Question

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 3062 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.

(c) Is the normal approximation to the binomial justified in this problem? Explain.

Answer 1

(a)  Let p represent the proportion of all Colorado physicians who provide some charity care. Let X be the event of success (provide charity care).

Given that X = 3062, Sample size (n) = 5616 .Then the point estimate for p is given by:   

(b) 99% confidence interval for p can be obtained by the formula:

where, at . from normal tables,

We get Z_0.995 = 2.576 2.58

Substituting the values in the formula for interval,

=

Lower limit = 0.528

Upper limit = 0.562

We are 99% confident that the true proportion of all Colorado physicians who provide some charity care lie between 0.528 and 0.532. In other words, if repeated samples were taken and the 99% confidence interval is computed for each sample, 99% of the time, the true proportion would lie in the interval (0.528,0.562).

If we take 100 additional samples of physicians, the proportion of all Colorado physicians who provide some charity care would fall in the range (0.528,0.562) 99 times out of 100.

(c) The normal approximation to the binomial is justified in this problem, since:

- The samples taken are independent (trials are independent) - Sample size (n) is sufficiently large - The sample proportion, , a pretty good estimate of the true proportion is close to 1/2 = 0.5.

.