Question

During a national debate on changes in health care, a cable news performs an opinion poll of 500 small business owners. It shows that 65% of the business owners do not approve of the changes. (a) What is the best estimate of the population proportion not in approval of these changes in health care? (b) Develop a 99% confidence interval around this population proportion. Comment on your result.

Answer 1

Solution :

Given that,

n = 500

(a)

= 65% = 0.65

1 - = 1 - 0.65 = 0.35

(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 = 2.575

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

= 2.576 * (((0.65 * 0.35) / 500)

= 0.055

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.65 - 0.055 < p < 0.65 + 0.055

0.595 < p < 0.705

(0.595 , 0.705)

A 99% confidence interval for population proportion p is : (0.595 , 0.705)