In: Statistics and Probability
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.
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 = Z0.005 = 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)