# 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

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 $99\mathrm{%}$ 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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Step 1
a)
Here,
The point estimate of the population proportion is calculated as follows:
$\stackrel{^}{p}=\frac{x}{n}$
$=\frac{3332}{5805}$
$=0.573987$
$\approx 0.5740$
Step 2
b)
Since the required confidence interval is $99\mathrm{%}$, the two-tailed z-critical value is 2.5758 using the Excel function $“=NORM.S.INV\left(0.995\right)\text{"}$.
The $99\mathrm{%}$ confidence interval for the population proportion is calculated as follows:
$CI=\stackrel{^}{p}±{z}_{\frac{\alpha }{2}}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}$
$=0.5740±\left(2.5758\right)\sqrt{0.5740\frac{1-0.5740}{5805}}$
$=0.5740±0.0167$
$=\left(0.5573,0.5907\right)$
$\approx \left(0.557,0.591\right)$
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:
$np=5805×\frac{3332}{5805}$
$=3332$
$>5$
$nq=5805×\left(1-\frac{3332}{5805}\right)$
$=2473$
$>5$
Thus, normal approximation to binomial is possible.
Yes,