Question

The following table compares the completion percentage and interception percentage of 5 NFL quarterbacks.

Completion Percentage	57	57	57	57	60
Interception Percentage	4.9	4.3	2.9	2.7	1.4

Step 1 of 5:

Calculate the sum of squared errors (SSE). Use the values b0=47.4019 and b1=−0.7667 for the calculations. Round your answer to three decimal places.

Step 2 of 5:

Calculate the estimated variance of errors, s2e . Round your answer to three decimal places.

Step 3 of 5:

Calculate the estimated variance of slope, s2b1 . Round your answer to three decimal places.

Step 4 of 5:

Construct the 98% confidence interval for the slope. Round your answers to three decimal places.

Step 5 of 5:

Construct the 90% confidence interval for the slope. Round your answers to three decimal places.

Answer 1

Let

Y=Interception Percentage

X=Completion Percentage

The regression model that we want to estimate is

where is the intercept and is the slope and is a random disturbance

The estimated regression line (using the values b0=47.4019 and b1=−0.7667) is

Step 1 of 5

The sum of square error SSE is

The calculation is below

Completion Percentage (X)	Interception Percentage (Y)	Predicted Y	Error-square
57	4.9	47.4019-0.7667*57=3.7	(4.9-3.7)^2=1.44
57	4.3	47.4019-0.7667*57=3.7	(4.3-3.7)^2=0.36
57	2.9	47.4019-0.7667*57=3.7	(2.9-3.7)^2=0.64
57	2.7	47.4019-0.7667*57=3.7	(2.7-3.7)^2=1
60	1.4	47.4019-0.7667*60=1.3999	(1.4-1.3999)^2=0
		SSE	3.4400

ans SSE = 3.44

Step 2 of 5

n=5 is the number of observations

The mean square error is

ans: The  estimated variance of errors,

step 3 of 5

Mean of X is

The sum of square X is

The  estimated variance of slope, is

ans: The  estimated variance of slope, is

step 4 of 5

98% confidence interval indicates a significance level of

The critical value of t using degrees of freedom n-2=5-2=3, for

Using t tables we get

The 98% confidence interval for the slope is

ans: the 98% confidence interval for the slope is [-2.579, 1.045]

step 5 of 5

90% confidence interval indicates a significance level of

The critical value of t using degrees of freedom n-2=5-2=3, for

Using t tables we get

The 90% confidence interval for the slope is

ans: the 90% confidence interval for the slope is [-1.706, 0.172]