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 575 randomly selected A
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Of the 575 randomly selected A
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Question 1182846: 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 575 randomly selected Americans surveyed, 371 were in favor of the initiative.
Round to 4 decimal places where possible.
a. With 90% confidence the proportion of all Americans who are in favor the new Green initiative is between ________ and ________.
b. If many groups of 575 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 is 371/575=0.6452
90% half-interval is z(0.95)*sqrt(p hat(1-p hat)/n)
This is 1.645*sqrt*(0.645*0.355/575)
=0.0328
So the 90% interval is (0.6124, 0.6780)
and 90% of these intervals conducted with the same sample size would be expected to contain the parameter and 10% would not. We just wouldn't know for sure which 90.
