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Owner: jkernen40
Created: Dec 14, 2017
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Scout.com Recruiting Rankings Project Part 4
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Data set 1. Scout.com Rankings of 2010 College Football Recrui   [Info]
To analyze this data, please sign in.

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. 

Result 1: real Z test   [Info]
One sample Z hypothesis test:
μ : Mean of variable
H0 : μ = 15
HA : μ ≠ 15
Standard deviation not specified.

Hypothesis test results:
VariablenSample MeanStd. Err.Z-StatP-value
Commits12019.30.482842898.9055883<0.0001

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

Result 2: Contingency table w/ Commits and Avg. Star Rating   [Info]
Contingency table results:
Rows: Avg Star Rating
Columns: Commits
5 to 1010 to 1515 to 2020 to 2525 to 30Total
1 to 1.5010001
1.5 to 248510229
2 to 2.5151110835
2.5 to 313512122
3 to 3.500710724
3.5 to 4004037
4 to 4.5010102
Total618324321120

Chi-Square test:
StatisticDFValueP-value
Chi-square2440.7995710.0175
Warning: over 20% of cells have an expected count less than 5.
Chi-Square 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. 

Result 3: One Way ANOVA w/ 3 variables   [Info]
Analysis of Variance results:
Data stored in separate columns.

Column statistics
ColumnnMeanStd. Dev.Std. Error
Commits12019.35.28927880.48284289
Avg Star Rating1202.51108330.63578020.058038526
Top 100 Recruits1200.7251.77263280.16181849

ANOVA table
SourceDFSSMSF-StatP-value
Columns225203.53812601.7691199.2961<0.0001
Error3573751.226810.507638
Total35928954.764

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: F-Stat = 1199.2961

Conclusion: With a P-value of <0.0001, we would reject the null for not meeting the .05 significance level

Extra Credit #1-T Stat

Result 4: real T test   [Info]
One sample T hypothesis test:
μ : Mean of variable
H0 : μ = 15
HA : μ ≠ 15

Hypothesis test results:
VariableSample MeanStd. Err.DFT-StatP-value
Commits19.30.482842891198.9055883<0.0001

Variable: Number of Commitments

Null: μ = 15

Alternative: μ  15

Statistic: T-Statistic = 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 #2-Contingency Table

Result 6: Contingency table (with data) EC   [Info]
Contingency table results:
Rows: Commits
Columns: 5 star Players
012345Total
5 to 106000006
10 to 15170000118
15 to 20245210032
20 to 25364200143
25 to 30152111121
Total98115213120

Chi-Square test:
StatisticDFValueP-value
Chi-square2015.6116120.7404
Warning: over 20% of cells have an expected count less than 5.
Chi-Square 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

Result 5: One Way ANOVA commits and avg star rating   [Info]
Analysis of Variance results:
Data stored in separate columns.

Column statistics
ColumnnMeanStd. Dev.Std. Error
Commits12019.35.28927880.48284289
Avg Star Rating1202.51108330.63578020.058038526

ANOVA table
SourceDFSSMSF-StatP-value
Columns116912.06316912.0631191.8008<0.0001
Error2383377.301814.190344
Total23920289.365

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 P-value of <0.0001, we would reject the null for not meeting the .05 significance level

Extra Credit #4-Further 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.

HTML link:
<A href="https://www.statcrunch.com/5.0/viewreport.php?reportid=74593">Scout.com Recruiting Rankings Project Part 4</A>

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By xg15
Dec 14, 2017

In hypothesis testing, we are testing the population mean.

Always Learning