Number 16 Regression and Predictions.in Section 10-2. In each case, find the regression equation, letting the first variable bethe predictor (x) variable. Find the indicated predicted value by following the predictionprocedure summarized in Figure 10-5.Exercises 13-28 use the same data sets as Exercises 13-2816. Altitude and Temperature At 6327 ft (or 6.327 thousand feet), the author recordedthe temperature. Find the best predicted temperature at that altitude. How does the resultcompare to the actual recorded value of 48°F?Altitude3101422283133Temperature573724-5-30-41-54

Name: Number 16 Regression and Predictions.in Section 10-2. In each case, fi
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Description: Number 16 Regression and Predictions.in Section 10-2. In each case, find the regression equation, letting the first variable bethe predictor (x) variable. Find the indicated predicted value by following the predictionprocedure summarized in Figure 10

Given Information : Altitude and Temperature : At 6327 ft (or 6.327 thousand feet) , the author recorded the temperature . Which is provided . The following data are passed: Altitude (X) Temperature (Y) 3 57 10 37 14 24 22 -5 28 -30 31 -41 33 -54 The independent variable is XX, and the dependent variable is YY. In order to compute the regression coefficients, the following table needs to be used: X Y X*Y X2 Y2 3 57 171 9 3249 10 37 370 100 1369 14 24 336 196 576 22 -5 -110 484 25 28 -30 -840 784 900 31 -41 -1271 961 1681 33 -54 -1782 1089 2916 Sum = 141 -12 -3126 3623 10716 Based on the above table, the following is calculated: Therefore, based on the above calculations, the regression coefficients (the slope mm, and the y-intercept nn) are obtained as follows: Therefore, we find that the regression equation is: Y = 72.4982 - 3.6843 XY = 72.4982 − 3.6843 X Given , Actual recorded value = 48 degree F Substitute , X = 48 Y = 72.4982 - 3.6843 * 48 Y = -104.3482 Hence , Temperature at 48 degrees F of altitude will be = -104.35