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Let X be a random variable of continuous type with probability density function
Based on single observation X, the most powerful; test of size α = 0.1, for testing H0 : θ = 1 against H1 : q = 2, rejects H0 if X < k. Then the value of k is
  • a)
    1
  • b)
    10/3
  • c)
    11/3
  • d)
    4
Correct answer is option 'B'. Can you explain this answer?
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Let X be a random variable of continuous type with probability density functionBased on single observation X, the most powerful; test of size α = 0.1, for testing H0 : θ = 1 against H1 : q = 2, rejects H0 if X < k. Then the value of k isa)1b)10/3c)11/3d)4Correct answer is option 'B'. Can you explain this answer?
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Let X be a random variable of continuous type with probability density functionBased on single observation X, the most powerful; test of size α = 0.1, for testing H0 : θ = 1 against H1 : q = 2, rejects H0 if X < k. Then the value of k isa)1b)10/3c)11/3d)4Correct answer is option 'B'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Let X be a random variable of continuous type with probability density functionBased on single observation X, the most powerful; test of size α = 0.1, for testing H0 : θ = 1 against H1 : q = 2, rejects H0 if X < k. Then the value of k isa)1b)10/3c)11/3d)4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let X be a random variable of continuous type with probability density functionBased on single observation X, the most powerful; test of size α = 0.1, for testing H0 : θ = 1 against H1 : q = 2, rejects H0 if X < k. Then the value of k isa)1b)10/3c)11/3d)4Correct answer is option 'B'. Can you explain this answer?.
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