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 5913 physicians in Colorado showed that 3035 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

Answer 1

Solution :

n = 5913

x = 3035

a ) = x / n = 3035 / 5913 = 0.513

1 - = 1 - 0.513 = 0.487

b ) At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.513 * 0.487) / 5913 )

= 0.017

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.513 - 0.017 < p < 0.513+ 0.017

lower limit = 0.496

upper limit = 0.530