Use the Stata output below. The data comes from students in an undergraduate economics course. The regression of interest is:\[\hat{f}inal = \hat{\beta}_0 + \hat{\beta}_1 atndrte + \hat{\beta}_2 hwrte + \hat{\beta}_3 priGPA + \hat{\beta}_4 ACT\]The Stata regression command is:`. reg final atndrte hwrte priGPA ACT`### Regression Output#### ANOVA Table| Source | SS | df | MS ||------------|-----------|-----|------------|| Model | 3094.70776 | 4 | 773.676939 || Residual | 11299.9465 | 669 | 17.832506 || Total | 15024.6543 | 673 | 22.324895 |#### Model Summary- Number of obs = 674- F(4, 669) = (Not provided)- Prob > F = (Not provided)- R-squared = (Not provided)- Adj R-squared = (Not provided)- Root MSE = (Not provided)#### Coefficients| Variable | Coef. | Std. Err. | t | P>|t| | [95% Conf. Interval] ||----------|-----------|-----------|-----------|----------|----------------------|| atndrte | 0.0138725 | 0.010863 | (Not provided) | (Not provided) | (Not provided) || hwrte | 0.0183476 | 0.018208 | (Not provided) | (Not provided) | (Not provided) || priGPA | 1.906347 | 0.3750236 | (Not provided) | (Not provided) | (Not provided) || ACT | 0.3990516 | 0.0535332 | (Not provided) | (Not provided) | (Not provided) || _cons | 9.225908 | 1.46863 | (Not provided) | (Not provided) | (Not provided) |### TaskConstruct a 95% confidence interval about \(\beta_

The objective is to construct 95% confidence interval about β3.

Answered: Use the Stata output below. The data…

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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Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 85 mm Hg. Use a significance level of 0.05 Right Arm 101 100 94 75 Left Arm 174 167 146 144 Click the icon to view the critical values of the Pearson correlation coefficient r The regression equation is (Round to one decimal place as needed.) 76 144 CRITS
Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right am blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 100 mm Hg Use a significance level of 0.05. Right Arm 103 102 95 78 Left Arm 175 169 147 146 144 m Click the icon to view the critical values of the Pearson correlation coetticient r The regression oquation is y =+O (Round to one decimal place as needed) Given that the systolic blood pressure in the right arm is 100 mm Hg, the best predicted systolic blood pressure in the left arm is mm Hg (Round to one decimal place as needed.)
A regression was run to determine if there is a relationship between hours of study per week (xx) and the test scores (yy). The results of the regression were: y=ax+b a=5.709 b=34.57 r2=0.641601 r=0.801 Use this to predict the score of a student who studies 9.5 hours per week, and please round your answer to a whole number.
The data show the bug chirps per minute at different temperatures. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted temperature for a time when a bug is chirping at the rate of 3000 chirps per minute. Use a significance level of 0.05. What is wrong with this predicted value? Chirps in 1 min 784 838 822 949 1188 813 Temperature (°F) 65.7 67.8 66.2 79.9 92.3 73.5 What is the regression equation? What is the best-predicted temperature for a time when a bug is chirping at the rate of 3000 chirps per minute? The best predicted temperature when a bug is chirping at 3000chirps per minute are? What is wrong with this predicted value? Choose the correct answer below. A. It is unrealistically high. The value 3000 is far outside of the range of observed values. B. The first variable should have been the dependent variable. C. It is only an…
A regression was run to determine if there is a relationship between hours of study per week (z) and the test scores (y)- The results of the regression were: y=ax+b a=5.701 b=31.89 r2=0.660969 r=0.813 Use this to predict the final exam score of a student who studies 3.5 hours per week, and please round your answer to a whole number. Question Help: Message instructor Submit Question rse here to search R
A regression was run to determine if there is a relationship between hours of study per week (xx) and the final exam scores (yy).The results of the regression were: y=ax+b a=6.22 b=21.6 r2=0.477481 r=0.691 Use this to predict the final exam score of a student who studies 8 hours per week, and please round your answer to a whole number.
Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 85 mm Hg. Use a significance level of 0.05 Right Arm 101 100 94 75 Left Arm 174 167 146 144 Click the icon to view the critical values of the Pearson correlation coefficient r The regression equation is (Round to one decimal place as needed.) 76 144 CRITS
Use the Stata output below. The data comes from students in an undergraduate economics course. The regression of interest is: final =B₁ + B₁ atndrte + ß₂hwrte + Â¸priGPA+ Â¹ ACT .reg final atndrte hwrte priGPA ACT Source Model 3094.70776 11929.9465 Residual Total final SS atndrte hwrte priGPA ACT _cons 15024.6543 df 773.676939 4 669 17.832506 MS 673 22.324895 Coef. Std. Err. .0138725 .0183476 .010863 1.906347 .3750236 .3990516 .0535332 9.225908 1.46863 Number of obs F(4, 669) Prob > F R-squared Adj R-squared Root MSE t P>|t| ||||||||||||| 674 [95% Conf. Interval] Approximate the p-value from the null hypothesis that all of the slope parameters are equal to zero.
The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 63 inches. Is the result close to the actual weight of 562 pounds? Use a significance level of 0.05. Chest_size_(inches) Weight_ (pounds)58 41450 31265 49959 45059 45648 260 What is the regression equation?^y = ____ + _____ x (round to one decimal place as needed.)What Is the best predicted weight of a bear with a chest size of 63 inches? ^y =____ pounds (round one decimal as needed)Is the result close to the actual weight of 452 pounds?(a) This result is very close to the actual weight of the bear.(b) This result is close to the actual weight of the bear.(c) This result is exactly…
Please do only the last three
Describe regression variation in terms of variation in Y.

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