Ch 14 #60 716DATA fileOnline EduCarolina? Explain.e. Do you believe the estimated regression equation developed in part (b) will providea good prediction of selling price of a particular house in Winston Salem, Northf. Would you be comfortable using the estimated regression equation developed in60. Online Education. One of the biggest changes in higher education in recent years haspart (b) to predict the selling price of a particular house in Seattle, Washington?been the growth of online universities. The Online Education Database is an independ-ent organization whose mission is to build a comprehensive list of the top accreditedonline colleges. The following table shows the retention rate (%) and the graduationWhy or why not?rate (%) for 29 online colleges.RetentionRate (%)6373784895681001005129334754457463456062676578753851696037GraduationRate (%)3333342525283236363636373738394144454647485051525355565761a. Develop a scatter diagram with retention rate as the independent variable.What does the scatter diagram indicate about the relationship between the twovariables?b. Develop the estimated regression equation.c. Test for a significant relationship. Use a = .05.d. Did the estimated regression equation provide a good fit?61. Machine Maintenance. Jensen Tire & Auto is in the process of deciding whether topurchase a maintenance contract for its new computer wheel alignment and balan-cing machine. Managers feel that maintenance expenso chould alated to usage.

Hi! Thank you for the question, As per the honor code, we are allowed to answer three sub-parts at a time so we are answering the first three as you have not mentioned which of these you are looking for. Please re-submit the question separately for the remaining sub-parts. The given data is Retention Rate (%) Graduation Rate (%) 7 25 51 25 4 28 29 32 33 33 47 33 63 34 45 36 60 36 62 36 67 36 65 37 78 37 75 38 54 39 45 41 38 44 51 45 69 46 60 47 37 48 63 50 73 51 78 52 48 53 95 55 68 56 100 57 100 61 a) We use excel to develop the scatter diagram. we perform the below steps. We select the entire data> Click on Insert> Click on Recommended Charts> Click on All Charts> Click on X Y (Scatter)> Select the 2nd chart of the scatterplot> Click on Ok. Now, Click on the "+" sign located in the top right corner of the scatterplot> From the pop-up, select Axis Titles. We rename the horizontal axis title as Retention Rate, the vertical axis title as Graduation Rate, and the chart title as Scatter Diagram. We get the below scatter diagram as output: From the above scatter diagram, we see that as the retention rate increases, the graduation rate also tends to increase as a whole. So, there is a positive correlation between the variables. b) The estimated regression equation is y^=b0+b1x, where x is the retention rate, y is the graduation rate, b0=y¯-b1x¯ and b1=∑(x-x¯)(y-y¯)∑(x-x¯)2. x y (x-x¯) (x-x¯)2 (y-y¯) (x-x¯)(y-y¯) 7 25 -50.4138 2541.5512 -16.7586 844.8647 51 25 -6.4138 41.1368 -16.7586 107.4863 4 28 -53.4138 2853.034 -13.7586 734.8991 29 32 -28.4138 807.344 -9.7586 277.2789 33 33 -24.4138 596.0336 -8.7586 213.8307 47 33 -10.4138 108.4472 -8.7586 91.2103 63 34 5.5862 31.2056 -7.7586 -43.3411 45 36 -12.4138 154.1024 -5.7586 71.4861 60 36 2.5862 6.6884 -5.7586 -14.8929 62 36 4.5862 21.0332 -5.7586 -26.4101 67 36 9.5862 91.8952 -5.7586 -55.2031 65 37 7.5862 57.5504 -4.7586 -36.0997 78 37 20.5862 423.7916 -4.7586 -97.9615 75 38 17.5862 309.2744 -3.7586 -66.0995 54 39 -3.4138 11.654 -2.7586 9.4173 45 41 -12.4138 154.1024 -0.7586 9.4171 38 44 -19.4138 376.8956 2.2414 -43.5141 51 45 -6.4138 41.1368 3.2414 -20.7897 69 46 11.5862 134.24 4.2414 49.1417 60 47 2.5862 6.6884 5.2414 13.5553 37 48 -20.4138 416.7232 6.2414 -127.4107 63 50 5.5862 31.2056 8.2414 46.0381 73 51 15.5862 242.9296 9.2414 144.0383 78 52 20.5862 423.7916 10.2414 210.8315 48 53 -9.4138 88.6196 11.2414 -105.8243 95 55 37.5862 1412.7224 13.2414 497.6939 68 56 10.5862 112.0676 14.2414 150.7623 100 57 42.5862 1813.5844 15.2414 649.0733 100 61 42.5862 1813.5844 19.2414 819.4181 Sum 1665 1211 0 15123.0336 0 4302.8963 For the given data, n=29,∑x=1665,∑y=1211,∑(x-x¯)2=15123.0336,∑(x-x¯)(y-y¯)=4302.8963. Now, x¯=∑xn=166529≈57.4138 y¯=∑yn=121129≈41.7586Now, b1=∑(x-x¯)(y-y¯)∑(x-x¯)2=4302.896315123.0336≈0.2845 and b0=y¯-b1x¯=41.7586-0.2845×57.4138≈25.4244 The estimated regression equation is y^=b0+b1x=25.42+0.28xc) The null and alternative hypotheses for the test of a significant relationship are H0:ρ=0H1:ρ≠0 y (y-y¯) (y-y¯)2 25 -16.7586 280.8507 25 -16.7586 280.8507 28 -13.7586 189.2991 32 -9.7586 95.2303 33 -8.7586 76.7131 33 -8.7586 76.7131 34 -7.7586 60.1959 36 -5.7586 33.1615 36 -5.7586 33.1615 36 -5.7586 33.1615 36 -5.7586 33.1615 37 -4.7586 22.6443 37 -4.7586 22.6443 38 -3.7586 14.1271 39 -2.7586 7.6099 41 -0.7586 0.5755 44 2.2414 5.0239 45 3.2414 10.5067 46 4.2414 17.9895 47 5.2414 27.4723 48 6.2414 38.9551 50 8.2414 67.9207 51 9.2414 85.4035 52 10.2414 104.8863 53 11.2414 126.3691 55 13.2414 175.3347 56 14.2414 202.8175 57 15.2414 232.3003 61 19.2414 370.2315 Sum 1211 0 2725.3111 So, ∑(y-y¯)2=2725.3111. The correlation coefficient is r=∑(x-x¯)(y-y¯)∑(x-x¯)2×∑(y-y¯)2=4302.896315123.0336×2725.3111≈0.6702 The observed value of the test statistic is t=rn-21-r2=0.6702×29-21-(0.6702)2≈4.692 As the test is a two-tailed test, so the p-value is p-value=2×P(T>|t|)=2×P(T>|4.692|)=2×P(T>4.692)≈2×0.000035≈0.0001 Here T follows the t-distribution with df=n-2=29-2=27 We use excel to find P(T>4.692). Given that α=0.05. Clearly, p-value=0.0001<0.05 So, we reject H0 at level α=0.05. We have enough evidence to conclude that there is a significant relationship between the variables.