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 573 randomly selected Americans surveyed, 369 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 573 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

a)

Given n = 573 and x = 369

P = x/n = 0.6441

90% CI

alpha = 0.1

Zc = 1.645 (Excel: NORM.S.INV(α/2))

CI = P +/- Zc * SQRT(P(1-P)/n)

= 0.6441 +/- 1.645 * SQRT(0.6441*(1-0.6441)/573)

CI = (0.6112, 0.6770)

With 90% confidence the proportion of all Americans who favor the new Green initiative is between 0.6112 and 0.6770

b)

If many groups of 573 randomly selected Americans were surveyed,
then a different confidence interval would be produced from each
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who favor the Green initiative and about **10%**
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