Linear regression Qn1. Consider a Linear Regression Model Y₁ = Be + Pix, + &; and the following dataentries4Observation Talk time consumpticMonthly incomeConstant1128431120481135561143611141671152701a)Derive the Ordinary Least Squares (OLS) estimator of B1 and Bo-b)Using information in the table above, calculate the values of the estimates in(a)Write down the fitted/predicted modeld)Given that the above model (y; = Bo + B₁x; + E₁) is a log-log model of 'Taiktime consumption' (regressand) and 'monthly income' (regressor). How wouldyou interpret the estimates of Bo and B₁?Using the information from the table and results from above questions,calculate and interpret the R-squared23456

Let yi, xi i=1,2,...,n be n observations. The simple linear regression model is given by…

Answered: Qn1. Consider a Linear Regression Model…

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter7: Distance And Approximation

Section7.3: Least Squares Approximation

Problem 31EQ

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If your graphing calculator is capable of computing a least-squares sinusoidal regression model, use it to find a second model for the data. Graph this new equation along with your first model. How do they compare?
Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?
Find the equation of the regression line for the following data set. x 1 2 3 y 0 3 4
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 following gives the number of accidents that occurred on Florida State Highway 101 during the last 4 months: Month Number of Accidents Jan 25 Feb 40 Mar 60 Apr 90 Using the least-squares regression method, the trend equation for forecasting is (round your responses to two decimal places): ŷ-0-0x Using least-squares regression, the forecast for the number of accidents that will occur in the month of May = accidents (enter your response as a whole number).
Interpret the least squares regression line of this data set. Meteorologists in a seaside town wanted to understand how their annual rainfall is affected by the temperature of coastal waters. For the past few years, they monitored the average temperature of coastal waters (in Celsius), x, as well as the annual rainfall (in millimetres), y. Rainfall statistics • The mean of the x-values is 11.503. • The mean of the y-values is 366.637. • The sample standard deviation of the x-values is 4.900. • The sample standard deviation of the y-values is 44.387. • The correlation coefficient of the data set is 0.896. The correct least squares regression line for the data set is: y = 8.116x + 273.273 Use it to complete the following sentence: The least squares regression line predicts an additional annual rainfall if the average temperature of coastal waters increases by one degree millimetres of Celsius.
A random sample of 65 high school seniors was selected from all high school seniors at a certain high school. The following scatterplot shows the height, in centimeters (cm), and the foot length, in cm, for each high school senior from the sample. The least-squares regression line is shown. The computer output from the least-squares regression analysis is also shown. Term Coef(SE) CoefT-ValueP-Value Constant 105.086.0017.510.000 Foot length 2.5990.23810.920.000 S=5.90181R–sq=65.42% (a) Calculate and interpret the residual for the high school senior with a foot length of 20cm and a height of 160cm. BoldItalicUnderlineSuperscriptSubscriptUndoRedoΩBullet listNumbered listImage (12 image limit) Edit imageView imageDelete image Question 2 (b) The standard deviation of the residuals is s=5.9. Interpret the value in context. BoldItalicUnderlineSuperscriptSubscriptUndoRedoΩBullet listNumbered listImage (12 image limit) Edit imageView imageDelete…
Compute the least-squares regression line for predicting the right foot temperature from the left foot temperature. Round the slope and y-Intercept values to four decimal places.
Suppose that Y is normal and we have three explanatory unknowns which are also normal, and we have an independent random sample of 12 members of the population, where for each member, the value of Y as well as the values of the three explanatory unknowns were observed. The data is entered into a computer using linear regression software and the output summary tells us that R-square is 0.85, the linear model coefficient of the first explanatory unknown is 7 with standard error estimate 2.5, the coefficient for the second explanatory unknown is 11 with standard error 2, and the coefficient for the third explanatory unknown is 15 with standard error 4. The regression intercept is reported as 28. The sum of squares in regression (SSR) is reported as 85000 and the sum of squared errors (SSE) is 15000. From this information, what is SSE/SST? (a) .2 (b) .13 (c) NONE OF THE OTHERS (d) .15 (e) .25
A prospective MBA student would like to examine the factors that impact starting salary upon graduation and decides to develop a model that uses program per-year tuition as a predictor of starting salary. Data were collected for 37 full-time MBA programs offered at private universities. The least squares equation was found Y = -13258.594 + 2.422X,, where X; is the program per-year tuition and Y; is the predicted mean starting salary. To perform a residual analysis for these data, the following results are obtained. of regression have been seriously violated. Residual index plot QQ Plot of Residuals Residuals Residuals 20000 20000 -20000 -20000 -40000 40000 TO 20 Index Normal Quantile Residuals vs. Program Per-Year Tuition ($) Residuals Predicted Values vs. Residuals Predicted Values 20000 140000- 120000- 100000 80000- -20000 60000 -40000 40000- 30000 50000 4000 Program Per-Year Tuition ($) 20000 60000 70000 -20000 20000 Residuals ..... a) To evaluate whether the assumption of linearity…
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).
The following gives the number of accidents that occurred on Florida State Highway 101 during the last 4 months: Month Jan Feb Mar Apr Number of Accidents 25 48 70 90 Using the least-squares regression method, the trend equation for forecasting is (round your responses to two decimal places): y = Using least-squares regression, the forecast for the number of accidents that will occur in the month of May = accidents (enter your response as a whole number).

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Qn1. Consider a Linear Regression Model Y₁ = Be + B₁x₁ +E; and the following data entries Observation Talk time consumption Monthly income Constant 1 128 43 1 2 120 48 2 125 56 1 1