Quiz 6 Name ANSWER KEYStat 217FORMULAS: 2nDFE DFSSMS  SSTSSMR 211bSEbt SHOW YOUR WORK!!!1. The principal at Super High School is interested in using the number of hours a high schooler studieseach week to predict the student’s academic performance, rated on a scale from 0 to 10. The MINITABoutput follows:Predictor Coef StdErr T PConstant -0.1872 0.8834 -0.212 0.83350Hours 0.6100 0.1879 ***** 0.00268S = *** R-Sq = ****% R-Sq(adj) = 21.9%Analysis of VarianceSource DF SS MS F P Regression 1 ****** 101.39 10.541 0.00268Residual Error 33 317.39 ****Total 34 418.78(a) (1/2 pt) Write out the Simple Linear Regression model. xy10 , where  ~ N(0,). (b) (1/2 pt) Write out the least-squares regression line for this problem.Perf = -.1872 + .61 Hours(c) (1/2 pt) How many students were used in this SLR?DFE = n – 2 = 33, so n = 35 students.(d) (1/2 pt) The Simple Linear Regression model assumes the error terms have a constant variance, 2. Givean unbiased estimate of 2.617.93339.3172DFESSEMSE(e) (1/2 pt) Calculate R2 .2421.78.41839.1012SSTSSMR(f) (1 pt) Interpret the value of R2 . 24% of the variability of academic performance is explained by the linear relationship with hours studied.(g) (1/2 pt) What is the correlation between number of hours studied and academic performance? 4920.2421.2 Rr(h) (1 ½ pts) Is there sufficient evidence to suggest that the slope, 1, is non-zero? Give the following regression output which help you answer this question: - the value of the test statistic - the distribution of the test statistic- the p-value The test statistic is 246.31879.61.11bSEbt, t~t(33), the p-value is .00268, and so we REJECT H0:-1=0. Thus, the evidence suggests that the slope is non-zero.5 15 25 35-505Fitted ValueResidualResiduals Versus the Fitted Values(response is Price)2. (1/2 pt) Does the evidence suggest that there is a linear relationship between number of hours studied and academic performance? Explain how you decided Yes or No. Since the slope is non-zero, then we conclude that the there is a linear relationship between academic performance and number of hours studied.3. (1 pt) Interpret the slope of the least squares regression line, b1 in terms of the problem.b1 = .61/1 . Thus, for each additional hour studied, academic performance increases on average by .61.4. (1 pt) William, a student at Super High, studies five hours a week. Based on the above regression output, what is Williams academic performance?.8 6.2)5(61.1872.ˆy Thus, we predict that William’s academic performance is 2.86.5. (1 pt) A new student is starting at Super High this year. She studies 7 hours per week. The principalis not sure whether to use a confidence interval (CI) or a prediction interval (PI) to estimate thisstudent’s academic performance at the end of the year. Do you recommend that the principal usea CI or a PI? Explain why your answer is correct.Use a Prediction Interval since we are predicting for a new student who studies 7 hours a week, not for amean for all students who study 7 hours a week.6. (1/2 pt) The constant variance assumption does not appear to be violated in the Residual versus Fitted Value plot shown on the right. Is there another assumption that appears to be violated for these data? Why or why not?Linearity appears to be violated due to the parabolic shape in the plot. 7. (1/2 pt) For the least squares regression line, A. SSE is the smallest when compared to all other lines fit to the