My dataset is a population because it represents all of the schools with football teams at a Division 1 level. Since it's a population and not a sample, I created this one sample Z hypothesis test based on the "Number of commitments" variable.
One sample Z hypothesis test:
μ : Mean of variable H_{0} : μ = 15 H_{A} : μ ≠ 15 Standard deviation not specified. Hypothesis test results:

Variable:Number of Commitments
Null: μ=15
Alternative: μ≠15
Statistic: Z stat=8.9055883
Conclusion:With a P value of <0.0001 failing to meet the significance level of 0.05, I reject the null hypothesis
Contingency table results:
Rows: Avg Star Rating Columns: Commits
ChiSquare test:
ChiSquare suspect. 
Variables: Commits and Average Star Rating
Null:Number of commitments and average star rating are independent of each other, meaning they dont have a relationship
Alternative:Number of commitments and average star rating have a relationship
Statistic: Chi Squared statistic =40.799571
Conclusion: With a P value of 0.0175 and a significance level of 0.05, we reject the null because the P value doesn't exceed the significance level.
Analysis of Variance results:
Data stored in separate columns. Column statistics
ANOVA table

Variables: Commitments, Average Star Rating, and 5 Star Recruits
Null: μ Commitments=μ Average Star Rating=μ 5 Star Recruits
Alternative: At least one variable does not equal the others
Statistic: FStat = 1199.2961
Conclusion: With a Pvalue of <0.0001, we would reject the null for not meeting the .05 significance level
Extra Credit #1T Stat
One sample T hypothesis test:
μ : Mean of variable H_{0} : μ = 15 H_{A} : μ ≠ 15 Hypothesis test results:

Variable: Number of Commitments
Null: μ = 15
Alternative: μ ≠ 15
Statistic: TStatistic = 8.9055883
Conclusion: Obviously I wouldnt normally run this test, as my dataset is a population, but the resulting data came out exactly the same as it did when I ran the Z test. As with the Z test, the P value of <0.0001 fails to meet the significance level of 0.05, so I reject the null hypothesis
Extra Credit #2Contingency Table
Contingency table results:
Rows: Commits Columns: 5 star Players
ChiSquare test:
ChiSquare suspect. 
Variables: Commits and 5 Star Players
Null:Number of commitments and number of 5 Star Players are independent of each other, meaning they dont have a relationship
Alternative:Number of commitments and number of 5 star players have a relationship
Statistic: Chi Squared statistic =15.611612
Conclusion: With a P value of 0.7404 and a significance level of 0.05, we fail to reject the null because the P value is higher.
Extra Credit #3 ANOVA test
Analysis of Variance results:
Data stored in separate columns. Column statistics
ANOVA table

Variables: Number of commitments and average star rating
Null: μ Commitments=μ Average Star Rating
Alternative: μ commitments ≠ μ average star rating
Statistic: F Stat= 1191.8008
Conclusion:With a Pvalue of <0.0001, we would reject the null for not meeting the .05 significance level
Extra Credit #4Further Research Needed
With some many variables included in my dataset, I could easily continue my research and really delve into the details and statistics of college football recruiting. If I were to continue my research, I would start comparing my dataset to the same variables from different years to see what data would fluctuate and how much it would deviate from year to year. Additional variables would have to be taken into effect, such as changes in coaching staffs who are doing the recruiting and total number of recruits. I would be curious to see which teams are consistently the best recruiters, year in and year out.
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Dec 14, 2017
In hypothesis testing, we are testing the population mean.