Question

The following table compares the completion percentage and interception percentage of 55 NFL quarterbacks.Completion Percentage Interception Percentage56 4.658 4.659 3.859 3.262 1.4Step 1 of 5 : Calculate the sum of squared errors (SSE). Use the values b0=36.9243b0=36.9243 and b1=−0.5681b1=−0.5681 for the calculations. Round your answer to three decimal places.Step 2 of 5: Calculate the estimated variance of errors, se2. 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 99% confidence interval for the slope. Round your answers to three decimal places.Step 5 of 5: Construct the 98% confidence interval for the slope. Round your answers to three decimal places.

Answer 1

Answer:

Step-by-step explanation:

Hello!

The two study variables you need to compare are:

Y: Completion percentage of an NFL quarterback

X: Interception percentage of an NFL quarterback

The following table compares the completion percentage and interception percentage of 55 NFL quarterbacks.

Completion Percentage 56 58 59 59 62

Interception Percentage 4.6 4.6 3.8 3.2 1.4

Sumatories:

∑X= 17.6

∑X²= 68.96

∑Y= 294

∑Y²= 17306

Step 1 of 5 : Calculate the sum of squared errors (SSE).

Use the values a= b0=36.9243 and b= b1=−0.5681 for the calculations. Round your answer to three decimal places.

In ANOVA the Squared sum of errors (SSE) represents the distance between each observation and the estimated regression line.

Symbolically:

SEE= ∑(Yi - Ÿi)²

The work formula is:

SSE= ∑($Y_i - a - bX_i$)²

Noe using the given data you can calculate it for a=36.9243 b=−0.5681

SSE= $(56 - 36.9243 - (-0.5681*4.6))^2 + (58 - 36.9243 - (-0.5681*4.6))^2 + (59 - 36.9243 - (-0.5681*3.8))^2 + (59 - 36.9243 - (-0.5681*3.2))^2 + (62 - 36.9243 - (-0.5681*1.4))^2$

SSE= 2860.055

Step 2 of 5: Calculate the estimated variance of errors, se2. Round your answer to three decimal places.

The formula to calculate the Se² is:

$S_e^2= \frac{1}{n-2}[sumY^2 - \frac{(sumY)^2}{n} - b^2(sumX^2) - \frac{(sumX)^2}{n} ]$

$S_e^2= \frac{1}{5-2}[17306 - \frac{(294)^2}{5} - (−0.5681)^2( 68.96) - \frac{(17.6)^2}{5} ]$

Se²= 5.510

Step 3 of 5: Calculate the estimated variance of the slope, s2b1. Round your answer to three decimal places.

To calculate the variance of the slope you can use the following formula:

$S_b^2= \frac{S_e^2}{sumX^2-(\frac{(sumX)^2}{n} )}$

$S_b^2= \frac{5.510}{68.96-(\frac{(17.6)^2}{5} )}$

Sb²= 0.786

Step 4 of 5: Construct the 99% confidence interval for the slope. Round your answers to three decimal places.

To calculate the 99%CI for the slope you have to use the student t statistic:

$t= \frac{b - \beta }{\frac{Sb}{\sqrt{n} } }$ ~tₙ₋₂

$t_{n-2; 1-\alpha /2} = t_{3; 0.995} =5.841$

b ± $t_{n-2; 1-\alpha /2} * (\frac{S_b}{\sqrt{n} })$

−0.5681 ± $5.84 * (\frac{0.887}{\sqrt{5} })$

[-2.88; 1.75]

Step 5 of 5: Construct the 98% confidence interval for the slope. Round your answers to three decimal places.

$t= \frac{b - \beta }{\frac{Sb}{\sqrt{n} } }$ ~tₙ₋₂

$t_{n-2; 1-\alpha /2} = t_{3; 0.99} = 4.541$

b ± $t_{n-2; 1-\alpha /2} * (\frac{S_b}{\sqrt{n} })$

−0.5681 ± $4.54 * (\frac{0.887}{\sqrt{5} })$

[-2.37;1.23]

I hope it helps!