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 5508 physicians in Colorado showed that 3089 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 :

Given that,

n = 5508

x = 3089

= x / n = 3086 /5508 = 0.561

1 - = 1 - 0.561 = 0.439

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 * (((0.561* 0.439) / 5508) = 0.017

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.561 - 0.017 < p < 0.561 + 0.017

0.544 < p < 0.578

The 99% confidence interval for the population proportion p is : ( 0.544, 0.578)

lower limit = 0.544

upper limit = 0.578