Answer already given. Need detailed solution please. Let X1,...,X5 be a random sample from Uniform(0, theta) population. Consider testing null hypothesis Ho : theta = 1 against H1: 0 =/= 1. Construct a 0.05 level test based on X(5) If the critical region is given by C {(x_1,...,x_n): x(5) >= 0.9998}, find the significance level. Answer: Reject Ho iff X(5) >= 0.995 or X(5) <= 0.478; 0.0001

Answer already given. Need detailed solution please. Let X1,...,X5 be a random sample from Uniform(0, theta) population. Consider testing null hypothesis Ho : theta = 1 against H1: 0 =/= 1. Construct a 0.05 level test based on X(5) If the critical region is given by C {(x_1,...,x_n): x(5) >= 0.9998}, find the significance level. Answer: Reject Ho iff X(5) >= 0.995 or X(5) <= 0.478; 0.0001

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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Just question 3 fill in the blank
I am having problems on understanidng where to start with my homework. i know i am supposed to stat with seperating the left side of the eqution but i am not sure if i did it correctly. Could you please look over my work and see if i did the problem correctly?. I have attached the question and my work.
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You may need to use the appropriate appendix table or technology to answer this question. Given are five observations for two variables, x and y. (Round your answers to two decimal places.) xi 1 2 3 4 5 yi 4 6 6 11 14 (a) Use sŷ* = s 1 n + (x* − x)2 Σ(xi − x)2 to estimate the standard deviation of ŷ* when x = 4. (b) Use ŷ* ± tα/2sŷ* to develop a 95% confidence interval for the expected value of y when x = 4. to (c) Use spred = s 1 + 1 n + (x* − x)2 Σ(xi − x)2 to estimate the standard deviation of an individual value of y when x = 4. (d) Use ŷ* ± tα/2spred to develop a 95% prediction interval for y when x = 4. to
You may need to use the appropriate appendix table or technology to answer this question. Given are five observations for two variables, x and y. (Round your answers to two decimal places.) xi 1 2 3 4 5 yi 4 8 5 11 15 (a) Use sŷ* = s 1 n + (x* − x)2 Σ(xi − x)2 to estimate the standard deviation of ŷ* when x = 5. (b) Use ŷ* ± t?/2sŷ* to develop a 95% confidence interval for the expected value of y when x = 5. to (c) Use spred = s 1 + 1 n + (x* − x)2 Σ(xi − x)2 to estimate the standard deviation of an individual value of y when x = 5. (d) Use ŷ* ± t?/2spred to develop a 95% prediction interval for y when x = 5. to I need all the parts, I attached a picture of the whole question in case the one I copied above was not clear enough.
DEx2)-Ex)(Exy) |t= (X- X)// Examus - cdn.student.uae.examus.net/?rldbqn=1&sessi... STAT-101 FEX_2021_2_Male State whether the question is true or false 4 - 30 -> 118:50 362 The degree of freedom of t-test for independent samples (where of & o are unknown and not assumed to be equal) with sample sizes n1=37 and n 43 is equal to 79. ce362d6 2dcf91 F-E <µ < i+E where E = Za/2• e362d ef917 Formulae Sheet STAT-101-2 d.f. = min (n– 1,n2 – 1) p – E
there is a claim that people living near major roads have higher blood levels of lead. 30 women living near the main road were randomly selected and the mean blood lead level was calculated as x̄1=15 and s1=5.0. in addition, 25 more women who did not live around the main road were sampled and calculated as x̄2=8.9 and s2=3.9. does this data obtained support the above claim? (test at α=0.05 significance level.)
You may need to use the appropriate appendix table or technology to answer this question. Given are five observations for two variables, x and y. (Round your answers to two decimal places.) X; 1 2 4 5 6 11 14 (x* - x)2 E(x, - x)2 (a) Use s Sye = SV 1 + to estimate the standard deviation of y* when x = 4. (b) Use ý* + taso* to develop a 95% confidence interval for the expected value of y when x = 4. to (x* – x)² I(x, - x)2 (c) Use s. Spred 1 1 +- + to estimate the standard deviation of an individual value of y when x = 4. n (d) Use ý* ±t/25pred to develop a 95% prediction interval for y when x = 4. to