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 5400 physicians in Colorado showed that 2927 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 all confidence intervals would include the true proportion of Colorado physicians providing at least some charity care.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.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.No; np < 5 and nq > 5.    Yes; np < 5 and nq < 5.Yes; np > 5 and nq > 5.

Answer 1

Solution:

Given:

n = sample size = Number of physicians in Colorado selected in sample = 5400

x = Number of physicians in Colorado provided at least some charity care = 2927

Part a) p = the proportion of all Colorado physicians who provide some charity care.

Find a point estimate for p.

Part b) Find a 99% confidence interval for p.

where

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

Thus

and

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.  

Since confidence interval is an interval estimate for population parameter.

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

n*p = 5400 *0.5420 = 2927 > 5 and n*q = n* (1-p) = 5400 * (1-0.5420) = 2473 > 5

Thus correct answer is:

Yes; np > 5 and nq > 5.