Construct an estimator which improves on A (you do not need to have a closedform for it, but you do need to prove that it is an estimator). Consider a random sample X1,..., X, with n > 3 from the exponential distributionwith parameter 1 > 0 and densityf(x; A) = e-=/A,x > 0.Consider the following two estimators of A:3EXi, Ã =nX(1).i=1

Given that X1,X2,......Xn with n>3 from the exponential distribution with parameter λ>0 and…

Answered: Consider a random sample X1,..., Xn…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter4: Equations Of Linear Functions

Section4.5: Correlation And Causation

Problem 18HP

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Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?
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Find the equation of the regression line for the following data set. x 1 2 3 y 0 3 4
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A coffee shop owner offers two brands of coffee, Brand "A" and Brand "B". The proportion of clients that prefer Brand "A" is 60%. Recently, the price of Brand "A" went up. The proportion of clients that buy brand "A" among the first 15 customers that enter the shop after the change in price is an estimator of the current proportion of clients that prefer Brand "A". Assume that the change in price has no effect on the preference of the clients. Then the mean squared error (MSE) of the estimator is equal to: (Give an answer of the form x.xxx, i.e. with 3 significance digits) (HINT: Carefully read the definition of MSE in Chapter 10 of the text book.)
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If the data is weight, the z-score for someone who is overweight would be what?? if the data is time spent watching TV, an individual with a z-score of zero would do what?? if the data is an annual salary in the US, and the population is all legally employed people in the US, the z-Score of people who make minimum wage would be what??
A researcher is interested in studying how the number of books parents read to their children can influence a child's reading skills in kindergarten. She randomly assigns families with 5 year old children to either read their child 3 books per day, 2 book per day, or 1 book per day for the first 2 months of kindergarten. They then measure the children's reading ability. How could the confound be fixed for this scenario? Be sure to tell me what technique you are using (constancy, repeated measures, randomization, elimination, or balancing), as well as how you would apply that technique to this specific scenario, and how that would fix this confound.
Please explain further the solution. Thank you!
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Consider a random sample X1,..., Xn with n> 3 from the exponential distribution with parameter > 0 and density f(x;A) = e-=/A, z>0. Consider the following two estimators of A: 3 X;, Ã =nX(1). 3 i=1