For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.
A random sample of 5616 physicians in Colorado showed that 3359 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 answers to three decimal places.) lower limit upper limit
Give a brief explanation of the meaning of your answer in the context of this problem.
99% of all confidence intervals would include the true proportion of Colorado physicians providing at least some charity care.
99% of the confidence intervals created using this method would include the true proportion of Colorado physicians providing at least some charity care.
1% of all confidence intervals would include the true proportion of Colorado physicians providing at least some charity care.
1% of the confidence intervals created using this method would include the true proportion of Colorado physicians providing at least some charity care.
(c)
Is the normal approximation to the binomial justified in this problem? Explain.
No; np < 5 and nq > 5.
Yes; np < 5 and nq < 5.
Yes; np > 5 and nq > 5.
No; np > 5 and nq < 5.
point estimate for p =0.5981
b)
99% confidence interval for p
lower limit =0.581 ,
upper limit= 0.615
99% of the confidence intervals created using this method would include the true proportion of Colorado physicians providing at least some charity care.
c)
Yes; np > 5 and nq > 5.
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