Do the warm Miami temperatures help the punter for the Miami Dolphins kick the b
ID: 3154491 • Letter: D
Question
Do the warm Miami temperatures help the punter for the Miami Dolphins kick the ball farther than the punter for the Green Bay Packers, who typically punts in frigid Wisconsin temperatures? That is, can a warm football in general be punted farther than a cold football? To investigate this question, from 20 similar footballs inflated with the same air pressure, 10 were randomly selected to be placed in a refrigerator for 1 hour and the other 10 were placed for 1 hour in a room with a temperature of 85 degrees . After being heated or cooled for 1 hour each ball was placed in an identical insulated container to keep its temperature as constant as possible. An assistant then randomly selected a ball and handed it to a punter from an NFL team who had volunteered for the project. The punter wore gloves so he could not feel the temperature of the ball he was punting. The yards traveled by each of the 20 punted balls was recorded (the punter was allowed to rest for 5 minutes after each set of 4 kicks so that punter fatigue would not be a factor). The distances in yards traveled by the 20 punts are shown below.
Warm Footballs 44 53 48 49 62 58 64 54 56 55
Cold Footballs 46 46 54 47 57 44 51 51 52 46
Do the data suggest that warm footballs can be punted farther than cold footballs?
Question 1. To answer this question use the data and your calculator or statcrunch to determine a 95% confidence interval for WC, where W is the mean distance traveled by warm footballs when they are punted and C is the mean distance traveled by cold footballs when they are punted (use 4 decimal places in your answer) lower limit of confidence interval upper limit of confidence interval
Explanation / Answer
Groups are independent,
SE(wbar-cbar)=sqrt [sw^2/nw+sc^2/nc], where wbar, cbar refres to mean yards travvelled in warm and colfd football, sw and cw refer to standard deviations corresponding to warm and cold foot ball, and nw and nc refer to corresponding sample size.
=sqrt [6.201254^2/10+4.221637^2/10]
=2.3723
df=15, t critical at alpha=0.05 is 2.131
95% C.I=(wbar-cbar)+-2.131*SE(wbar-cbar)
=(54.3-49.4)+-2.131*2.3723
=-0.1554 to 9.9553
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