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Problem 3. (20 points) An engineer would like to use simple linear regression to investigate the relationship between roadway surface temperature in % (x) and pavem...

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Problem 3. (20 points) An engineer would like to use simple linear regression to investigate the relationship between roadway surface temperature in % (x) and pavement deflection (y). A sample of n-20 pairs of observations was obtained and the following summary quantities were calculated: Zyi 12.75,Zy? =8.86,Ex;1478,Ex} =143,215.8,and Exv;-1083.67 (all summations are for i = 1...n)-Find the least-squares estimates of the slope and the intercept_ Assess the utility of the model by formulating and

Problem 3. (20 points) An engineer would like to use simple linear regression to investigate the relationship between roadway surface temperature in % (x) and pavement deflection (y). A sample of n-20 pairs of observations was obtained and the following summary quantities were calculated: Zyi 12.75,Zy? =8.86,Ex;1478,Ex} =143,215.8,and Exv;-1083.67 (all summations are for i = 1...n)- Find the least-squares estimates of the slope and the intercept_ Assess the utility of the model by formulating and testing an appropriate hypothesis regarding the slope parameter: Use &-0.05. Report the p-value corresponding to this test: What is your conclusion? {C) What change in mean pavement deflection would be expected for 5F increase in surface temperature? (d) What is the predicted value of pavement deflection when the surface temperature is 859F? What is the respective 95% prediction interval?



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A study of the tipping habits of restaurant-goers was completed.The data for two of the variables $-x,$ the amount of the restaurant check, and $y,$ the amount left as a tip for the servers-were used to construct a scatter diagram. a. Do you expect the two variables to show a linear relationship? Explain. b. What will the scatter diagram suggest about linear correlation? Explain. c. What value do you expect for the slope of the line of best fit? Explain. d. What value do you expect for the $y$ -intercept of the line of best fit? Explain.The data are used to determine the equation for the line of best fit: $\hat{y}=0.02+0.177 x$. e. What does the slope of this line represent as applied to the actual situation? Does the value 0.177 make sense? Explain.f. What does the $y$ -intercept of this line represent as applied to the actual situation? Does the value 0.02 make sense? Explain. g. If the next restaurant check was for $\$ 30 dollars. what would the line of best fit predict for the tip? h. Using the line of best fit, predict the tip for a check of $\$ 31 dollas. What is the difference between this amount and the amount in part g for a $\$ 30$ check? Does this difference make sense? Where do you see it in the equation for the line of best fit?

In question # eight. We have a problem about linear regression and we're trying to figure out what's going to happen when a point is added to this scatter plot. Now the information that were given about the relationship that we're seeing in the scatter plot is that the slope of this least squares regression line would be .377 R squared, which is the coefficient of determination would be 95%, which is very strong. And S which represents the standard deviation of the residuals is .64. Now, when I take a look at this relationship and if I were to just sketch police squares regression line, I would want to imagine where the point that they're giving me would be on this line. The point that they give us is 35 14. And if you actually follow the slope of the data and take a look at the picture, that point would probably be pretty close to the line. So there's a couple things that happen when you add a point that's very close to the least squares regression line in a data set that's already existing. The fact that it's close to the line means that the slope shouldn't change very much because it's following the trend of the data. The rest of the data follows that same pattern. That slope will not change now for our squared R squared is supposed to tell us about the strength of the relationship. And if I take a look at that point, because it's continuing the trend of the relationship, I can confidently say that R squared would actually increase because this point being added so close to the least squares regression line is only going to make our squared more, going to get stronger, it's going to make it higher. So I would predict that this would go up now, s which represents the standard deviation of the residuals essentially. Typically the average distance away from this least squares regression line that all of these points are at this point, since I think it's going to be pretty close to the least squares regression line and that distance between that point at least great regression line, it's residual would be small. That would make me think that the standard deviation of the residuals would decrease a little bit since that distance would be small and therefore we lower that average distance. So, if you take a look at the answer choices, the only answer choice that shows the slope staying the same, Which is 0.377, the r squared value increasing which there's an answer choice that shows r squared becoming 97% and the standard deviation of the residuals s Decreasing to 0.617 is answer choice be so that's the correct answer.

This is problem # 22. We are given a set of data regarding the length of the right humerus and right tibia in rats first, we will write our least squares regression line. So using our data, we get the following for our five key points. We have 198.96. 286.04. We get 7,120.6. We have 4952.4 and we have 10239 0.9. So this allows us to find our equation for regression line. Using these equations and we get the following, we have white hat equal to 1.438 x -0.01. So while we cannot interpret the Y intercept as a length of zero for a bone does not make sense. We can say that an increase to the length of the right humerus causes an increase to the correlates with an increase in the length of the right tibia, specifically By 1438 cm or in this case millimeters. So we can now use this to determine the residual. If the length is 26.11. So given this input of X equal to 26.11, we get a predicted value of 37 54. Given our data point of 37-96, this gives us a residual equal to negative or posited. In this case positive, zero point for two. So in this case the length of the tibia is slightly above average. Finally, We can estimate the length of a humorous at 25.31 cm and can determine that the length of the right tibia should be 36 39 cm mm.


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