Question

A newsgroup is interested in constructing a 90% confidence interval for the difference in the proportions of Texans and New Yorkers who favor a new Green initiative. Of the 582 randomly selected Texans surveyed, 428 were in favor of the initiative and of the 569 randomly selected New Yorkers surveyed, 456 were in favor of the initiative.

. With 90% confidence the difference in the proportions of Texans and New Yorkers who favor a new Green initiative is between (round to 3 decimal places) and (round to 3 decimal places).

b. If many groups of 582 randomly selected Texans and 569 randomly selected New Yorkers 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 the difference in the proportions of Texans and New Yorkers who favor a new Green initiative and about percent will not contain the true population difference in proportions.

Answer 1

Here, , n1 = 582 , n2 = 569
p1cap = 0.7354 , p2cap = 0.8014

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.7354 * (1-0.7354)/582 + 0.8014*(1-0.8014)/569)
SE = 0.0248

For 0.9 CI, z-value = 1.64
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.7354 - 0.8014 - 1.64*0.0248, 0.7354 - 0.8014 + 1.64*0.0248)
CI = (-0.107 , -0.025)

With 90% confidence the difference in the proportions of Texans and New Yorkers who favor a new Green initiative is between -0.107 and -0.025

b. If many groups of 582 randomly selected Texans and 569 randomly selected New Yorkers 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 the difference in the proportions of Texans and New Yorkers who favor a new Green initiative and about 10 percent will not contain the true population difference in proportions.