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TWO SAMPLE PROPORTION z
Amos Tversky and Thomas Gilovich in their study on the “Hot Hand” in basketball found that in a random sample of games, Larry Bird hit a second free throw on 48 of 53 attempts after the first free throw was missed, and hit a second free thrown in 251 of 285 attempts after the first free throw was made. Is there sufficient evidence to say that the probability that Bird will make a second free throw is different depending on whether or not he made the first throw?

Respuesta :

A = he hits first free throw
B = he hits second free throw

P ( B|A')   = 48/53
P (B|A)    = 251/285 = 0.881

P(A) = 285/(53 + 285) = 285/338  = 0.843
P (B) = (251 + 48)/338  = 299/338 = 0.885

Hope this helps

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