A random sample of 5805 physicians in Colorado showed that 3332 provided at least some charity care (i.e., treated poor people at no cost). a) Let p r

lwfrgin 2020-11-27 Answered
A random sample of 5805 physicians in Colorado showed that 3332 provided at least some charity care (i.e., treated poor people at no cost).
a)
Let p represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for p. (Round your answer to four decimal places.)
b)
Find a 99% confidence interval for p. (Round your answer to three decimal places.)
lower limit =?
upper limit =?
c)
Is the normal approximation to the binomial justified in this problem? Explain
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Expert Answer

Brighton
Answered 2020-11-28 Author has 103 answers

Step 1
a)
Here, x=3332 and n=5805.
The point estimate of the population proportion is calculated as follows:
p^=xn
=33325805
=0.573987
0.5740
Step 2
b)
Since the required confidence interval is 99%, the two-tailed z-critical value is 2.5758 using the Excel function =NORM.S.INV(0.995)".
The 99% confidence interval for the population proportion is calculated as follows:
CI=p^±zα2p^(1p^)n
=0.5740±(2.5758)0.574010.57405805
=0.5740±0.0167
=(0.5573,0.5907)
(0.557,0.591)
Lower limit =0.557
Upper limit =0.591
Step 3
c)
The normal approximation to binomial is possible only if,
np>5
nq>5
Checking the conditions for normality:
np=5805×33325805
=3332
>5
nq=5805×(133325805)
=2473
>5
Thus, normal approximation to binomial is possible.
Yes, np>5 and nq>5

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