SOLUTION: 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 566 randomly selected Am
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-> SOLUTION: 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 566 randomly selected Am
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Question 1182849: 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 566 randomly selected Americans surveyed, 389 were in favor of the initiative.
Round to 2 decimal places where possible
b. If many groups of 566 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 by Boreal(15235) (Show Source):
You can put this solution on YOUR website! p hat=389/566=0.687
90% half-interval is z*sqrt(p*(1-p)/n)=1.645*sqrt(0.687*0.313/566)
=0.0321
the interval tis (0.66, 0.72)
For b. about 90% will contain the interval and about 10% will not.
