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.

## ## Call:## lm(formula = y ~ I(x^1) + I(x^2) + I(x^3) + I(x^4) + I(x^5) + ## I(x^6), data = df)## ## Residuals:## Min 1Q Median 3Q Max ## -2.9282 -0.6652 0.0071 0.6308 3.7435 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -0.05348 0.20758 -0.258 0.797272 ## I(x^1) 1.10952 0.35957 3.086 0.002675 ** ## I(x^2) 0.92462 0.51810 1.785 0.077583 . ## I(x^3) 1.21361 0.30649 3.960 0.000147 ***## I(x^4) 1.12671 0.27192 4.144 7.53e-05 ***## I(x^5) 0.93841 0.05348 17.549 < 2e-16 ***## I(x^6) 0.97314 0.03658 26.605 < 2e-16 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 1.2 on 93 degrees of freedom## Multiple R-squared: 0.999, Adjusted R-squared: 0.9989 ## F-statistic: 1.553e+04 on 6 and 93 DF, p-value: < 2.2e-16

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?

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)

Suppose we have:a response variable Y andan explanatory factor X with four levels.We ran the following code in R:mymodel <- lm(Y ~ X)Which of the following statements is FALSE.Group of answer choicesThe t value column of summary(mymodel) gives the t-statistic to test the null hypothesis that the true value of the regression coefficient is zero.By default, the estimated value of the intercept is the fitted value for the first level of X.The value of the fourth regression coefficient in the summary(mymodel) output is the estimated difference between the expected values of Y at level 4 and level 3 of X.mymodel has four regression coefficients.The Pr(>|t|) values in summary(mymodel) are the p-values for the null hypothesis that the true value of the regression coefficient is zero.

The director of admissions of a small college selected 120 students at random from the new freshman class in a study to determine whether a student’s grade point average (GPA) at the end of the freshman year (Y) can be predicted from the ACT test score (X). So the director would like to fit a simple linear regression on Y and X.Based on the summary statistics below, please calculate the estimate of .=2666=2385.328=368.886=1183.379 = 93.928