document.write( "Question 1202134: SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 86% confidence interval to 15 points, how many students should the administrator sample? Make sure to give a whole number answer.
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Algebra.Com's Answer #837128 by Theo(13342)\"\" \"About 
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mean is 1500
\n" ); document.write( "standard deviation is 300
\n" ); document.write( "confidence interval is .86
\n" ); document.write( "standard error = standard deviation / square root of sample size = s/sqrt(n)
\n" ); document.write( "s is the standard error
\n" ); document.write( "n is the sample size
\n" ); document.write( "they want to limit the margin of error to less than or equal 15 points.
\n" ); document.write( "z-score formula is z = (x-m)/s
\n" ); document.write( "x is the sample mean
\n" ); document.write( "m is the population mean
\n" ); document.write( "s is the standard error
\n" ); document.write( "(x-m) is the margin of error
\n" ); document.write( "with (x-m) equal to 15, the formula becomes:
\n" ); document.write( "z = 15 / s
\n" ); document.write( "at 86% confidence interval, the critical z-score is plus or minus 1.475791028.
\n" ); document.write( "the formula becomes:
\n" ); document.write( "1.475791028 = 15 / s
\n" ); document.write( "s = standard error = standard deviation / square root of sample size = 300/ sqrt(n).
\n" ); document.write( "the formula becomes:
\n" ); document.write( "1.475791028 = 15 / (300 / sqrt(n))
\n" ); document.write( "simplify to get:
\n" ); document.write( "1.475791028 = 15 / 300 * sqrt(n)
\n" ); document.write( "solve for sqrt(n) to get:
\n" ); document.write( "sqrt(n) = 1.475791028 * 300 / 15 = 29.51582055.
\n" ); document.write( "when sqrt(n) = 29.51582055, s = 300 / 29.51582055 = 10.16404065.
\n" ); document.write( "when s = 10.16404065, the z-score formula becomes 1.475791028 = (x-m) / 10.16404065
\n" ); document.write( "solve for (x-m) to get:
\n" ); document.write( "(x-m) = 1.475791028 * 10.16404065 = 15.
\n" ); document.write( "margin of error = 15 is confirmed to be accurate using the normal distribution calculator at https://www.hackmath.net/en/calculator/normal-distribution
\n" ); document.write( "the results are shown below.
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\n" ); document.write( "the mean was input as 1500.
\n" ); document.write( "the standard error was input as 10.16404065.
\n" ); document.write( "the square root of n was equal to 29.51582055.
\n" ); document.write( "you can see that the margin of error was plus or minus 15 because the 86% confidence interval was from 1485 to 1515.
\n" ); document.write( "solve for n to get n = sqrt(n) squared = 871.183663.
\n" ); document.write( "round that to the next higher integers to get n = 872.
\n" ); document.write( "that would be the minimum sample size for the margin of error to be less than 15.\r
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