Question

8. Run a regression analysis on the following bivariate set of data with y as the response variable.

x	y
27.2	68.2
28.1	66.7
28.7	64.6
30.2	66.4
33.7	69.5
31.8	68.3
30.4	67.8
28.6	65.5
32.5	69.4
34.8	67.9
33.3	67.1
28	66.1

- Find the correlation coefficient and report it accurate to three decimal places.
r =

- What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.)
r² = _____%

- Based on the data, calculate the regression line (each value to three decimal places)

y = ____ x + _____

- Predict what value (on average) for the response variable will be obtained from a value of 33.4 as the explanatory variable. Use a significance level of α=0.05α0.05 to assess the strength of the linear correlation.

What is the predicted response value? (Report answer accurate to one decimal place.)
y = ____

Answer 1

X Values
∑ = 367.3
Mean = 30.608
∑(X - M_x)² = SS_x = 71.969

Y Values
∑ = 807.5
Mean = 67.292
∑(Y - M_y)² = SS_y = 24.849

X and Y Combined
N = 12
∑(X - M_x)(Y - M_y) = 24.541

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

r = 24.541 / √((71.969)(24.849)) = 0.580

r^2=0.336

So r^2=33.6%

Sum of X = 367.3
Sum of Y = 807.5
Mean X = 30.6083
Mean Y = 67.2917
Sum of squares (SS_X) = 71.9692
Sum of products (SP) = 24.5408

Regression Equation = ŷ = bX + a

b = SP/SS_X = 24.54/71.97 = 0.341

a = M_Y - bM_X = 67.29 - (0.34*30.61) = 56.855

ŷ = 0.341X + 56.855

For x=33.4, ŷ = (0.341*33.4) + 56.855=68.2