4-Consider the following data set representing the relationship between the independentvariable x and the dependent variable y:x 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0y: 1.5 2.1 2.8 3.6 4.5 5.1 6.2 7.0 8.1 9.5a. Use the method of linear least squares to derive the best-fit line y=mx+by for the givendata. Calculate the slope m and the intercept b manually. Show all steps in yourcalculations, including how you compute the necessary sums (e.g., sums of x, y, x², xy).b. Write a MATLAB script to perform linear least squares regression on the same data set.Use the polyfit function to find the coefficients of the best-fit line and plot the data pointsalong with the regression line.

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Answered: 4- Consider the following data set…

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter5: Inner Product Spaces

Section5.CR: Review Exercises

Problem 62CR

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Cellular Phone Subscribers The table shows the numbers of cellular phone subscribers y in millions in the United States from 2008 through 2013. Source: CTIA- The Wireless Association Year200820092010201120122013Number,y270286296316326336 (a) Find the least squares regression line for the data. Let x represent the year, with x=8 corresponding to 2008. (b) Use the linear regression capabilities of a graphing utility to find a linear model for the data. How does this model compare with the model obtained in part a? (c) Use the linear model to create a table of estimated values for y. Compare the estimated values with the actual data.
Consider the following data: Fit a line, y = x1 + x2t, to the data using the least squares approach.
a) interpret the scatter plot. Give a reasonable estimate for the linear correlation r. b) use technology to find the equation of the least squares regression line describing the relationship between Year (t) and Global Temperature (G). Around to 0.0001. c) Plot the regression line on the scatterplot below. Clearly label three points including (t,G) on the LSRL. d) Clearly interpret the slope, intercept, and R^2 of the linear model on the context of the problem statement. Report with proper units. -slope: -intercept: -R^2: e) use the model to predict the Global Temperature in the year 2030. f) compute and mark the residual for the data point (2016, 14.80 degrees C) circled on the scatterplot. Data attached below
The accompanying data represent the weights of various domestic cars and their gas mileages in the city. The linear correlation coefficient between the weight of a car and its miles per gallon in the city is r= - 0.977. The least-squares regression line treating weight as the explanatory variable and miles per gallon as the response variable is -0.0061x +41.3297. Complete parts (a) and (b) below. V = Click the icon to view the data table. (a) What proportion of the variability in miles per gallon is explained by the relation between weight of the car and miles per gallon? The proportion of the variability in miles per gallon explained by the relation between weight of the car and miles per gallon is %. (Round to one decimal place as needed.) (b) Interpret the coefficient of determination. % of the variance in is by the linear model. (Round to one decimal place as needed.) Data Table Full data set Miles per Miles per Weight (pounds), x Weight (pounds), x Car Car Gallon, y Gallon, y 1…
A researcher wishes to determine whether the rate of water flow (in liters per second) over an experiment soil bed can be used to predict the amount of soil washed away (in kilograms). The researcher measures the amount of soil washed away for various flow rates, and from these data calculates the least squares regression line to be: amount of eroded soil = 1.3(flow rate) + 0.4 One of the flow rates used by the researcher was 0.3 liters per second and for this flow rate the amount of eroded soil was 0.8 kilograms. These values were used in determining the equation of the regression line. Determine the residual corresponding to these values.
The accompanying data represent the weights of various domestic cars and their gas mileages in the city. The linear correlation coefficient between the weight of a car and its miles per gallon in the city is r= - 0.984. The least-squares regression line treating weight as the explanatory variable and miles per gallon as the response variable is y = - 0.0066x + 43.3954. Complete parts (a) and (b) below. Click the icon to view the data table. (a) What proportion of the variability in miles per gallon is explained by the relation between weight of the car and miles per gallon? Data Table The proportion of the variability in miles per gallon explained by the relation between weight of the car and miles per gallon is %. (Round to one decimal place as needed.) (b) Interpret the coefficient of determination. Full data set % of the variance in is by the linear model. Miles per Miles per Weight (pounds), x Weight (pounds), x Car Car (Round to one decimal place as needed.) Gallon, y Gallon, y…
Consider the data points (2,0), (−2, −1) and (0, −2). Which one of the following is the least squares line of best fit for this data? O O y = x y = 1 / y = 4x 1 1
The following data represent the number of flash drives sold per day at a local computer shop andtheir prices.Price (x) Units Sold (y)$34 336 432 635 530 938 240 1 a. Develop a least squares regression line and explain what the slope of the line indicates.b. Compute the coefficient of determination and comment on the strength of relationship betweenx and y.
An owner of a home in the Midwest installed solar panels to reduce heating costs. After installing the solar panels, he measured the amount of natural gas use y (in cubic feet) to heat the home and outside temperature x (in degree-days, where a day’s degree-days are the number of degrees its average temperature falls below 65F) over a 23-month period. He then computed the least squares regression line for predicting y from x and found it to be, ŷ=85+16x How much, on average, does gas used increase for each additional degree-day? The predicted amount of gas used when the outside temperature is 20 degree- days.
The accompanying data represent the weights of various domestic cars and their gas mileages in the city. The linear correlation coefficient between the weight of a car and its miles per gallon in the city is r= - 0.972. The least-squares regression line treating weight as the explanatory variable and miles per gallon as the response variable is y= - 0.0070x + 44.4405. Complete parts (a) and (b) below. Click the icon to view the data table. ..... (a) What proportion of the variability in miles per gallon is explained by the relation between weight of the car and miles per gallon? The proportion of the variability in miles per gallon explained by the relation between weight of the car and miles per gallon is %. (Round to one decimal place as needed.) (b) Interpret the coefficient of determination. % of the variance in is by the linear model. Data Table (Round to one decimal p Full data set gas mileage Miles per Weight (pounds), x Weight (pounds), x Miles per Gallon, y Car Car Gallon, y…
An article gave a scatter plot, along with the least squares line, of x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. The simple linear regression model provides a very good fit to data on rainfall and runoff volume (n = 15) given below. The equation of the least squares line is y = −1.495 + 0.83011x, r2 = 0.978, and s = 4.96. x 5 12 14 17 23 30 40 47 55 67 72 81 96 112 127 y 4 10 13 15 15 24 26 45 38 46 54 69 81 99 101 (a) Use the fact that sy hat = 1.36 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.) , m3 Does the resulting interval suggest that precise information about the value of runoff for this future observation is available? Explain your reasoning. Yes, precise information is available because the resulting interval is very wide. Yes, precise information is available because the resulting interval is very…
The following data represents the rate of return of the stock exchange (x) and the rate of return of a certain stock (y) The equation of the least squares regression line is y = 0.9764x +0.5975 and the standard error of the estimate is 2.3518. Answer parts (a)-(d). 0.23 1.21 -0.57 -0.59 5.89 0.94 1.40 3.86 3.25 -2.53 1.89 -1.91 -2.01 3.00 0.90 O 1.78 0.06 5.90 - 2.58 0.06 5.79 - 1.57 *.... (a) What is the mean rate of return for the stock if the rate of return of the S&P 500 is 424%? y= % (Round to one decimal place as needed.) (b) Construct a 90% confidence interval about the mean rate of return of y if x4.24%. Lower Bound Upper Bound (Round to one decimal place as needed.) (c) Predict the rate of retun of the stock if the rate of retum on the S&P 500 for a randomly selected month is 4.24%. y=% (Round to one decimal place as needed.) (d) Construct a 90% prediction interval about the rate of return of y if x 4.24%. Lower Bound Upper Bound (Round to one decimal place as needed.)

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