Question

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 593 randomly selected Americans surveyed, 350 were in favor of the initiative. Round answers to 4 decimal places where possible.

a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between ? and ? .

b. If many groups of 593 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About ? percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about ? percent will not contain the true population proportion.

Answer 1

Solution :

Given that,

n = 593

x = 350

a) Point estimate = sample proportion = = x / n = 350 / 593 = 0.5902

1 - = 1 - 0.5902 = 0.4098

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 (((0.5902 * 0.4098) / 593)

= 0.0332

A 90% confidence interval for population proportion p is ,

± E

= 0.5902  ± 0.0332

= ( 0.5570, 0.6234 )  

b. If many groups of 593 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About 90 percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about 10 percent will not contain the true population proportion