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 525 randomly selected Texans surveyed, 400 were in favor of the initiative and of the 569 randomly selected New Yorkers surveyed, 474 were in favor of the initiative. Round to 3 decimal places where appropriate. If the assumptions are met, we are 90% 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 525 randomly selected Texans and 569 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

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**Answer:**

p1 = 400/525 = 0.7619

p2 = 474/569 = 0.833

CI = -0.07114 +- 1.645(sqrt(0.7619*0.2381/525 + 0.833*0.167/569)

= -0.07114+-0.03995593

= **-0.1111<
p< -0.03118**

b)

**Blank 1 -
90%**

**Blank 2 -
10%**

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