A group of environmental researchers is studying the relationship between the nu

Question

A group of environmental researchers is studying the relationship between the number X of angiosperm (flowering plant) species found and the number Y of rodent species found on 10-meter-square plots in the Meadowlands. Surveying 117 plots they find x = 1 846, y Y = 667, x2-3 1, 2 5 5yY2-4 105, X. Y = 10, 863 3. a. Find the means, variances, and covariance. (Use the 17/ , correction.) Then find the regression coefficients and the correlation coefficient, using X as a predictor of Y. Give the equation of the regression line Characterize the relationship. How would I expect the number of rodent species to increase as the number of angiosperm species increase? Solve for the regression coefficients and correlation coefficient using Y as a predictor of X. b. c. d. i. Report how the coefficients changed, if at all ii. If they changed, why would you expect that? If not, why would you expect that? iii. Using the Analysis of Variance F-test (Explained and Unexplained Variation) for X as a predictor ofy, determine if we can say that significance Variation Total e. Q at the 5% level of df Sum Average F-Ratio value lained Residual

Explanation / Answer

Q.3

(a) X = 1846 , Y = 667, X2 = 31255 ; Y2 = 4105 , XY = 10863

Mean (X) = x = 1846/ 117 = 15.7778

Mean (Y) = 667/117 = 5.700

Var(X) = (1/n(n-1)) [nX2 - (X)2] = (1/116 * 117) ( 117 * 31255 - 18462) = 18.3554

Var(Y) = (1/n(n-1)) [nY2 - (Y)2]  = (1/116 * 117) * ( 117 * 4105 - 6672) = 2.608

Covaraince = 1/ (n(n-1) [nXY - XY] =  (1/116 * 117) * [ 117* 10863 - 1846 *667] = 2.9243

Correlation Coefficient r = [n(xy) - (x)((y)] / sqrt [ (n (x2 ) - (x)2) ( n (y2 ) - (y)2]

r = [117 * 10863 - 1846 *667]/ sqrt [ (117 * 31255 - 18462) * (117 * 4105 - 6672)]

r = 39689/ sqrt [249116 * 35396]

r = 39689/ 93902.96 = 0.4227

here if the regresssion line is y^ = a + bx

then

a = [(y) (x2 ) - (x) (xy)]/ [ n (x2 ) - (x)2 ]  

a = [ 667 * 31255 - 1846 * 10863] / [ 117 * 31255 - 18462]

a = 793987/ 249119 = 3.1872

b = [ n(xy) - (x)((y)]/ [ n (x2 ) - (x)2 ]

b = [ 117 * 10863 - 1846 * 667]/ [ 117 * 31255 - 18462]

b = 39689/249119 = 0.159

(b) Linear regression line is y^ = 0.159x + 3.1872

(c) If there is an increase of one angiosperm species increase, there will be an increase of 0.159 rodent species will increase.

(d) Here now we will calculate the inverse relationship between the two variables.

now let say X = a' + B'Y

so a' =  [(x) (y2 ) - (y) (xy)]/ [ n (y2 ) - (y)2 ]  

a' = [1846 * 4105 - 667 * 10863] / [ 117 * 4105 - 6672]

a' = 332209/ 35396 = 9.3855

b' = [ n(xy) - (x)((y)]/ [ n (y2 ) - (y)2 ]

b' = 39689 / 35396 = 1.121

here y^ = 1.121x + 9.3855

(ii) Here the coefficient would have changed as we can see. It is because the axis have changed and the paremaeters of X and Y are also changed and the correlation coefficient value is not equal to 1.

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