Question

In: Statistics and Probability

The table below gives the completion percentage and interception percentage for five randomly selected NFL quarterbacks....

The table below gives the completion percentage and interception percentage for five randomly selected NFL quarterbacks. Based on this data, consider the equation of the regression line, y^=b0+b1x, for using the completion percentage to predict the interception percentage for an NFL quarterback. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.

Completion Percentage 58 60 61 62 65
Interception Percentage 5 4.5 3.5 3 2.5

a) Find the estimated slope. Round your answer to three decimal places.

b) Find the estimated y-intercept. Round your answer to three decimal places.

c) Determine if the statement "All points predicted by the linear model fall on the same line" is true or false.

d) Find the estimated value of y when x=55.9. Round your answer to three decimal places.

e) According to the estimated linear model, if the value of the independent variable is increased by one unit, then the change in the dependent variable y^ is given by?

f) Find the value of the coefficient of determination. Round your answer to three decimal places.

Solutions

Expert Solution

a.

Sum of X = 306
Sum of Y = 18.5
Mean X = 61.2
Mean Y = 3.7
Sum of squares (SSX) = 26.8
Sum of products (SP) = -10.2

Regression Equation = ŷ = bX + a

b = SP/SSX = -10.2/26.8 = -0.381

b. a = MY - bMX = 3.7 - (-0.38*61.2) = 26.993

ŷ = -0.381X + 26.993

c.

Hence "All points predicted by the linear model fall on the same line" is False

d. For x=55.9, ŷ = -0.381*55.9+ 26.993=5.695

e. If the value of the independent variable is increased by one unit, then the change in the dependent variable y^ is given by Slope and its value is -0.381

f. First we will find r

X Values
∑ = 306
Mean = 61.2
∑(X - Mx)2 = SSx = 26.8

Y Values
∑ = 18.5
Mean = 3.7
∑(Y - My)2 = SSy = 4.3

X and Y Combined
N = 5
∑(X - Mx)(Y - My) = -10.2

R Calculation
r = ∑((X - My)(Y - Mx)) / √((SSx)(SSy))

r = -10.2 / √((26.8)(4.3)) = -0.950

So r^2=-0.950^2=0.903


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