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 506 randomly selected Texans surveyed, 436 were in favor of the initiative and of the 525 randomly selected New Yorkers surveyed, 459 were in favor of the initiative.

a. 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 506 randomly selected Texans and 525 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

Solution:

Given that n = 506, x = 436

a. Then the 90% confidence interval for the true proportion (p) is
p̂ ± Zα/2 * sqrt(p̂(1-p̂)/n) where,
p̂ = x/n = 436/506 = 0.8617
q-hat = 1- p̂ = 1- 0.8617 = 0.1383 and
Level of significance α = 1-0.90 = 0.10
Z_α/2 = Z_0.05 = 1.645
Confidence interval CI = p̂ ± Zα/2 * sqrt(p̂(1-p̂)/n)
= 0.8617 ± 1.645 *sqrt(0.8617 (1-0.8617 )/506)
= 0.8617 ± 0.0252
= [0.837, 0.887]
With 90% confidence the difference in the proportions of Texans and New Yorkers who favor a new Green initiative is between 0.837 and 0.887

b. If many groups of 506 randomly selected Texans and 525 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.