Consider the following generic fixed effects model, along with a two-period (t = 1, 2), randomly sampled panel data set with dependent variable y andindependent variable x:Yit = Bo + 5o d24 + B1rit + ai +UitwhereYit = value of y for individual i at timetd2; = binary variable equal to 1 in the second time period (t = 2), and o otherwise (t = 1)Cit = value of a for individual i at time ta; = unobserved (time-invariant) effectUj = idiosyncratic errorTaking the first difference of the model (that is, subtracting the regression equation for t=1 from the regression equation for t=2) yields the followingfirst-differenced equation:Note: Let A denote the change from t=1 to t = 2.Ay; = Yi2 – Yi1=SoAu;You now plan to use OLS to estimate your first-differenced equation, in order to obtain the first-differenced estimator. Suppose that ait is a binaryvariable equal to 1 if the individual was born male, and o otherwise.True or False: The first-differenced estimator cannot be estimated in this case.O TrueO False

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Answered: Consider the following generic fixed…

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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A student at a junior college conducted a survey of 20 randomly selected full-time students to determine the relation between the number of hours of video game playing each week, x, and grade-point average, y. She found that a linear relation exists between the two variables. The least-squares regression line that describes this relation is y = -0.0552x +2.9446. (a) Predict the grade-point average of a student who plays video games 8 hours per week. The predicted grade-point average is 2.50 (Round to the nearest hundredth as needed.) (b) Interpret the slope. For each additional hour that a student spends playing video games in a week, the grade-point average will by points, on average. increase decrease Enter your answer in the answer box and then click Check Answer. parts remaining Clear All Check Answer P Type here to search TOUC Panasonic CF-54 ED 10 F8 #5 F6 F3 8. 2 FOA F1 Es & 7 0 8 8 99 2 A2 10 161 1 A1 Y 15 + 1/ ID 4圣 %U4 12
rou have a dummy variable and an interaction term in the same regression model, the coefficient estimate of the in riable allows for changes in the and the coefficient estimate of the interaction term allows for changes in t O Error term; slope O Slope; intercept OSlope; error term O Intercept; slope
An SRS of apartment listings in a large northeastern city comparing monthly rent ($) versus size (〖ft〗^2) yields following computer output:
. A manufacturing facility uses the following model to estimate the weekly number of data chips that can be produced where: Z = number of chips produced X = number of Manufacturing Specialists Y = number of Quality Control Specialists = f(x, y) = 14,000x+400y+22x²y-13y³
A study was conducted to assess the relationship between students’s score in final exam (y) and number of hours spent for exam (x) in each day. Data on a random sample 20 students were obtained and a regression model was estimated; and the least squares estimates obtained are: intercept a=28.5 and slope b=4.3 with SE(b)=Sb=0.017. The SS are: TSS=2540 and ESS=850. ****** QA) What is the difference between exam score obtained by two students one who studied 5 hours and the other who studied 9 hours per day. QB) In the above Question 1, find 95% CI for the slope and interpret it. In the above Question 1, find and interpret the coefficient of determination (r-square value).
Suppose that researchers obtain a random sample of adults ages 18 – 40 and collect data on the following variables: shoe size – in inches age – in years height – in inches forearm length – in inches Suppose further that a multiple linear regression model is fit to the resulting data set using R Studio and that the following output is obtained from it. Use this output to answer the question that follows: > summary(lm(shoesize ~ age + height + forearm, data = measures)) Coefficients: (Intercept)ageheightforearm Estimate10.14882 0.06045 -0.02108 -0.06479 Std. Error 4.49245 0.06838 0.06350 0.06847 t value2.259 0.884 -0.332 -0.946 Pr(>|t|) 0.0264 0.3792 0.7408 0.3467 Residual standard error: 1.719 on 85 degrees of freedomMultiple R-squared: 0.01983, Adjusted R-squared: -0.01477 F-statistic: 0.5731 on 3 and 85 DF, p-value: 0.6342 Using the information from above, fill in the blanks for the least-squares regression equation. Input all values to 5…
Suppose that researchers obtain a random sample of adults ages 18 – 40 and collect data on the following variables: shoe size – in inches age – in years height – in inches forearm length – in inches Suppose further that a multiple linear regression model is fit to the resulting data set using R Studio and that the following output is obtained from it. Use this output to answer the question that follows: > summary(lm(shoesize ~ age + height + forearm, data = measures)) Coefficients: (Intercept)ageheightforearm Estimate10.14882 0.06045 -0.02108 -0.06479 Std. Error 4.49245 0.06838 0.06350 0.06847 t value2.259 0.884 -0.332 -0.946 Pr(>|t|) 0.0264 0.3792 0.7408 0.3467 Residual standard error: 1.719 on 85 degrees of freedomMultiple R-squared: 0.01983, Adjusted R-squared: 0.01477 F-statistic: 0.5731 on 3 and 85 DF, p-value: 0.6342 What is the estimate for the standard deviation of the residuals? 1.719 0.01983 -0.946 0.6342
A researcher collected data on ( X, Y ), and wanted to test if the true relationship is linear. Therefore, she collected replicated values of X. Here is a summary of the data, and some results: Xi No. of replicates yi SSE = 312. 1 3 9 10 5 11 8. 3 2 12 14 4 4. 15 13 3 9 6 6 5 11 10 7 1 8 4 8 3 7 5 At significance level a = 0.05, carry out the test of Lack-of-Fit for linearity of the model. No need to show an ANOVA Table. Calculations: Test statistic: T = Critical value = Conclusion: Since we 2.
Suppose we conduct a study on the time spent studying for an exam and exam scores. We survey students in introductory statistics and find the explanatory variable (time, in hours) ranges from 0 to 16 and exam scores (in points, out of 100 maximum) range from 5 to 100. The correlation between time spent studying and exam scores is 0.80. We fit a simple linear regression model to these data and get the following least squares line: score = 4 + 6*(time). You pick a random student from the course and observe that the number of hours she studied for the exam is 1 standard deviation above the mean time spent studying. How many standard deviations above or below the mean exam score would you expect her to score?
The Update to the Task Force Report on Blood Pressure Control in Children [12] reported the observed 90th per-centile of SBP in single years of age from age 1 to 17 based on prior studies. The data for boys of average height are given in Table 11.18. Suppose we seek a more efficient way to display the data and choose linear regression to accomplish this task. age sbp 1 99 2 102 3 105 4 107 5 108 6 110 7 111 8 112 9 114 10 115 11 117 12 120 13 122 14 125 15 127 16 130 17 132 Do you think the linear regression provides a good fit to the data? Why or why not? Use residual analysis to justify your answer. Am I supposed to run a residual plot and QQ-plot for this question?
https://drive.google.com/file/d/1-aOh_O4aiPAqBldOcC70gIxAYx5PCMnx/view?usp=sharing
The manufacturer of Beanie Baby dolls used quarterly price data for 2012/-2020/V (t = 1, ..., 36) and the regression equation Pt= a + bt+c₁D1 t + c2 D2 + + c3 D3 t to forecast doll prices in the year 2021. Pt is the quarterly price of dolls, and D1, D2t, and D3+ are dummy variables for quarters I, II, and III, respectively. DEPENDENT VARIABLE: PT R-SQUARE P-VALUE ON F 0.0001 OBSERVATIONS: F-RATIO 76.34 STANDARD 36 0.9078 PARAMETER VARIABLE ESTIMATE ERROR T-RATIOP-VALUE INTERCEPT 24.0 6.20 3.87 0.0005 T 0.8 0.240 3.33 0.0022 D1 -8.0 2.60 -3.08 0.0043 D2 -6.0 1.80 -3.33 0.0022 D3 -4.0 0.60 -6.67 0.0001 The estimated quarterly increase in price is and the estimated annual increase in price is Multiple Choice O O $1.50; $6.00 $1.40; $4.00 $0.60; $2.40 $0.80; $3.20 None of the choices are correct.

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