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EXAMPLE 4 Find a margin of error MEDIA SURVEY In a survey of 1011 people, 52% said that television is their main source of news. a.What is the margin of error for the survey? b.Give an interval that is likely to contain the exact percent of all people who use television as their main source of news.

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EXAMPLE 4 Find a margin of error SOLUTION a.Use the margin of error formula. Margin of error Write margin of error formula. + – 1011 1 = Substitute 1011 for n. 0.031 + – Use a calculator. ANSWER The margin of error for the survey is about 3.1%. + – + – n 1 =

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EXAMPLE 4 Find a margin of error b.To find the interval, subtract and add 3.1% to the percent of people surveyed who said television is their main source of news (52%). 52% – 3.1% 52% + 3.1% ANSWER It is likely that the exact percent of all people who use television as their main source of news is between 48.9% and 55.1%. = 48.9% = 55.1%

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EXAMPLE 5 Standardized Test Practice SOLUTION Use the margin of error formula. Margin of error Write margin of error formula. Substitute for margin error. 0.05 + – n = 400 Solve for n. 0.0025= 1 n Square each side. 0.05 + – = + – 1 n + – 1 n = There were 400 people surveyed. ANSWER The correct answer is C. A BCD

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GUIDED PRACTICE for Examples 4 and 5 INTERNET: In a survey of 1202 people, 11% said that they use the Internet or e-mail more than 10 hours per week. What is the margin of error for the survey? How many people would need to be surveyed to reduce the margin of error to 2%? + – 3. The margin of error for the survey is about 2.9%. + – 2500 people need to be surveyed. ANSWER

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