Explanation Given: D = [ 0.2 0.4 0.4 ] ; P = [ 0.7 0.1 0.2 0.3 0.3 0.4 0.4 0.5 0.1 ] Explanation: The initial distribution, D , and the transition matrix, P , are provided. The distribution after the first experiment, D 1 , is the product of the initial distribution and the transition matrix. D 1 = [ 0.2 0.4 0.4 ] [ 0.7 0.1 0.2 0.3 0.3 0.4 0.4 0.5 0.1 ] = [ 0.42 0.34 0.24 ] The distribution after the two repetitions of P , D 2 , is the product of the distribution after the first occurrence of the experiment, D 1 , and the transition matrix. D 2 = [ 0.42 0.34 0.24 ] [ 0.7 0.1 0.2 0.3 0.3 0.4 0.4 0.5 0.1 ] = [ 0.492 0.264 0.244 ] The long-term distribution would correspond to the equilibrium vector. A vector v is said to be an equilibrium vector for a transition matrix P , if the following condition holds: v P = v Let the vector v be defined as follows, so that the multiplication holds. [ v 1 v 2 v 3 ] Then, v P = v [ v 1 v 2 v 3 ] [ 0.7 0.1 0.2 0.3 0.3 0.4 0.4 0.5 0.1 ] = [ v 1 v 2 v 3 ] Simplify the left-hand side to obtain, [ 0.7 v 1 + 0.3 v 2 + 0.4 v 3 0.1 v 1 + 0.3 v 2 + 0.5 v 3 0.2 v 1 + 0.4 v 2 + 0.1 v 3 ] = [ v 1 v 2 v 3 ] This gives rise to the following system of equations, 0.7 v 1 + 0.3 v 2 + 0.4 v 3 = v 1 0.1 v 1 + 0.3 v 2 + 0.5 v 3 = v 2 0.2 v 1 + 0.4 v 2 + 0.1 v 3 = v 3 This can be simplified to, − 0.3 v 1 + 0.3 v 2 + 0.4 v 3 = 0 0.1 v 1 − 0.7 v 2 + 0.5 v 3 = 0 0.2 v 1 + 0.4 v 2 − 0.9 v 3 = 0 The system involving these 4 equations can be solved with the help of Gauss Jordan method. The augmented matrix for the system is: [ − 0.3 0.3 0.4 0 0.1 − 0.7 0.5 0 0.2 0.4 − 0.9 0 1 1 1 1 ] − 1 0.3 R 1 → R 1 gives, [ 1 − 1 − 4 3 0 0

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ISBN: 9780321979438

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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(a) Test the hypothesis. Consider the hypothesis test Ho = : against H₁o < 02. Suppose that the sample sizes aren₁ = 7 and n₂ = 13 and that $² = 22.4 and $22 = 28.2. Use α = 0.05. Ho is not ✓ rejected. 9-9 IV (b) Find a 95% confidence interval on of 102. Round your answer to two decimal places (e.g. 98.76).

Let us suppose we have some article reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are 6, the sample mean force for lifting was 6.2 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.4 pounds with standard deviation 0.3 pounds. Assume that the standard deviations are known. Suppose that you wanted to detect a true difference in mean force of 0.25 pounds on the hands for these two activities. Under the null hypothesis, 40 = 0. What level of type II error would you recommend here? Round your answer to four decimal places (e.g. 98.7654). Use a = 0.05. β = i What sample size would be required? Assume the sample sizes are to be equal.…

= Consider the hypothesis test Ho: μ₁ = μ₂ against H₁ μ₁ μ2. Suppose that sample sizes are n₁ = 15 and n₂ = 15, that x1 = 4.7 and X2 = 7.8 and that s² = 4 and s² = 6.26. Assume that o and that the data are drawn from normal distributions. Use απ 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) for a true difference in means of 3? (c) Assuming equal sample sizes, what sample size should be used to obtain ẞ = 0.05 if the true difference in means is - 2? Assume that α = 0.05. (a) The null hypothesis is 98.7654). rejected. The P-value is 0.0008 (b) The power is 0.94 . Round your answer to four decimal places (e.g. Round your answer to two decimal places (e.g. 98.76). (c) n₁ = n2 = 1 . Round your answer to the nearest integer.

Chapter 10 Solutions

Finite Mathematics (11th Edition)

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