Suppose the random variable, X, follows an exponential distribution with mean 0 (0 ≤0 <∞). Let X₁, X2,..., Xn be a random sample of size n from the population of X.(a) Find the rejection region for a most powerful test of Ho: 0 = 0o versus H₁ : 0 =0 <00.(b) Is the above test uniformly most powerful? Explain your answer with argument.(c) Show (or argue) that the distribution of Σ₁1X₁ = ²nX is x²n (which can beused to find the constant associated with the rejection region of the above test).

In this problem, we are considering a random variable X that follows an exponential distribution…

Answered: Suppose the random variable, X, follows…

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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Let X equal the number of female children in a three-child family. We shall use a chi-square goodness-of-fit statistic to test the null hypothesis that the distribution of X is b(3, 0.5) (a) Define the test statistic and critical region, using an a = 0.05 significance level. (b) Among students who were taking statistics, 52 came from families with three children. For these families, x = 0, 1, 2, and 3 for 5, 17, 24, and 6 families, respectively. Calculate the value of the test statistic and state your conclusion, considering how the sample was selected.
Many people believe that the daily change in price of a company's stock in the stock market is a random variable with mean 0 and variance o² during the middle 2 months between 2 earning report dates when there is not much news released. That is, if Yn represents the closing price of the stock on the nth trading day of that period, then Yn = Yn-1+Xn, for 1 ≤ n ≤ 60, where X₁, X2,..., X60 are independent and identically distributed random variables with mean 0 and variance o². Suppose that the stock price at the beginning of the 2-month period is 100. If o = 2, what is the approximate probability that the stock's closing price on the 30th trading day will exceed 110, i.e., what is P(Y30 > 110)
Suppose the random variable, X, follows a geometric distribution with parameter 0 (0< 0 < 1). Let X₁, X2, X, be a random sample of size n from the population of X. (e) Show how the mle ofis related to its mme. Explain if it is always true. (f) Verify if X, is a sufficient statistic for or not. (g) Justify if = = is an unbiased estimator for 6.
For a Normal random variable X, with E(X)= 3, Var(X) = 4 Must use Standard Normal Distribuiton Table to compute the following (a) (6points) P(X4.82) (round your answer to 4 decimal places) 4
A normal random variable X is given with a mean = 22 and a standard deviation o = 17. Give an approximate value of the z such that: Pr(X ≤ 1) = 0.375 Give approximate values of an z and y such that: Pr(y ≤X ≤1) = 0.431. NOTE: For this problem, both z AND y must be FINITE so you can't just take y to -00 and solve the problem akin to what you did in (a). Your strategy here should be to find z and y such that Pr(X ≤1) - Pr(X ≤y) gets as close to 0.431 as you want. Read the Z table and play with the arithmetic!
Let X₁, X₂, ..., Xn be a random sample of size n from a geometric distribution for which p is the probabil- ity of success. (a) Use the method of moments to find a point estimate for p. (b) Explain intuitively why your estimate makes good sense.
An ordinary (fair) coin is tossed 3 times. Outcomes are thus triple of “heads” (h) and tails (t) which we write hth, ttt, etc. For each outcome, let R be the random variable counting the number of tails in each outcome. For example, if the outcome is hht, then R (hht)=1. Suppose that the random variable X is defined in terms of R as follows X=6R-2R^2-1. The values of X are given in the table below. A) Calculate the values of the probability distribution function of X, i.e. the function Px. First, fill in the first row with the values X. Then fill in the appropriate probability in the second row.
If a discrete random variable X has the following probability distribution: X -2, - 1, 0, 1, 2 P(X) 0.2, 0.3, 0.15, 0.2, 0.15 Use this to find the following: (a) The mean of X and E[X^2]. (b) The probability distribution for Y = 2X^2 + 2 (i.e, all values of Y and P(Y )). (c) Using part (b) (i.e, the probability distribution forY ), find E[Y ]. (d) Using part (a), verify your answer in part (c) for E[Y ]. **Note: Please do not just copy from Chegg!
Suppose the lifespan (in months) of a smartphone battery can be modeled as a continuous random variable with CDF F(x) = 1 − e-x/3 x ≥ 0 What is the probability that the battery lasts between 12 to 15 months?
Let X1, X2, X3, ..., X, be a random sample from a distribution with known variance Var(X,) = o², and unknown mean EX, = 0. Find a (1 – a) confidence interval for 0. Assume that n is large.
You obtain 336 successes in a sample of size n1=391n1=391 from the first population. You obtain 419 successes in a sample of size n2=472n2=472 from the second population. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.What is the test statistic for this sample
Suppose that a medical test has a 85% chance of detecting a disease if the person has it (P(PT|D)=0.85) and a 90% chance of correctly indicating that the disease is absent if the person really does not have the disease (P(NT|Dc=0.90). Suppose that 95% of the population does not have the disease P(Dc)=0.95 What is the probability that a randomly chosen person will test negative P(NT)?

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