By Nicholas Lim, Kyle Maw, and Thomas Murray

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Presentation transcript:

By Nicholas Lim, Kyle Maw, and Thomas Murray

The Project For this project we set out to determine if the par for a mini-golf course is accurate. We chose to test the accuracy of the Monster Mini-Golf Course in Fairfield, New Jersey. To get accurate results we played to games of 18 holes to insure that luck did not play a factor.

The Test For our Data we decided to use a Chi-Square Goodness of Fit test to see how well par compared to our scores. We will use 17 degrees of Freedom as there are 18 holes on a mini golf course.

State: We want to perform a test of the hypotheses: H。: Monster Mini Golf’s stated par for this course is correct / p = 46 Ha : Monster Mini Golf’s stated par for this course is incorrect / p ≠ 46

Plan: If conditions are met we will perform a chi squared test for goodness of fit. Random - randomly selected mini golf course 10% - 18 holes of mini golf is less than 10% of all mini golf holes Large - Expected values are less than 5 so we will proceed with caution

Do: MAW Using Technology, the calculator’s X2 GOF - Test gives P = .9999 Df = 17 Hole: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Par/Expected: Observed #1: Observed #2:

Conclude: MAW Because my p-value, .9999, is greater than alpha = 0.05, I fail to reject the null hypothesis. I do not have convincing evidence that Monster Mini-Golf’s claim of par is incorrect.

Do: Murray Expected Total: 46 Observed 1 Total: 53 Hole: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Par/Expected: Observed #1: Observed #2: Expected Total: 46 Observed 1 Total: 53 Observed 2 Total: 53 Chi-Square: 10.333333 P value: .889 Degrees of Freedom: 17

Conclude: Murray Since my P-Value of .889 was bigger than the alpha of .05 I fail to reject the null hypothesis. I do not have convincing evidence that Monster Mini-Golf’s claim of par is incorrect.

Do: Lim Hole: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Par/Expected: Observed #1: Observed #2: Using Technology, the calculator’s X2 GOF - Test give X2 = 3.5142 P = .99978 Df = 17

Conclude: Lim Because my P-value, 0.99978, is greater than alpha which equals .05, we fail to reject the null hypothesis. We do not have convincing evidence that the monster mini golf’s claim of par is incorrect.

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Summary Thus in based on our information we cannot reject Monster Mini-Golf’s standards for Par. This means that the designated standard is fair to all golfers regardless of how good or bad they might be. Perhaps we just need to improve our mini-golf abilities.