8.2.1. Let X have the pmf f(x; 0) = 0² (1-0)¹-2, 2 = 0, 1, zero elsewhere. Wetest the simple hypothesis Ho: 0= against the alternative composite hypothesisH₁0

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Answered: 8.2.1. Let X have the pmf f(x; 0) = 0…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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3. An individual who has automobile insurance from a certain company is ran- domly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is 1 p(y) | .60 .25 .10 .05 What is the variance of Y? Α. 1.1 В. .60 C. .79 D. 1 E. None of the above 4. Based on the pmf in problem 3, what is the probability that an individual was cited between 1 and 3 times (inclusive)? A. .25 В. .05 С. .30 D. .40 E. None of the above
4.65 Let X1, X2, ..., Xn be a random sample from a N(u, 16) population, where u is an unknown parameter. Test Ho : u = 10 versus H1 : u+10 at a = 0.05 using the test statistic X. (a) Find the decision rule for the test. (b) Draw the power curves for n =9 and n= 100 on the same set of axes.
10. Let X1,..., X„ be an iid sample from f(x | 0) = 0x°-1, positive. Find the MLE and MOM estimators for 0. 0 < I < 1, where the parameter 0 is
1. In each instance indicate whether a Type I error, a Type II error, or no error was committed by the researcher. Note that Ho: µ1 − µ2 = 0 is the standard null hypothesis (i.e., no mean group difference) for a two-group t-test, although other versions of Ho are possible such as Ho: µ1 − µ2 = -.5 (i.e., the mean difference does not differ from -0.5). H0 H1 true value. Researcher's Decision Type of Error? (a) µ1 − µ2 = 0 µ1 − µ2 ≠ 0 2.33 reject H0 _______________ (b) µ1 − µ2 = 0 µ1 − µ2 ≠ 0 -1.27 fail to reject H0. ________________ (c) µ1 − µ2 = -.5. µ1 − µ2 ≠ -.5 -.5 fail to reject H0 ________________
7.4.15-T Question Help v A humane society claims that less than 73% of households in a certain country own a pet. In a random sample of 600 households in that country, 420 say they own-a pet. At a = 0.10, is there enough evidence to support the society's claim? Complete parts (a) through (c) below. (a) Identify the claim and state Ho and H Identify the claim in this scenario. Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a decimal. Do not round.) O A. More than % of households in the country own a pet. O B. Less than % of households in the country own a pet. O C. The percentage households in the country that own a pet is not %. O D. % of households in the country own a pet. Click to select and enter your answer(s) and then click Check Answer. 4 parts remaining Clear All Check Answer 1:06 AM P Type here to search 2/21/2021 on fs 10 144 esc @ %23 %24 tbacksoace 3. 6. 7. 8. 9. W R Y tab H. K
Question 5.3 Y, Y2, . Yn are n iid observations of a continuous random variable named Y One observation will be the test statistic in the following hypothesis test: Hạ: Y-Gamma(a = 5,8 = 4) H: Y-EXP(B = 20) H, will be rejected iff Y > k, where k is a known constant. 5.3.a Set up (but do not attempt to evaluate) a definite integral that would find the probability of Type 1 error. 5.3.b Set up (but do not attempt to evaluate) a definite integral that would find the probability of Type 2 error.
4.54 Figure 4.16 shows a randomization distribution for testing Ho 1=2 versus Ha statistic used for each sample is D = x, – x2. In each case, use the distribution to decide which value is closer to the p-value for the observed difference in : H1 # µz. The sample means. (a) The p-value for D = x - I = -2.9 is closest to: %3D 0.01 or 0.25? (b) The p-value for D = x, – x, = 1.2 is closest to: %3D 0.30 or 0.60? 90 80 70 60 50 40 30 20 10 무 -3 -2 -1 1 2 3 Differences Figure 4.16 Randomization distribution for Exercises 4.54 to 4.58 Frequency

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8.2.1. Let X have the pmf f(x; 0) = 0 (1 - 0) ¹-2, x = 0, 1, zero elsewhere. We test the simple hypothesis Ho: 0= against the alternative composite hypothesis H₁0