Using a linear-regression equation to predict values between actual data points is anexample ofa) extrapolationOb) least-squares fitOc) interpolationd) residuals31

It is given that the statement about to the regression analysis. Here, need to find out the correct…

Answered: Using a linear-regression equation to…

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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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?
What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars) and x = size of the diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE Interpret the standard deviation of the regression model. a) We expect most of the sampled diamond prices to fall within $1117.56 of their least squares predicted values. b) We can explain 89.25% of the variation in the sampled diamond prices around their mean using the size of the diamond in a linear model. c) For every 1-carat increase in the size of a diamond, we estimate that the price of the diamond will increase by $1117.56. d) We expect most of the sampled diamond prices to fall within $2235.12 of their least squares predicted values.
A regression was run to determine if there is a relationship between the happiness index (y) and life expectancy in years of a given country (x).The results of the regression were: ˆy=a+bxa=0.313b=0.145 (a) Write the equation of the Least Squares Regression line of the formˆy= + x(b) Which is a possible value for the correlation coefficient, r? -0.666 -1.772 1.772 0.666 (c) If a country increases its life expectancy, the happiness index will increase decrease (d) If the life expectancy is increased by 4 years in a certain country, how much will the happiness index change? Round to two decimal places.Use the regression line to predict the happiness index of a country with a life expectancy of 65 years. Round to two decimal places.
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.
when a regression is used as a method of predicting dependent variables from one or more independent variables. How are the independent variables different from each other yet related to the dependent variable?
The model developed from sample data that has the form of Yhat = bo +bjX is known as the multiple regression model with two predictor variables. (True or False) O True O False
A regression was run to determine if there is a relationship between the happiness index (y) and life expectancy in years of a given country (x). The results of the regression were: ý=a+bx a=-1.692 b=0.117 (a) Write the equation of the Least Squares Regression line of the form (b) Which is a possible value for the correlation coefficient, r? O -0.649 O 1.32 O 0.649 O-1.32 (c) If a country increases its life expectancy, the happiness index will O increase O decrease (d) If the life expectancy is increased by 2.5 years in a certain country, how much will the happiness index change? Round to two decimal places. (e) Use the regression line to predict the happiness index of a country with a life expectancy of 75 years. Round to two decimal places.
Help please!!
Show that an interaction term of a dummy variable and a regressor changes the slope of a regression line..
The number of pounds of steam used per month at a plant is thought to be related to the average monthly ambient temperature (in ◦F). The past year’s usages and temperatures follow.
The following result perspective in RapidMiner shows a multiple linear regression model. Based on the diagram, the model for our dependent variable Y is Predicted Y= (Insulation *0.420)+(Temperature *0.071)+(Avg_Age*0.065)+(Home_Size *0.311)+7.589 Attribute Insulation Temperature Avg Age Home Size (Intercept) O True O False Coefficient 3.323 -0.869 1.968 3.173 134.511 Std. Error 0.420 0.071 0.065 0.311 7.589 Std. Coefficient 0.164 -0.262 0.527 0.131 ? Tolerance 0.431 0.405 0.491 0.914 ? t-Stat 7.906 -12.222 30.217 10.210 17.725

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Using a linear-regression equation to predict values between actual data points is an example of a) extrapolation Ob) least-squares fit Oc) interpolation d) residuals