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During the NFL’s 2014 AFC championship game officials measured the air pressure

ID: 3314660 • Letter: D

Question

During the NFL’s 2014 AFC championship game officials measured the air pressure on game balls following a tip that one team’s balls were under inflated. We found that the 11 balls measured for the New England Patriots had a mean psi of 11.10 (well below the legal limit) and a standard deviation of 0.40. Patriot supporters could argue that the under inflated balls were due to the elements and other outside effects. To test this the officials also measured 4 balls from the opposing team (Indianapolis Colts) to be used in comparison and found a mean psi of 12.63, with a standard deviation of 0.12. There is no significant skewness or outliers in the data. Use the t-distribution to determine if the average air pressure in the New England Patriot’s balls was significantly less than the average air pressure in the Indianapolis Colt’s balls.

Let group 1 and group 2 be the air pressure in the New England Patriot’s balls and the air pressure in the Indianapolis Colt’s balls, respectively.

Calculate the relevant test statistic.

Round your answer to two decimal places.

the absolute tolerance is +/-0.5

the absolute tolerance is +/-0.0007

Explanation / Answer

The statistical software output for this problem is:

Two sample T summary hypothesis test:
1 : Mean of Population 1
2 : Mean of Population 2
1 - 2 : Difference between two means
H0 : 1 - 2 = 0
HA : 1 - 2 < 0
(without pooled variances)

Hypothesis test results:

Hence,

t - statistic = -11.36

P - value = 0.0000

Difference Sample Diff. Std. Err. DF T-Stat P-value 1 - 2 -1.53 0.13470507 12.923704 -11.358148 <0.0001
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