Damage Y in M 80 60 40 25 70 10 100 5 50 90 Distance X in Km 20 15 12 8 16 6 30

Damage y in m 80 60 40 25 70 10 100 5 50 90 distance

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Damage (Y) in ($M) 80 60 40 25 70 10 100 5 50 90 Distance (X) in (Km) 20 15 12 8 16 6 30 2 10 25 Using a linear regression equation of the form ࠵? = ࠵? + ࠵?࠵? + ࠵? i) Find the values for ( ࠵? ) and ( ࠵? ) ii) Interpret your result. To solve the questions you need to look at the formula above and know what you need to include in the table. Y X XY X 2 Y 2 80 20 1600 400 6400 60 15 900 225 3600 40 12 480 144 1600 25 8 200 64 625 70 16 1120 256 4900 10 6 60 36 100 100 30 3000 900 10000 5 2 10 4 25 50 10 500 100 2500 90 25 2250 625 8100 530 144 10120 2754 37850 NB: The last column figures are the summations.
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∑ ࠵? = 530 ; ∑ ࠵? = 144 ; ∑ ࠵?࠵? = 10120 ; ∑ ࠵? 7 = 2754 ; ࠵? : = EFG HG = 53 ; ࠵? : = HII HG = 14.4 and ࠵? = 10 i) From the slope equation ࠵? = ࠵? ∑ ࠵?࠵? − ∑ ࠵? ∑ ࠵? ࠵? ∑ ࠵? 7 − (∑ ࠵?) 7 ࠵? = 10(10120) − (144)(530) 10(2754) − (144) 7 = 3.66 From ࠵? = ࠵? N − ࠵?࠵? N ࠵? = 53 − (3.66)(14.4) = 0.3 This means that the regression equation becomes ࠵? = 0.3 + 3.66࠵? Note: the error term does not appear again after estimation. ii) For the interpretation of the slope. The slope (b) is 3.66 Slope: For any 1 kilometer increase in distance from the fire station, the amount of fire damage increases by 3.66 (in millions) dollars This is because there is a positive (+) relationship between the distance (X) and amount of damage caused. OR A unit change in X will lead to a change in Y by 3.66 units in the same direction (because the sign is positive). For the intercept: When distance is zero (when fire occurs at fire station) the damage is approximately zero (0.3 dollars)
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NB: Assuming I want to predict the amount of damage caused when the distance from the fire station is 40km, I just substitute for ࠵? = 40 into the estimated equation. ࠵? = 0.3 + 3.66(40) = 146.7 ࠵?࠵?࠵?࠵?࠵?࠵?࠵? Coefficient of Determination (R 2 ) It is a measure of the proportion of variability in the dependent variable “explained” by the regression relationship. Thus, it gives the proportion of the in the dependent variable that is explained by the variation in the independent variable. The coefficient of determination has a value that lies between 0 and 1 or 0 to 100%. It is obtained by the formula ࠵? 7 = W ∑ ;XY ∑ =;Z<; : [ ∑ ; [ Z<; : [ OR ࠵? 7 = [∑(=Z= :)(;Z; :)] [ ∑(=Z= :) [ ∑(;Z; :) [ Where ࠵? : = ∑ ; < is the mean value of Y and ࠵? : = ∑ = < is the mean value of X For example, an ࠵? 7 with the value 0.63 is read as “about 63% of the variations in Y is as a result of variations in X. You could also say “Variations in X explains 63% of the variations in Y”. Another example, if you have an ࠵? 7 value of 0.28 which is same as 28%, it is read as “Variations in X explains (or account for) 28% of the variations in Y” or “28% of the variations in Y is as a result of variations in X”. From the example above, A recent research by the Alabama fire services found that, the extent of damage caused during fire outbreak depends on the distance between fire station and the place of occurrence. A sample has been collected from various locations within the Alabama Metropolis to confirm this research. The table below summarizes the data. Damage (Y) in ($M) 80 60 40 25 70 10 100 5 50 90 Distance (X) in (Km) 20 15 12 8 16 6 30 2 10 25 Calculate the coefficient of determination and interpret it.
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Solution Y X XY X 2 Y 2 80 20 1600 400 6400 60 15 900 225 3600 40 12 480 144 1600 25 8 200 64 625 70 16 1120 256 4900 10 6 60 36 100 100 30 3000 900 10000 5 2 10 4 25 50 10 500 100 2500 90 25 2250 625 8100 530 144 10120 2754 37850 NB: The last column figures are the summations.
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