Question

A newsgroup is interested in constructing a 99% confidence interval for the difference in the proportions of Texans and New Yorkers who favor a new Green initiative. Of the 517 randomly selected Texans surveyed, 363 were in favor of the initiative and of the 526 randomly selected New Yorkers surveyed, 464 were in favor of the initiative. Round to 3 decimal places where appropriate.

If the assumptions are met, we are 99% confident that the difference in population proportions of all Texans who favor a new Green initiative and of all New Yorkers who favor a new Green initiative is between _and _?.

If many groups of 517 randomly selected Texans and 526 randomly selected New Yorkers were surveyed, then a different confidence interval would be produced from each group. About _ % of these confidence intervals will contain the true population proportion of the difference in the proportions of Texans and New Yorkers who favor a new Green initiative and about _ %will not contain the true population difference in proportions?

Answer 1

Pooled proportion, p = (363 + 464) / (517 + 526) = 0.7929051

Standard error of difference in proportions, SE =

= 0.0250957

Z value for 99% confidence interval is  2.576

Sample proportions are,

p1 = 363/517 = 0.7021277

p1 = 464/526 = 0.8821293

Point estimate of difference in proportions, p1 - p2 = 0.7021277 - 0.8821293 = -0.180

99% confidence interval for the difference in the proportions of Texans and New Yorkers who favor a new Green initiative

(-0.180 - 2.576 * 0.0250957 , -0.180 + 2.576 * 0.0250957)

(-0.245 , -0.115)

About 99 % of these confidence intervals will contain the true population proportion of the difference in the proportions of Texans and New Yorkers who favor a new Green initiative and about 1 %will not contain the true population difference in proportions