Question

of 11 ID: MST.SLR.REE.06.0020 [1 point] The following table shows the average petrol price and the number of online shopping orders over a given month: Petrol Price and Online Shopping Average Petrol Price Number of Online per gallon over a month ($) Shopping Orders 3.0325 3,776 1.7175 1,738 1,927 1.8925 4.61 1.9475 2.4825 2.1675 2.69 4,854 2,852 2,356 2,032 2,881 3.5625 4,404 1.49 2,364 The relationship between the average petrol price and the number of online shopping orders in a given month is proposed to follow the simple regression equation below: show variables û = bo + b2x Calculate the percentage of variability in the number of online shopping orders that is not explained by the average petrol price. Give your answer as a percentage to 1 decimal place. Percentage = %

Answer 1

(a)

Sum of X = 25.5925
Sum of Y = 29184
Mean X = 2.5593
Mean Y = 2918.4
Sum of squares (SS_X) = 8.2831
Sum of products (SP) = 8539.2705

Regression Equation = ŷ = bX + a

b = SP/SS_X = 8539.27/8.28 = 1030.92214

a = M_Y - bM_X = 2918.4 - (1030.92*2.56) = 280.01251

ŷ = 280.012+ 1030.92X

(b)

X Values
∑ = 25.592
Mean = 2.559
∑(X - M_x)² = SS_x = 8.283

Y Values
∑ = 29184
Mean = 2918.4
∑(Y - M_y)² = SS_y = 10480416.4

X and Y Combined
N = 10
∑(X - M_x)(Y - M_y) = 8539.27

R Calculation
r = ∑((X - M_y)(Y - M_x)) / √((SS_x)(SS_y))

r = 8539.27 / √((8.283)(10480416.4)) = 0.9165

r = 0.9165

The value of R², the coefficient of determination, is 0.8399

proportion = 1 - R²,

proportion = 1 - 0.8399

proportion = 0.1601 (multiple by 100 to convert percentage)

proportion = 16.0%

ANSWERED

PLEASE RATE ME POSITIVE

THANKS