A random sample of 5589 physicians in Colorado showed that 3359 provided at least some charity care (i., treated poor people at no cost).
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Question

A random sample of 5589 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 the confidence intervals created using this method would include the true proportion of Colorado physicians providing at least some charity care.

99% 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.

1% of all confidence intervals 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.

Yes; np > 5 and nq > 5.

No; np < 5 and nq > 5.

 Yes; np < 5 and nq < 5.

No; np > 5 and nq < 5.

Top Answer

a) p = 0.6010 b) lower limit=0.584 upper limit= 0.618 99% of the... View the full answer

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