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 5535 physicians in Colorado showed that 3386 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.

1% of all confidence intervals 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.

(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.    No; np < 5 and nq > 5.Yes; np > 5 and nq > 5

Answer 1

Solution :

Given that,

n = 5535

x = 3386

Point estimate = sample proportion = = x / n = 3386/5535=0.6117

1 -   = 1-0.6117 =0.3883

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

  Margin of error = E = Z/2   * (((( * (1 - )) / n)

= 2.576* (((0.6117*0.3883) /5535 )

= 0.017

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.6117-0.017 < p < 0.6117+ 0.017

0.595< p < 0.629

The 99% confidence interval for the population proportion p is :lower limit= 0.595,upper limit= 0.629