Exhibit 12-1A regression analysis resulted in the following information regarding a dependent variable (v) and an independent variable (x).n=10Ex=55Σι= 55Ex²=385Ev=385Στ= 220Refer to Exhibit 12-1. The least squares estimate of b₁ equals

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Answered: A regression analysis resulted in the…

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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Biologist Theodore Garland, Jr. studied the relationship between running speeds and morphology of 49 species of cursorial mammals (mammals adapted to or specialized for running). One of the relationships he investigated was maximal sprint speed in kilometers per hour and the ratio of metatarsal-to-femur length. A least-squares regression on the data he collected produces the equation ŷ = 37.67 + 33.18x where x is metatarsal-to-femur ratio and y is predicted maximal sprint speed in kilometers per hour. The standard error of the intercept is 5.69 and the standard error of the slope is 7.94. Construct a 96% confidence interval for the slope of the population regression line. Give your answers precise to at least two decimal places. contact us help 6:42 PM povecy polcy terms of use careers A E O 4») 18 -క90.4 58 12/14/2020 a 17 |耳即 delets prt sc insert 112 19 18 + 16 backspace f5 fA
two variables have at least squares regression line of 28.77 + 0.346 7x. Calculate the value of y that is predicted for an x value of 35.
A least squares regression model performed to predict the selling price of houses found the following equation: Pricê = 169.328+35.3Area + 0.718Lotsize - 6543Age where Price is in dollars, Area is in square feet, Lotsize is in square feet, and Age is in years. The R2 is 0.92. One of the following interpretations is correct. Which is it? Explain why. a.) Each year a house Ages, it is worth $6543 less. b.) Every extra square foot of Area is associated with an additional $35.30 in average price, for houses with a given Lotsize and Age. c.) Every dollar in price means Lotsize increases 0.718 square feet. d.) This model fits 92% of the data points exactly.
Use the least squares regression model to estimate the maximum sustained wind speed in a hurricane when the pressure reading is 950 mb. Interpret the slope of the least squares regression model in the context of wind speed and atmospheric pressure. Determine the coefficient of determination and interpret its meaning in the context of wind speed and atmospheric pressure.
Find the least squares approximating squares error for the points (1.1), (4.3). line and the least (2, 2), (3.2), and
Do not show any work on this question. For these ordered pairs: (0, 0.1), (1, 1), (2,2.4),(4,3.7), and (5, 5.7): Find (f ,y) and plot it as well as the five given points. Find the equation ofthe least squares regression line. Round off the x coefficient and the constant to the nearest onethousandth. Graph the equation together with the points above. If needed do not use a Pvalue to answer any problem. No credit will be given if a P value is used. Use only criticalvalues.
An article gave a scatter plot, along with the least squares line, of x = rainfall volume (m³) and y data on rainfall and runoff volume (n = runoff volume (m³) for a particular location. The simple linear regression model provides a very good fit to 15) given below. The equation of the least squares line is y = -2.364 + 0.84267x, ² 0.976, and s = 5.21. = x 5 12 14 17 23 30 40 47 55 67 72 81 96 112 127 y 3 9 12 14 14 24 27 45 38 46 52 71 81 100 101 (a) Use the fact that s = 1.43 when rainfall volume is 40 m³ to predict runoff in a way that conveys information about reliability and precision. (Calculate a 95% PI. Round your answers to two decimal places.) Ŷ 28.25 1x ) m³ Does the resulting interval suggest that precise information about the value of runoff for this future observation is available? Explain your reasoning. OYes, precise information is available because the resulting interval is very wide. 34.46 Yes, precise information is available because the resulting interval is very…

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A regression analysis resulted in the following information regarding a dependent variable (v) and an independent variable (x). n= 10 Ex=55 Σ1=55 Ex²=385 Στα 385 Στις 220 Refer to Exhibit 12-1. The least squares estimate of by equals