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Margin of Error proportion

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In a local school district it was found that in a random sample of 1550 students, 785 of those students brought their own lunch to school. What is the margin of error for the 99% confidence interval estimate for the true percentage of students who bring their lunch to school. Answer as a percent rounded to the nearest tenth.

Apr 12th, 2015

the proportion of students in the sample who brought their own lunch is 785/1550 = 0.5065

the margin of error for the 99% confidence interval is  0.033 or 3.3%

Apr 12th, 2015

If you need to know how I got that answer, let me know.

Apr 12th, 2015

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Apr 12th, 2015
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Apr 12th, 2015
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