Exercise 14. Let X₁, X2, X3 be independent Bernoulli trials, i.e. the X; are i.i.d. random variableswith P(X₂ = 1) = p and P(X; = 0) = 1-p. We want to make inference about p, from observationsx1, x2, x3 = {0, 1}.-Assume we already know that either p = 0.3 or p = 0.8, and other values of p are not possible.In this case we want to decide whether the hypothesis Ho: p = 0.3 or the alternative H₁: p = 0.8is more likely. Consider the test which rejects Ho if and only if s := ³-₁ ; ≥ 2. For this test,determine the probabilities of type I and type II errors.

Given that We have to find probability of type 1 and type 2 error:

Answered: Exercise 14. Let X₁, X2, X3 be…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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6. A Bernoulli trial can be modeledd as a random variable X that can take on only two values: X=0 and X= 1, which are murnally exclusive. If P(X 1) 3 p. what is (a) PCX = 0) ? G Var(N) Earropy of N
Let x be a random variable that represents hemoglobin count (HC) in grams per 100 milliliters of whole blood. Then x has a distribution that is approximately normal, with population mean of about 14 for healthy adult women. Suppose that a female patient has taken 10 laboratory blood tests during the past year. The HC data sent to the patient's doctor are as follows. 15 17 17 20 15 12 13 16 15 12 State the null and alternate hypotheses. H0: ? = 14; H1: ? < 14 H0: ? > 14; H1: ? = 14 H0: ? < 14; H1: ? = 14 H0: ? = 14; H1: ? ≠ 14 H0: ? = 14; H1: ? > 14 What is the value of the sample test statistic? (Round your answer to three decimal places.)
b) Suppose that Y₁, Y₂,..., Y₁0 is a random sample from a bernoulli (P) distribution. It is desired that the null hypothesis Ho: P = 0.5 against the alternative hypothesis H₁: P = 0.1 at a level of significance. If Ho is rejected when Σı Y = 1. i) Determine the level of significance. ii) Find the power of the test.
2
The density function of X is given by f(x): and b. (a + bx² = 0 otherwise 0≤x≤1. If E[X]=, find a
7)
In Exercises 1-4, X₁, X₂, , Xn are i.i.d. experiments of a random variable X and X1, X2,,n are the sample outcomes, which we often abbreviate by the shorthand no- tation , and is the sample mean. Exercise 1 Let X~ Uniform([a, b]). Let â and 6 be the MLE estimators of a and b, respectively. (i) Show that and use this to show that ã = â E(ô - â) - = 1 1 ¹ ₁ (b − a), b = b + ₁₂ ² ₁ (b − a) – â), (ô n 1 n 1 - are unbiased estimators of a and b, respectively. (ii) Compute the MLE estimates a, b of a, b for the following sample data ☎¹: n 1 (b − a) n+ 1 - -2.45 0.65 -0.35 -2.38 -1.26 0.58 -1.52 -1.33 -0.42 -1.39 0.01 -1.87 -0.67 -0.34 -0.33 -1.45 0.02 (iii) Compute the unbiased estimates a, b of a, b for the same data above. ã,
6.STATISTICAL INFERENCEQuestion 3 a ii)
I dont understand what my professor is asking me to do here...
Suppose the Total Sum of Squares for a completely randomized design with 5 treatments and 30 total measurements (six per treatment) is equal to 500. In each of the following cases, conduct an F-test of the null hypothesis that the mean responses for the five treatments are the same. Use α = .10 b. SST is 50% of SS(Total)
3. The paper "Effects of Caffeine on Repeated Sprint Ability, Reactive Agility Time, Sleep and Next Day Performance"† describes an experiment in which male athlete volunteers who were considered low caffeine consumers were assigned at random to one of two experimental groups. Those assigned to the caffeine group drank a beverage which contained caffeine 1 hour before an exercise session. Those in the no-caffeine group drank a beverage that did not contain caffeine 1 hour before an exercise session. That night, participants wore a device that measures sleep activity. The researchers reported that there was no significant difference in mean sleep duration for the two experimental groups. In the context of this experiment, explain what it means to say that there is no significant difference in the group means. A) The difference in mean sleep duration for the caffeine and no-caffeine groups could have occurred by chance just due to the random assignment, assuming there is a real difference…
8. Assume we have two independent normal population random variables, X and Y. We 15 having have samples from each population with n = nx 81.31, j construct one-sided 95% confidence intervals for ux 9 and m = ny %3D 78.61 and their sample variances, s 60.76, s, 48.24. Then x =

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