Question

During a national debate on changes to health care, a cable news service performs an opinion poll of 630 small-business owners. It shows that 78% of small-business owners do not approve of the changes. (Round your answers to 4 decimal places.)

1. Develop a 98% confidence interval for the proportion opposing health care changes. Confidence interval for the proportion: ___ and ___
2. Do majority approve? Yes or No

Answer 1

Solution :

Given that,

n = 630

Point estimate = sample proportion = =0.78

1 -   = 1-0.78 =0.22

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.78*0.22) / 630)

E = 0.0384

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.78-0.0384 < p < 0.78+0.0384

0.7416< p < 0.8184

The 98% confidence interval for the population proportion p is : 0.7416, 0.8184, yes