In: Statistics and Probability
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 |
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 _