Question

In: Statistics and Probability

Exercise 1. Defense and wins. The data in the table below shows the wining percentage and...

Exercise 1. Defense and wins. The data in the table below shows the wining percentage and the yards per game allowed in the 2014-15 season for a random sampling of teams. At α = 0.05, can you support the claim that there is a correlation between the winning percentage and the yards allowed?

Team

Winning percentage

Total yards

Baltimore Ravens

0.625

336.9

Cleveland Browns

0.438

366.1

Denver Broncos

0.750

305.2

Jacksonville Jaguars

0.188

370.8

New England Patriots

0.750

344.1

Oakland Raiders

0.188

357.6

Pittsburgh Steelers

0.688

353.4

Solutions

Expert Solution

The correlation coefficient between the wining percentage and total yard is obtained using the formula,

Team Winning percentage, Y Total yards, X X^2 Y^2 XY
Baltimore Ravens 0.625 336.9 113501.6 0.390625 210.5625
Cleveland Browns 0.438 366.1 134029.2 0.191844 160.3518
Denver Broncos 0.75 305.2 93147.04 0.5625 228.9
Jacksonville Jaguars 0.188 370.8 137492.6 0.035344 69.7104
New England Patriots 0.75 344.1 118404.8 0.5625 258.075
Oakland Raiders 0.188 357.6 127877.8 0.035344 67.2288
Pittsburgh Steelers 0.688 353.4 124891.6 0.473344 243.1392
Sum 3.627 2434.1 849344.6 2.251501 1237.968

Now, the hypothesis test for significance of the correlation coefficient is performed in following steps,

Step 1: The null and alternative hypotheses are,

Step 2: The significance level,

The critical value for the t-statistic for degree of freedom = n-2=5

Step 3: The t-statistic is obtained using the formula,

Step 4:

Since the t-statistic is less than t-critical value, the null hypothesis is not rejected. Hence there is no significant correlation is between percentage of wining and yard per game allowed.

ΠΣΧΥΗΣΧΣΥ) Γ vΠΙΣ ΣΧΕΙn ΣΥ ΣΥ

X2-849344.6, X 2434.1 Y 3.627 Y2 2.251501 XY 12237.968

ΙΣΧΥ ΣΧ ΣΥ) 162.7068 -0.70284 ν.ΣΧ ΣΧΡΙΙΣΥΣΥΡΙ 231.49856

Ho population correlation coefficient, p 0

HA population correlation coefficient, p0

0.05

teritical = 2.571 for 2-tailed

rVn 2 t V1-2

-0.70284 7- 2 -2.20933 1-(-0.70284)

t-statistic2.20933<teritical 2.571 _


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