Consider the following computer output for a linear regression of height onweight:The regression equation isHeightPredictorConstantWeightCoefStdev0.15930.007385t-ratio13.3480-0.1560000.0000.00083.81-21.12s = 0.03873 R-sq = 98.2% R-sq(adj) = 98.0%The equation of the regression line is:A. Height = 13.35 + .03873 (Weight)B. Height = 13.35 + .1593 (Weight)O C. Height = 13.35 - .156 (Weight)D. Height = .1593 + .007 (Weight)O E. Height = - .156 + 13.35 (Weight)

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Answered: Consider the following computer output…

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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I need help to interpret the slope of the scatter plot and to do the linear regression equation
We collected teacher ratings for 25 courses taught by an instructor over a six-vear period. The students' ratings of the instructor are on a scale of 1 to 9. We found that The linear regression equation is: Average Rating = 7,88 -0.068 Numher of Students 1. Interpret the slope of this model including units: The average rating decreases per each additional student 0.068 ed teacher rating for a class size of 15 students using the given 7.88 prediction equation is Next page CS Scanned with CamScanner
Fit a linear regression model for the following data.
A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y). The results of the regression were: y=ax+b a=-1.307 b=32.392 r²=0.675684 r=-0.822 Use this to predict the number of situps a person who watches 12.5 hours of TV can do (to one decimal place)
A researcher records age in years (x) and systolic blood pressure (y) for volunteers. They perform a regression analysis was performed, and a portion of the computer output is as follows: ŷ = 4.5+ 14.4x Coefficients (Intercept) x Estimate 4.5 Ho: B₁ = 0 H₁: B₁ > 0 Ho: B₁ = 0 Ha: B₁ <0 14.4 Ho: B₁ = 0 Ha: B₁ #0 Std. Error Test statistic 2.9 4.7 1.55 3.06 P-value Specify the null and the alternative hypotheses that you would use in order to test whether a linear relationship exists between x and y. 0.07 0
The arm span and foot length were both measured (in centimeters) for each of 20 students in a biology class. The computer output displays the regression analysis. Predictor Coef SE Coef t-ratio Constant Arm span -7.611 2.567 2.965 0.046 0.186 0.035 5.377 0.000 S = 1.61 R-Sq = 63.0% R-Sq (Adj) = 64.9% Which of the following is the best interpretation of the standard deviation of the residuals? The typical arm span is 161 centimeters. The typical foot length is 16.1 centimeters. The typical distance between the observed and predicted arm spans is 1.61 centimeters. The typical distance between the observed and predicted foot lengths is 1.61 centimeters. Op
From the results shown above, write the regression equation.
Q1: [Regression and Correlation] The index of biotic integrity (IBI) is a measure of water quality in streams. As a data analyst you must monitor, track, and predict changes in water quality. You want to create a simple linear regression model that will allow you to predict changes in IBI in forested area. The following table conveys sample data from a coastal forest region and gives the data for IBI and forested area in square kilometers. Let forest area be the predictor variable (x) and IBI be the response variable (y). IBI X 47 72 21 72234 19 58 49 Forest Area Y 38 59 27 24 63 49 45 Required a. Q2: [Naïve Bayes] a. Write the pseudo-code of the following Naïve Bayes algorithm b. Consider the given dataset that classifies animals into two distinct classes. The classes are labeled as 'cat' and 'dog'. Use Naive Bayes classifier to figure out the class (cat/dog) of an instance if it has the following values of attributed Plot Scatter graph b. Determine the regression equation c. Compute…
An automobile rental company wants to predict the yearly maintenance expense (Y) for an automobile using the number of miles driven during the year () and the age of the car (, in years) at the beginning of the year. The company has gathered the data on 10 automobiles and run a regression analysis with the results shown below:. Summary measures Multiple R 0.9689 R-Square 0.9387 Adj R-Square 0.9212 StErr of Estimate 72.218 Regression coefficients Coefficient Std Err t-value p-value Constant 33.796 48.181 0.7014 0.5057 Miles Driven 0.0549 0.0191 2.8666 0.0241 Age of car 21.467 20.573 1.0434 0.3314 Use the information above to estimate the annual maintenance expense for a 10 years old car with 60,000 miles.
The trip rate (y) and the corresponding household sizes (x) from a sample are shown in Table Q2. Apply the regression method to find the trip rate for a household size of 10. Table Q2
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A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y). The results of the regression were: y=ax+b a=-0.666 b=36.053 r2=0.552049 r=-0.743 Use this to predict the number of situps a person who watches 4.5 hours of TV can do (to one decimal place)

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