Question

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From 1998 to 2008, there were 11,590 NBA regular season games. Home teams won 7,021 games while losing 4,569 games.

a. Use these statistics to construct a 95% confidence interval for the proportion of NBA games won by the home team.

​​​​​​​b. (.59688 , .61468)

I am 95% confident that the true confidence interval is between (.59688 , .61468).

c. ​​​​​​​Technical question: Is this an independent data set?

d.​​​​​​​ Should we trust our margin of error here?

Answer 1

a.
TRADITIONAL METHOD
given that,
possible chances (x)=7021
sample size(n)=11590
success rate ( p )= x/n = 0.60578
I.
sample proportion = 0.60578
standard error = Sqrt ( (0.60578*0.39422) /11590) )
= 0.00454
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.00454
= 0.0089
III.
CI = [ p ± margin of error ]
confidence interval = [0.60578 ± 0.0089]
= [ 0.59688 , 0.61468]
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DIRECT METHOD
given that,
possible chances (x)=7021
sample size(n)=11590
success rate ( p )= x/n = 0.60578
CI = confidence interval
confidence interval = [ 0.60578 ± 1.96 * Sqrt ( (0.60578*0.39422) /11590) ) ]
= [0.60578 - 1.96 * Sqrt ( (0.60578*0.39422) /11590) , 0.60578 + 1.96 * Sqrt ( (0.60578*0.39422) /11590) ]
= [0.59688 , 0.61468]
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interpretations:
1. We are 95% sure that the interval [ 0.59688 , 0.61468] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
b.
95% confidence interval for the proportion of NBA games won by the home team is [ 0.59688 , 0.61468]
c.
yes,
an independent data set because it is random sample
d.
margin of error = 0.0089