Question

111 96 68 76 66 91 72 66 58 108 79 91 116 84 80 The data show the time intervals after an eruption (to the next eruption) of a certain geyser. Find the regression equation, letting the first variable be the independent (*) variable. Find the best predicted time of the interval after an eruption given that the current eruption has a height of 121 feet. Use a significance level of 0.05. Height (19) Interval after (min) 79 Click the icon to view the critical values of the Pearson correlation coefficient. What in the regression equation? 9=0.x (Round to two decimal places as needed)

Answer 1

from above table we calculate

X Y X*Y X2 y2 96 68 6528 9216 4624 111 80 8880 12321 6400 76 66 5016 5776 4356 91 72 6552 8281 5184 66 58 3828 4356 3364 108 79 8532 11664 6241 116 84 9744 13456 7056 91 79 7189 8281 6241 Sum = 755 586 56269 73351 43466

let m be the slope and n be the y-intercept

n 755 X = = 94.375 n 8 n 586 Y Y; - 73.25 n 8 2 SSXX 73351 – 755? /8 = 2097.875 Σκ' ! (Σκ) Ση» ! (Σ») 2 SSYY = ΣΥ? = 43466 – 586 /8 = 541.5 η 1 SSXY ΣΧ;Υ; (Σκ) (ΣΥ) 56269 – 755 x 586/8 = 965.25 n i=1

m = SSXY SSxx 965.25 2097.875 0.4601 n=Y - X.m = ·m = 73.25 94.375 x 0.4601 = 29.8273 Therefore, we find that the regression equation is: Y = 29.8273 +0.4601X

sorrelation coeff 89, xy 12 x 9998) 965.25 = 0.9056 (2097.815 541.5)/2 df - m-22.8-2-6 and critical value for significance lull 0.05 is 0.7076 which is less than son' is significant. af

As the cosulation coefficiunt is greater than. the critical value so aan use the kogu ussion line for prediction. sou when y=+21 feel 29.8273 +0.460 1 x (121) 85.9994 height I a 85.5 mun