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How big is the home field advantage in the National Football League (NFL)? To investigate, we select a sample of 80 games from the 2011 regular season1 and find the home team scored an average of 25.16 points with a standard deviation 10.14 points. In a separate sample of 80 different games, the away team scored an average of 21.75 points with a standard deviation of 10.33 points. Use this summary information to find a 90%c onfidence interval for the mean home field advantage, µH- µA, in points scored.

The 90% confidence interval is______ to________ .

Respuesta :

kendrich

Answer:

0.74 to 6.06

Step-by-step explanation:

The groups are independnet,

SE(xh bar-xa bar)=sqrt [sh^2/nh+sa^2/na]=sqrt [10.1^2/80+10.3^2/80]=1.61

At df=157, the t critical is 1.65

90%c.i=(xh bar-xa bar)+-tcritical SE(xh bar-xa bar)

=(25.2-21.8)+-1.65*1.61

=0.74 to 6.06

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