In linear models the Variance-Covariance matrix plays key role when estimating the parameters.State the variance-covariance structure of Population-averaged models: ( generalized estimatingequations-GEE) for B and Subject-specific models: Likelihood methods based on yi - MVN (Xy,V)to estimate y and parameters in V,Show that for OLS Y = X'B + ɛ the estimator B can be derived to B = (X'X)¯'x'Y with%3DVar(ß) = (X'X)-'o?

In linear models, the variance-covariance matrix plays a key role when estimating the parameters.…

Answered: In linear models the…

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter1: Functions

Section1.2: The Least Square Line

Problem 8E

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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?
The concentration of dissolved solids and the turbidity of stream are measured simultaneously for five separately days selected days selected at random throughout a year. the data are as follows Do mg/L 400 550 700 800 500 Turbidity JTU 5 30 32 58 20 since turbidity is easier to measure, a regression equation may be used to predict the concentration of dissolved solids on the basis of known turbidity. Assume that the variance of dissolved solid concentrations is constant with turbidity.
Show the best fitted line on scatter diagram and Find the predicted value for each y using the exposure time and the equation obtained in part b (b. Find the equation of regression line between radiation doses on exposure time .usingleast square method)
Two regression lines of a sample are X+6 Y=6 and 3X+2Y=0. Find the correlation coefficient.
A regression and correlation analysis resulted in the following information regarding a dependent variable (y) and an independent variable (x). Σx = 90 Σ(y - )(x - ) = 466 Σy = 170 Σ(x - )2 = 234 n = 10 Σ(y - )2 = 1434 SSE = 505.98 The least squares estimate of the slope or b1 equals a. .923. b. 1.991. c. -1.991. d. -.923.
Interpret the least squares regression line of this data set. Meteorologists in a seaside town wanted to understand how their annual rainfall is affected by the temperature of coastal waters. For the past few years, they monitored the average temperature of coastal waters (in Celsius), x, as well as the annual rainfall (in millimetres), y. Rainfall statistics • The mean of the x-values is 11.503. • The mean of the y-values is 366.637. • The sample standard deviation of the x-values is 4.900. • The sample standard deviation of the y-values is 44.387. • The correlation coefficient of the data set is 0.896. The correct least squares regression line for the data set is: y = 8.116x + 273.273 Use it to complete the following sentence: The least squares regression line predicts an additional annual rainfall if the average temperature of coastal waters increases by one degree millimetres of Celsius.
Calculate the equation of the regression line and calculate the correlation coefficient
Use the least squares regression line of this data set to predict a value. Meteorologists in a seaside town wanted to understand how their annual rainfall is affected by the temperature of coastal waters. For the past few years, they monitored the average temperature of coastal waters (in Celsius), x, as well as the annual rainfall (in millimetres), y. Rainfall statistics • The mean of the x-values is 11.503. • The mean of the y-values is 366.637. • The sample standard deviation of the x-values is 4.900. • The sample standard deviation of the y-values is 44.387. • The correlation coefficient of the data set is 0.896. The least squares regression line of this data set is: y = 8.116x + 273.273 How much rainfall does this line predict in a year if the average temperature of coastal waters is 15 degrees Celsius? Round your answer to the nearest integer. millimetres
BUS – 173 (Applied Statistics) # Determining if the following statement are True and False. If False, write the correct statement. Also Solve the MCQ 01. The possible range of coefficient of determination (r^2) is 0 to 1 (a) True (b) False 02. The possible range of coefficient of correlation (r) is -1 to +1 (a) True (b) False 03. In the simple linear regression model, b represents the (a) least square estimate of A (b) least square estimate of B (c) Random error term (d) Not given

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