Consider the regression modely = a + ßx+u,where x is an endogenous variable.a) Write x₁ = x + ê; and show that = x.b) Suppose there are k valid instruments: Z1, Z2, ..., Zk.Let x₁= πo+f1zil + π2Zi2+...+îkzik. Show that the 2SLS estimator is equivalentto an IV estimator using as an instrument:Σ₁₁(&i- Ñ)(vi- V) _ Σ₁(Ri—Ñ)(Yi-Y)Ei=1(xi-x)² 21(®i- Đ)(xi-x)=[Hint: OLS first order conditions imply №₁ ê¡Â¡ = 0 and Σ¼_₁ ê¡= 0.]c) Suppose there is only one instrument z and ✰; = fo+ fizi. Show that the 2SLSestimator is numerically identical to the IV estimator:=Σ₁ (Îi—Ñ)(vi-V) – Σ₁(²₁-7)(vi-V)Σ=1(ât− x)2 Σ₁ (Z₁-7)(xi-x)*n[Hint: Use the result in (b) and plug in â; and ☎ = ñîo+â¦Z.]

Answered: Image /qna-images/answer/f3298933-9260-4d47-9456-2f6a63589a48.jpg

Answered: Consider the regression model y = a +…

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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ANSWER THE FOLLOWING QUESTION.
Heights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from heights of 133 to 188 cm and weights of 40 to 150 kg. Let the predictor variable x be the first variable given. The 100 paired measurements yield x = 167.54 cm, y = 81.35 kg, r=0.186, P-value = 0.064, and y = - 109 + 1.12x. Find the best predicted value of ŷ (weight) given an adult male who is 180 cm tall. Use a 0.10 significance level. The best predicted value of y for an adult male who is 180 cm tall is (Round to two decimal places as needed.) kg.
Heights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from heights of 138 to 188 cm and weights of 40 to 150 kg. Let the predictor variable x be the first variable given. The 100 paired measurements yield x = 167.61 cm, y = 81.52 kg, r=0.271, P-value=0.006, and y = -103 +1.18x. Find the best predicted value of ŷ (weight) given an adult male who is 155 cm tall. Use a 0.10 significance level. The best predicted value of y for an adult male who is 155 cm tall is (Round to two decimal places as needed.) kg.
find the least squares regression line equation from the given data.
Heights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from heights of 132 to 193 cm and weights of 39 to 150 kg. Let the predictor variable x be the first variable given. The 100 paired measurements yield x = 167.59 cm, y = 81.52 kg, r= 0.416, P-value = 0.000, and y = - 102 + 1.13x. Find the best predicted value of y (weight) given an adult male who is 147 cm tall. Use a 0.05 significance level. The best predicted value of y for an adult male who is 147 cm tall is kg. (Round to two decimal places as needed.)
Heights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from heights of 137 to 189 cm and weights of 37 to 150 kg. Let the predictor variable x be the first variable given. The 100 paired measurements yield x = 167.50 cm, y =81.41 kg, r=0.232, P-value = 0.020, and y = - 109 + 1.17x. Find the best predicted value of y (weight) given an adult male who is 145 cm tall. Use a 0.01 significance level. The best predicted value of y for an adult male who is 145 cm tall is kg. (Round to two decimal places as needed.)
Consider the following population model for household consumption: cons = a + b1 * inc+ b2 * educ+ b3 * hhsize + u where cons is consumption, inc is income, educ is the education level of household head, hhsize is the size of a household. Suppose a researcher estimates the model and gets the predicted value, cons_hat, and then runs a regression of cons_hat on educ, inc, and hhsize. Which of the following choice is correct and please explain why. A) be certain that R^2 = 1 B) be certain that R^2 = 0 C) be certain that R^2 is less than 1 but greater than 0. D) not be certain
Show calculations or explanation for each question. a) Which of the following techniques is used to predict the value of one variable on thebasis of other variables?a. Correlation analysisb. Coefficient of correlationc. Covarianced. Regression analysis b) In the least squares regression line, y^=3-2x the predicted value of y equals:a. 1.0 when x = −1.0b. 2.0 when x = 1.0c. 2.0 when x = −1.0d. 1.0 when x = 1.0 c) In the simple linear regression model, the y-intercept represents the:a. change in y per unit change in x.b. change in x per unit change in y.c. value of y when x = 0.d. value of x when y = 0.
)A county real estate appraiser wants to develop a statistical model to predict the appraised value of 3) houses in a section of the county called East Meadow. One of the many variables thought to be an important predictor of appraised value is the number of rooms in the house. Consequently, the appraiser decided to fit the simple linear regression model: E(u) = Bo + Bix, where y = appraised value of the house (in thousands of dollars) and x = number of rooms. Using data collected for a sample of n = 73 houses in Fast Meadow, the following results were obtained: y = 73.80 + 19.72x What are the properties of the least squares line, y = 73.80 + 19.72x? A) Average error of prediction is 0, and SSE is minimum. B) It will always be a statistically useful predictor of y. C) It is normal, mean 0, constant variance, and independent. D) All 73 of the sample y-values fall on the line.
A year-long fitness center study sought to determine if there is a relationship between the amount of muscle mass gained y(kilograms) and the weekly time spent working out under the guidance of a trainer x(minutes). The resulting least-squares regression line for the study is y=2.04 + 0.12x A) predictions using this equation will be fairly good since about 95% of the variation in muscle mass can be explained by the linear relationship with time spent working out. B)Predictions using this equation will be faily good since about 90.25% of the variation in muscle mass can be explained by the linear relationship with time spent working out C)Predictions using this equation will be fairly poor since only about 95% of the variation in muscle mass can be explained by the linear relationship with time spent working out D) Predictions using this equation will be fairly poor since only about 90.25% of the variation in muscle mass can be explained by the linear relationship with time spent…

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