Consider the following output for a linear model for estimating the salary of employees at a company (in $) by the number of years employed at the company. > technitron.lm = lm(salary ~ yrs.empl, data = technitron.df)> newdata = list(yrs.empl = c(6, 8))> predict(technitron.lm, newdata, interval = "confidence") fit lwr upr1 35037.47 32582.25 37492.692 37251.91 35024.31 39479.50> predict(technitron.lm, newdata, interval = "prediction") fit lwr upr1 35037.47 20396.88 49678.062 37251.91 22647.76 51856.05Give the appropriate interval for the average salary of employees who have been employed for 6 years.Group of answer choices(32582.25, 37492.69)(20396.88, 49678.06)(35024.31, 39479.50)(22647.76, 51856.05)

Using a sample of recent university graduates, you estimate a simple linear regression using the initial annual salary as the dependent variable and the graduate's weighted average mark (WAM) as the explanatory variable. If the regression model has an estimated intercept of 3200 and an estimated slope coefficient of 550, what is the predicted starting salary of a student with a WAM of 67?

The annual profits for a company are given in the following table, where x represents the number of years since 2012, and y represents the profit in thousands of dollars. Write the linear regression equation that represents this set of data, rounding all coefficients to the nearest hundredth. Using this equation, find the projected profit (in thousands of dollars) for 2021, rounded to the nearest thousand dollars.Years since 2012 (x) Profits (y)(in thousands of dollars)00 12212211 13713722 13313333 15815844 16716755 175175

(Enter only a number in the response box, not the units)    The yield of wool in a particular farm is modeled by a linear function  P(t)=90t+3,450P𝑡=90𝑡+3,450 , where  P(t)𝑃𝑡  is measured in kg and  t𝑡  in years.     What is the initial production?     P(0)=𝑃0=    kg.     What is the prediction for the production in 8 years? P(8)=𝑃8=     kg.         The value in 8 years is   P(8)=4,056𝑃8=4,056  kg.     What is the absolute error of this model at  t=8𝑡=8 ?  Absolute error =     kg.     What is the percentage error? Round the answer to one decimal place.               Percentage error =     %

Consider the following summary output for estimating the exam mark of a student based on their Stage I grade (A, B or C).> stage1.lm = lm(Exam ~ Stage1, data = course.df)> summary(stage1.lm)Call:lm(formula = Exam ~ Stage1, data = course.df)Residuals: Min 1Q Median 3Q Max -44.755 -9.352 1.818 9.420 29.478 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 68.755 2.003 34.331 < 2e-16 ***Stage1B -17.233 2.878 -5.988 1.64e-08 ***Stage1C -29.912 2.804 -10.666 < 2e-16 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1Residual standard error: 14.02 on 143 degrees of freedomMultiple R-squared: 0.4444, Adjusted R-squared: 0.4367 F-statistic: 57.2 on 2 and 143 DF, p-value: < 2.2e-16Give a point estimate of the mean exam mark for students who obtained an A in Stage 1. Give this to 1dp.

Consider the following summary output for estimating the exam mark of a student based on their Stage I grade (A, B or C).> stage1.lm = lm(Exam ~ Stage1, data = course.df)> summary(stage1.lm)Call:lm(formula = Exam ~ Stage1, data = course.df)Residuals: Min 1Q Median 3Q Max -44.755 -9.352 1.818 9.420 29.478 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 68.755 2.003 34.331 < 2e-16 ***Stage1B -17.233 2.878 -5.988 1.64e-08 ***Stage1C -29.912 2.804 -10.666 < 2e-16 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1Residual standard error: 14.02 on 143 degrees of freedomMultiple R-squared: 0.4444, Adjusted R-squared: 0.4367 F-statistic: 57.2 on 2 and 143 DF, p-value: < 2.2e-16Give a point estimate of the mean exam mark for students who obtained a C in Stage 1. Give this to 1dp.