document.write( "Question 930949: a random sample of 50 people was asked to keep a record of the amount of time spend watching tv in in a specified week. if the sample mean was 24.4 hours and the sample standard deviation was 7.4 hours, give a 95% confidence interval estimate for the average time spent watching tv by all members of the population that week. what is the margin of error? \n" ); document.write( "
Algebra.Com's Answer #565353 by ewatrrr(24785)\"\" \"About 
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sample mean was 24.4 hours and the sample standard deviation was 7.4 hours
\n" ); document.write( "ME = 1.96(7.4/sqrt(50) = 2.05 rounded
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\n" ); document.write( "24.4 ± 2 the confidence Interval estimate
\n" ); document.write( "(22.4, 26.4)
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