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9090 contiden u in the peratage of a ntify th Wayn dnd Paul' sucessfully cordet word. Interpret the'interval! within the context 'of opplication. or To not need to comment' bn the valdet of the confidence interval. where

9090 contiden u in the peratage of a ntify th Wayn dnd Paul' sucessfully cordet word. Interpret the'interval! within the context 'of opplication. or To not need to comment' bn the valdet of the confidence interval. where

MATLAB: An Introduction with Applications
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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Confidence Interval Test Statistic Degrees of Freedom CHAPTER 6 Distribution Probability Density Function Variance (U-L)? Mean (1 1 Uniform fx) = U-L L+U n, +n, -2 L<X<U 1 2 12 Exponential fx) =elu n n, When; s s 3 区, X-H N(0,1) Normal: If X - N(µ,0²), then Z= (E-i)-D, Exponential: P(X<a) = 1-e", P(X>a) = e", P(a <X<b) = ew _ e bla min(ni-1, nz-1) Approximation: Vn, n, When > 3o < If X-B(n, p) and np > 5, and n(1-p) > 5 then binomial is approximately normal with u= np and o = np(1-p) CНАРТER 7 P(1- p) p-Po P.(1-P.)/n N/A X-u X-H- N(0,1). a Nn If X - N(H,0), then X- N(4,0/n) and Z=- If X follows a non-normal distribution with mean u and variance o and n230 , then X-u (1-P) P(1- -») (P,- P.) N/A. ( – Þ2)± Z«/2 n2 X - N(µ,0²/n) approximately and Z= - N(0,1). n, n, P-p n P(l- p), then Z= d-do to = If P- N(0,1). d+ ta/2m-1 n-1 p(l–p) Sa/Vn CHAPTERS 8-10 (Za/20 p(1-p) M. n, +n, – 2 n-1 X+x2 Test Statistic Degrees of Freedom Confidence Interval N/A o Vn n-1 In (E-)- D N/A Z, = V n, n, MacBook Air
Transcribed Image Text:Confidence Interval Test Statistic Degrees of Freedom CHAPTER 6 Distribution Probability Density Function Variance (U-L)? Mean (1 1 Uniform fx) = U-L L+U n, +n, -2 L<X<U 1 2 12 Exponential fx) =elu n n, When; s s 3 区, X-H N(0,1) Normal: If X - N(µ,0²), then Z= (E-i)-D, Exponential: P(X<a) = 1-e", P(X>a) = e", P(a <X<b) = ew _ e bla min(ni-1, nz-1) Approximation: Vn, n, When > 3o < If X-B(n, p) and np > 5, and n(1-p) > 5 then binomial is approximately normal with u= np and o = np(1-p) CНАРТER 7 P(1- p) p-Po P.(1-P.)/n N/A X-u X-H- N(0,1). a Nn If X - N(H,0), then X- N(4,0/n) and Z=- If X follows a non-normal distribution with mean u and variance o and n230 , then X-u (1-P) P(1- -») (P,- P.) N/A. ( – Þ2)± Z«/2 n2 X - N(µ,0²/n) approximately and Z= - N(0,1). n, n, P-p n P(l- p), then Z= d-do to = If P- N(0,1). d+ ta/2m-1 n-1 p(l–p) Sa/Vn CHAPTERS 8-10 (Za/20 p(1-p) M. n, +n, – 2 n-1 X+x2 Test Statistic Degrees of Freedom Confidence Interval N/A o Vn n-1 In (E-)- D N/A Z, = V n, n, MacBook Air
Question H has played ord le 90 times and sfullydontifued the corcect word Talayne, Sucessfully identifiethe correct word F5 tmes. They Played these game dndeğendently of each bfhes. Computsa 9070confidlenca interval tor the differenc where Wayn dnd faul' sudessfully cordet word. Interpretthe' interval within the context 'of opplication. You do not need to comment bn the Validit of the confidence interval:
Transcribed Image Text:Question H has played ord le 90 times and sfullydontifued the corcect word Talayne, Sucessfully identifiethe correct word F5 tmes. They Played these game dndeğendently of each bfhes. Computsa 9070confidlenca interval tor the differenc where Wayn dnd faul' sudessfully cordet word. Interpretthe' interval within the context 'of opplication. You do not need to comment bn the Validit of the confidence interval:
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