2.1Suppose that {x} is a stationary process with constant mean μ₁ = E(x₁) = μ and autocorrelationfunction Pk. Define x = ¹x, to be the sample mean of x₁,x₂,...‚¤‚n 1=1Prove that var(x)=%⁰nQUESTION 2n-11+2-2 (1-4) ₂ Pk where yo = var(x).k=1

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Answered: a stationary process with constant mean…

MATLAB: An Introduction with Applications

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Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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N= 150 observations were collected on a time series that was identified as a AR(2) time series. The following statistics were computed from the data. Mean - 45.0 Variance 15.6 Autocorrelation function (up to lag 5) r1 = 0.80, r2 = .50, rз = .26, r4 = -.10, rs = 0.08: == Estimate the parameters of the model using the method of moments.
A population has a mean µ = 161 and a standard deviation o = 20. Find the mean and standard deviation of the sampling distribution of sample means with sample size n = 56. The mean is = μx (Round to three decimal places as needed.) and the standard deviation is o- = X 3 X
A normal distribution has a mean of μμ= 54 and a standard deviation of σσ=6 What is the probability of randomly selecting a score GREATER than 51? A normal distribution has a mean of μμ= 54 and a standard deviation of σσ=6 What is the probability of selecting a sample of n = 4 scores with a mean more than M = 51? A normal distribution has a mean of μμ= 54 and a standard deviation of σσ=6 What is the probability of selecting a sample of n = 36 scores with a mean more than M = 51?
What is the variance and standard deviation for the number of times the SAT is taken? x Frequency (f) fx f*x2= (f*x) x (x) (4) = (3)x(1) 5152727-(2547473)(2547473)/1518859=880035.3688/1518859=0.5794 (1) (2) (3)=(1)x(2) (4)=(3x1) Population Variance 1 721,769 721,769 721,769 2 601,325 1,202,650 2,405,300 3 166,736 500,208 1,500,624 4 22,299 89,196 356,784 5 6,730 33,650 168,250 Total 1,518,859 2,547,473 5,152,727 880035.3688 0.5794 n=1,518,859 ∑fx = 2,547,473 The population variance is 0.5794 The Standard Deviation? 0.76 Can you show me the steps to reach the standard deviation of .76 ?
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A random sample of 41 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.09 years, with sample standard deviation s = 0.83 years. However, it is thought that the overall population mean age of coyotes is μ = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use α = 0.01. Solve the problem using the critical region method of testing (i.e., traditional method). (Round your answers to three decimal places.) test statitics= critical=
Vehicle speeds at a certain highway location are believed to have approximately a normal distribution with mean µ = 50 mph and standard deviation σ = 5 mph. The speeds for a randomly selected sample of n = 25 vehicles will be recorded. (a) Give numerical values for the mean and standard deviation of the sampling distribution of possible sample means for randomly selected samples of n = 25 from the population of vehicle speeds. Mean = s.d.(x) = (b) Use the Empirical Rule to find values that fill in the blanks in the following sentence. For a random sample of n = 25 vehicles, there is about a 95% chance that the mean vehicle speed in the sample will be between and mph. (c) Sample speeds for a random sample of 25 vehicles are measured at this location, and the sample mean is 59 mph. Given the answer to part (b), explain whether this result is consistent with the belief that the mean speed at this location is μ = 50 mph. A sample mean of 59 mph (when n = 25) --Select--- be consistent with…

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a stationary process with constant mean μ, = E(x₁) = μ and auto ==x, to be the sample mean of x₁, X2 ..., Xµµ n =l k r(x)=. Yo x) = ² = [1 + 2 Σ (1-4) 0₂ ] n k=1 Pk where % = var(x,).