Below are the 3-point shooting performances for some of the top 3-point shooters
ID: 3021835 • Letter: B
Question
Below are the 3-point shooting performances for some of the top 3-point shooters in the NBA during the 2014-2015 NBA season.
(1)Player (2)Field Goals Attempted (3)Field Goals Made
(1)Stephen Curry (2)646 (3)286
(1)KyleKorver (2)449 (3)221
(1)James Harden (2)555 (3)208
(1)J.J. Redick (2)458 (3)200
(1)Klay Thompson (2)545 (3)239
(1)Damian Lillard (2)572 (3)196
(1)Trevor Ariza (2)555 (3)194
(1)Danny Green (2)457 (3)191
(1)Robert Covington (2)446 (3)167
(1)Wesley Matthews (2)445 (3)173
For a given player, the outcome of a particular shot can be modeled as a Bernoulli random variable:if Yi is the outcome of shot i, then Yi = 1 if the shot is made and Yi =0 if the shot is missed. Let p denote the probability of making any particular 3-pt shot attempt.The natural estimator of p is W= FGM/FGA.
a) Estimate p for Stephen Curry
b) Find the standard deviation of the estimator W in terms of p and the number of shot attempts n.
c) The asymptotic distribution of (W-p)/se(W) is standard normal, where se(W)= (W(1 W)/n)^1/2. Use this fact to test H0:p=0.5 against H1:p<0.5 for Stephen Curry. Use a 1% significance level.
Explanation / Answer
a)
W(Curry) = 286/646 = 0.442724458 [ANSWER]
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b)
s(P) = sqrt(p(1-p)/n) [answer]
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c)
Formulating the null and alternatuve hypotheses,
Ho: p >= 0.5
Ha: p < 0.5
As we see, the hypothesized po = 0.5
Getting the point estimate of p, W,
W = x / n = 286/646 = 0.442724458
Getting the standard error of p^, sp,
sp = sqrt[po (1 - po)/n] = 0.019672237
Getting the z statistic,
z = (W - po)/sp = -2.911491059
As this is a 1 tailed test, then, getting the p value,
p = 0.001798541
significance level = 0.01
As P < 0.01, we REJECT THE NULL HYPOTHESIS.
Hence, there is significant evidence that p < 0.5 for Stephen Curry. [CONCLUSION]
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