutually independent random variables, with each X; having the Geometric (p) distribution. Before proceeding further, let us recall the following definition. Definition A collection of random variables {Y₁, Y2,..., Yr} defined on the same sample space, with S, denoting the support of Y; for 1≤ i ≤r, is said to be mutually independent if, V (S1, S2, , Sr) EX S, the Cartesian product of (S₁, S2, ..., Sr), 1 = 1 are independent. r P(‚ñ‚ (Y₁ = $i)) = [[P(Y₁ = 81). 1=1 Remember that if {Y₁, Y2,, Yr} is a collection of mutually independent random variables, then for any 1 ≤ m ≤r, the random variables U = g(Y₁, Y2₂, •, Ym) and V = h(Ym+1, Ym+2,, Yr)
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