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- The following fictitious table shows kryptonite price, in dollar per gram, t years after 2006. t= Years since 2006 0 1 2 3 4 5 6 7 8 9 10 K= Price 56 51 50 55 58 52 45 43 44 48 51 Make a quartic model of these data. Round the regression parameters to two decimal places.Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?XYZ Corporation Stock Prices The following table shows the average stock price, in dollars, of XYZ Corporation in the given month. Month Stock price January 2011 43.71 February 2011 44.22 March 2011 44.44 April 2011 45.17 May 2011 45.97 a. Find the equation of the regression line. Round the regression coefficients to three decimal places. b. Plot the data points and the regression line. c. Explain in practical terms the meaning of the slope of the regression line. d. Based on the trend of the regression line, what do you predict the stock price to be in January 2012? January 2013?
- Question 4: A company studied the productivity of their employees on a new information system. They were interested in if the age (X) of data entry operators influenced the number of completed entries made per hour (Y). If the regression equation is = 14.374 + 0.145x. The SD of age is = 14.04, and the SD of the number of completed entries made per hour is = 2.61. What is the correlation coefficient between age and productivity? How to interpret the correlation? How to interpret the slope? If a data entry operator is 40 years old, what is the predict productivity using the regression equation?Question 27 In March, Maggie collected the daily sales of ice-cream in Montreal (Y) and the outdoor temperature for each day (X). She computed the sample correlation coefficient between X and Y to be 0.1, and used the data to build a simple linear regression model Y = b + cX.Question 4 The following table displays the mathematics test scores for a random sample of college students, along with their final SY16C grades. Fit the regression line y = a+bx to the data and interpret the results. Use the regression equation to determine the SY16C grade for a college student who scored 60 on their achievement test. What would their SY16C grade be? Mathematics test (x) SY16C grades (y) 1 39 65 2 43 78 3 21 52 4 64 82 5 57 92 6 47 89 7 28 73 8 75 98 9 34 56
- Which of the following statements is NOT true? a.If the correlation coefficent, rr, is negative, then the slope of the regression line must always be negative. b.For variables x and y, the value of the correlation coefficent will not change c.if the units of x and y are changed, (i.e., if both x and y are changed from say inches to meters) d.If r≠0r≠0 , then a significant linear correlation will exist between the two variables e.The value of the correlation coefficient must be within the interval: −1≤r≤1Question 23: Run a regression analysis on the following bivariate set of data with y as the response variable. x y 103.5 90.7 58.3 62.9 90.6 70.6 65.6 63.1 71.3 63.3 72.9 66.5 49.2 48.6 69.7 70.9 94.1 88.4 67.4 60.2 72.4 72.8 72.8 68 Find the correlation coefficient and report it accurate to three decimal places.r = What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.)r² = %Based on the data, calculate the regression line (each value to three decimal places)y = x + Predict what value (on average) for the response variable will be obtained from a value of 96.4 as the explanatory variable. Use a significance level of α=0.05α=0.05 to assess the strength of the linear correlation.What is the predicted response value? (Report answer accurate to one decimal…The grades of a class of 9 students on a midterm report (x) and on the final examination (y) are as follows: Give the following: a. linear regression line and equation b. computation of the coefficient of determination ?^2 c. Computation of the coefficient of correlation ? d. Estimate the final examination grade of a student who received a grade of 85 on the midterm report.