document.write( "Question 926883: it is known that the variance of a population equals 1936. A random sample of 121 has been taken from the population. there is a 0.95 probability that the sample mean will provide a margin of error of? \n" ); document.write( "

Algebra.Com's Answer #562544 by Theo(13342) $\"\"$ $\"About$

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variance = 1936
\n" ); document.write( "standard deviation = square root of variance = 44
\n" ); document.write( "standard error = standard deviation divided by square root of sample size.
\n" ); document.write( "sample size = 121
\n" ); document.write( "square root of sample size = 11
\n" ); document.write( "standard error = 44/11 = 4
\n" ); document.write( "critical z factor for a confidence interval of .95 is equal to plus or minus 1.96.
\n" ); document.write( "margin of error equals critical z factor * standard error = 4 * 1.96 = 7.84
\n" ); document.write( "confidence interval equals mean plus or minus margin of error equals mean plus or minus 7.84.
\n" ); document.write( "here's a reference that has an example that will help you to understand better if you need to.
\n" ); document.write( "http://stattrek.com/estimation/margin-of-error.aspx\r
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