7.4.9. Let X₁,..., Xn be iid with pdf f(x; 0) = 1/(30), 0 < x < 20, zero else-where, where @ > 0.(a) Find the mle 6 of 0.(b) Is e a sufficient statistic for 0? Why?(c) Is (n+1)/n the unique MVUE of 0? Why?

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Answered: (a) Find the mle 0 of 0. (b) Is a…

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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Find a recent journal article(peer reviewed scientific publication) in a field of interest to you which contains references to p-values. I will help you find an article if you want, tell me what you’re interested in. How large is the p-value? Do you think the results of this study are strong? How might they be stronger? Is there anything the authors should have included?
Suppose you wish to test the hypothesis H0:P=0.9 versus H1:P≠0.9 based on a random sample of size 100 and number of successes 70. Find the value of the test statistic. (Note: use the continuity correction).
The degrees of freedom is critical to computing the p-value for the chi-square goodness of fit. We will discuss the p-value in the next lab. The degrees of freedom for the Chi-Square goodness of fit is equal to (the number of non-zero rows in the table) - 1. The number of non-zero rows is the number of expected values that are non-zero. For example: Phenotype ... Expected ... Disease ... 234 ... Wild-type ... 6743 ... has two rows where the Expected value is non-zero. So the degrees of freedom is 2-1 = 1. Another example: Phenotype ... Expected ... Disease ... 2374 ... Wild-type ... 0 ... has one row where the Expected value is non-zero. So the degrees of freedom is 1-1 = 0. With 0 degrees of freedom, the chi-square test cannot be calculated. A third example (sex-linked): Phenotype ... Expected ... Disease-Male ... 2374 ... Disease-Female ... 9456 Wild-type-Male ... 1001 ... Wild-type-Female ... 235 ... has four rows where the Expected values are…
Stoaches are fictional creatures, sometimes mistaken for mome raths. Suppose a two-tailed t-test is conducted to test for a difference between the population means of adult female stoach weights and adult male stoach weights. The female sample has 13 weights, and the male sample has 7 weights. The t-statistic for the test is -0.20542. What is the p-value? (Give your answer to 4 decimal places.) p-value:
VI- The Decision which supports economist claim, for the test of hypothesis, at the level α=0.50 and the z statistic is (- 1.69 ), to test the hypothesis, H0⋅μ0=μ1againstHA⋅μ0notequalμ1 is ………....………………..
What is the critical t score if you are testing a one-tailed hypothesis, α level of .05 with df = 36? a. 1.671 b. 2.026 c. 2.028 d. 1.697
You are performing a left-tailed test with test statistic z=−2.04, find the p-value to 4 decimal places.Note: You will need to use the z and t sheet to get this answer.
J Suppose that Shoe Sices Of American women have a bell Shaped elu standard with a mean of 8,37 and a eleviation of 1.49, using the empirical Rule what percentage of american women have Shoe Sizes that are no more than 9.su? do not Roundl your answer
10.
You are conducting a multinomial hypothesis test for the claim that the 4 categories occur with the following frequencies:Ho : pA=0.2 pB=0.3; pC=0.1; pD=0.4Complete the table. Category ObservedFrequency ExpectedFrequency SquaredResidual A 38 B 41 C 10 D 57 What is the chi-square test-statistic for this data?χ2=Report all answers accurate to three decimal places.VERY IMPORTANT NOTE: Answers are reported to 3 decimal places, but you should use all the decimal places to calculate the final chi-square test statistic...DO NOT use the rounded number you report in the table.
"On average, which value is expected for the F-ratio if the null hypothesis is true?" F = 0 F = k 1 F = N k F = 1

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(a) Find the mle 0 of 0. (b) Is a sufficient statistic for 0? Why?