document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1139622>Question 1139622</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can someone help with explain how to properly solve this equation?<BR>\n" );
document.write( "People were polled on how many books they read the previous year. Initial survey results indicate that s=11.8\r<BR>\n" );
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document.write( "How many subjects are needed to estimate the mean number of books read the previous year within four books with 95​% ​confidence?\r<BR>\n" );
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document.write( "How many subjects are needed to estimate the mean number of books read the previous year within two books with 95​% ​confidence?<BR>\n" );
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document.write( "How many subjects are needed to estimate the mean number of books read the previous year within four books with 99​% ​confidence? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #760137 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=760137\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> i think it works like this.\r<BR>\n" );
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document.write( "i believe what you called s is the standard deviation.\r<BR>\n" );
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document.write( "i will assume that's true, since i don't think i can solve the problem otherwise.\r<BR>\n" );
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document.write( "you would use the z-score for 95% confidence level and 99% confidence level.\r<BR>\n" );
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document.write( "z-score for 95% confidence level is plus or minus 1.96.\r<BR>\n" );
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document.write( "z-score for 99% confidence level is plus or minus 2.576.\r<BR>\n" );
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document.write( "these are both rounded to 3 decimal digits, which is usually the detail you want to get to.\r<BR>\n" );
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document.write( "the z-score formula is z = (x - m) / s.\r<BR>\n" );
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document.write( "z is the z-score<BR>\n" );
document.write( "x is the actual score<BR>\n" );
document.write( "m is the actual mean<BR>\n" );
document.write( "s is the standard error, not to be confused with the standard deviation.\r<BR>\n" );
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document.write( "the standard error formula is s = standard deviation / square root of sample size.\r<BR>\n" );
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document.write( "in algebraic notation that i use, this would be shown as s = sd / sqrt(n).\r<BR>\n" );
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document.write( "with a standard deviation of 11.8, this becomes s = 11.8 / sqrt(n).\r<BR>\n" );
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document.write( "when you want to estimate the number of books within plus or minus 4 at 95% confidence level, the formula for z-score becomes as follows:\r<BR>\n" );
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document.write( "z = (x - m) / s becomes 1.96 = 4 / s\r<BR>\n" );
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document.write( "we are working with the high side of the z-score.<BR>\n" );
document.write( "since the normal distribution is symmetric, we don't really need to do the low side also, because finding n for the high side will also get you the same number for the low side as i will show you below.\r<BR>\n" );
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document.write( "the margin of error is equal to (x - m) which is 4 in this case.\r<BR>\n" );
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document.write( "our formula has become 1.96 = 4 / s.\r<BR>\n" );
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document.write( "since s = 11.8/sqrt(n), then the formula becomes 1.96 = 4 / (11.8/sqrt(n)).\r<BR>\n" );
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document.write( "this becomes 1.96 = 4 * sqrt(n) / 11.8\r<BR>\n" );
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document.write( "solve for sqrt(n) to get sqrt(n) = 1.96 * 11.8 / 4 = 5.782.\r<BR>\n" );
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document.write( "square both sides of sqrt(n) = 5.782 and you get n = 33.431524.\r<BR>\n" );
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document.write( "normally you would then round up to the next highest integer, but i'll leave it as is because that gets your answer to be a margin of error of exactly 4.\r<BR>\n" );
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document.write( "if you were to also work with the low side z-score, then the formula for z-score becomes as follows:\r<BR>\n" );
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document.write( "-1.96 = -4 * sqrt(n) / 11.8\r<BR>\n" );
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document.write( "solve for sqrt(n) to get sqrt(n) = -1.96 * 11.8 / -4 which becomes sqrt(n) = 5.782 which is the same as we got when working with the high side z-score.\r<BR>\n" );
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document.write( "bottom line is, if we want the margin of error to be plus or minus 4 books at 95% confidence interval, then we need a sample size of 33.431524.\r<BR>\n" );
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document.write( "from what i can see, the mean can be anything as long as the difference between the mean and the actual number of books is 4 books and the standard deviation is 11.81.\r<BR>\n" );
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document.write( "for example:\r<BR>\n" );
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document.write( "assume the mean is 10 books and the standard deviation is 11.8 and the sample size is 33.431524 which makes the standard error equal to 2.040816327.\r<BR>\n" );
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document.write( "at 95% confidence level, z = (x - m) / s becomes 1.96 = (x - 10) /  2.040816327.\r<BR>\n" );
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document.write( "solve for x to get x = 1.96 * 2.040816327 + 10 = 14 which is a difference of 4 from the mean.\r<BR>\n" );
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document.write( "on the low side, z = (x - m) / s becomes -1.96 = (x - 10) / 2.040816327.\r<BR>\n" );
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document.write( "solve for x to get x = -1.96 * 2.040816327 + 10 = 6 which is a difference of 4 from the mean.\r<BR>\n" );
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document.write( "assume the mean is 5000 books and the standard deviation is 11.8 and the sample size is 33.431524 which makes standard error equal to 2.04081637.\r<BR>\n" );
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document.write( "at 95% confidence level, z = (x - m) / s becomes 1.96 = (x - 5000) / 2.040816327.\r<BR>\n" );
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document.write( "solve for x to get x = 1.96 * 2.040816327 + 5000 = 5004 which is a difference of 4 from the mean.\r<BR>\n" );
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document.write( "as long as the standard deviation is 11.8 and the sample size is 33.431524, the standard error will be 2.040816327 and you will get a margin of error of plus or minus 4 at 95% confidence level.\r<BR>\n" );
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document.write( "if you want the margin of error to be plus or minus 2 at 95% confidence level, then the z-score formula of 1.96 = 4 / (11.8/sqrt(n)) becomes:\r<BR>\n" );
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document.write( "1.96 = 2 / (11.8/sqrt(n)).\r<BR>\n" );
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document.write( "it's the same formula except you are replacing the difference of 4 with a difference of 2.\r<BR>\n" );
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document.write( "this becomes 1.96 = 2 * sqrt(n) / 11.8.\r<BR>\n" );
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document.write( "solve for sqrt(n) to get sqrt(n) = 1.96 * 11.8 / 2 = 11.564.\r<BR>\n" );
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document.write( "square both sides to get n = 133.726096.\r<BR>\n" );
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document.write( "the standard error becomes 11.8 / sqrt(133.726096) = 1.020408163.\r<BR>\n" );
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document.write( "the mean can be anything as long as the standard error is 1.020408163, while the standard error will always be 1.020408163 with a standard deviation of 11.8 and a sample size of 133.726096.\r<BR>\n" );
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document.write( "for example, suppose the mean is 500.\r<BR>\n" );
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document.write( "then the z-score formula for the high side is 1.96 = (x - 500) / 1.020408163.\r<BR>\n" );
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document.write( "solve for x to get x = 1.96 * 1.020408163 + 500 = 502\r<BR>\n" );
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document.write( "on the low side, the z-score formula becomes -1.96 = (x - 500) / 1.020408163.\r<BR>\n" );
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document.write( "solve for x to get x = -1.96 * 1.020408163 + 500 = 498.\r<BR>\n" );
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document.write( "to summarize the results:\r<BR>\n" );
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document.write( "at 95% confidence interval, .....\r<BR>\n" );
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document.write( "if you want the margin of error to be plus or minus 4, the sample size needs to be equal to 33.431524.\r<BR>\n" );
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document.write( "if you want the margin of error to be plus or minus 2, the sample size needs to be equal to 133.726096\r<BR>\n" );
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document.write( "if you want to find the margin of error to be plus or minus 4 at 99% confidence level, then you do the same procedure using the z-score for 99% confidence level, which is plus or minus 2.576.\r<BR>\n" );
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document.write( "you only need to work with the high side z-score because the sample size will be the same whether or not you work with the low side z-score because the normal distribution curve is symmetric about the mean.\r<BR>\n" );
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document.write( "once again, your formula for standard error is s = standard deviation / square root of sample size which is shown as s = sd / sqrt(n).\r<BR>\n" );
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document.write( "your standard deviation is equal to 11.8 as before.<BR>\n" );
document.write( "your z-score now becomes plus or minus 2.576<BR>\n" );
document.write( "your margin of error is plus or minus 4.\r<BR>\n" );
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document.write( "working with the high side z-score, the z-score formula of z = (x - m) / s becomes 2.576 = 4 / s.\r<BR>\n" );
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document.write( "since s = 11.8 / sqrt(n), this formula becomes 2.576 = 4 / (11.8 / sqrt(n)) which then becomes 2.576 = 4 * sqrt(n) / 11.8\r<BR>\n" );
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document.write( "solve for sqrt(n) to get sqrt(n) = 2.576 * 11.8 / 4 = 7.5992.\r<BR>\n" );
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document.write( "square both sides of that equation to get n = 57.74784064.\r<BR>\n" );
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document.write( "when n = 57.747840964, s = 11.8 / sqrt(57.74784064) = 1.552795031.\r<BR>\n" );
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document.write( "once again, the mean can be anything as long as the z-score is 2.576 and the standard error is 1.552795031 which was derived from a standard deviation of 11.8 and a sample size of 57.74784064.\r<BR>\n" );
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document.write( "for example, assume the mean is 349 books.\r<BR>\n" );
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document.write( "z-score formula becomes 2.576 = (x - 349) / 1.552795031.\r<BR>\n" );
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document.write( "solve for x to get x = 2.576 * 1.552795031 + 349 = 353, a difference of 4 from the mean.\r<BR>\n" );
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document.write( "low side z-score formula becomes -2.576 = (x - 349) / 1.552795031.\r<BR>\n" );
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document.write( "solve for x to get x = -2.576 * 1.552795031 + 349 = 345, a difference of 4 from the mean.\r<BR>\n" );
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document.write( "to summarize the results again, adding the final result using 99% confidence level, your solution should be, if i did this correctly:\r<BR>\n" );
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document.write( "at 95% confidence interval, .....\r<BR>\n" );
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document.write( "if you want the margin of error to be plus or minus 4, the sample size needs to be equal to 33.431524.\r<BR>\n" );
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document.write( "if you want the margin of error to be plus or minus 2, the sample size needs to be equal to 133.726096\r<BR>\n" );
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document.write( "at 99% confidence interval, .....\r<BR>\n" );
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document.write( "if you ant the margin of error to be plus or minus 4, the sample size needs to be equal to 57.74784064.\r<BR>\n" );
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document.write( "once again, you would probably round to the next higher or lower integer since the sample size has to be a whole number.\r<BR>\n" );
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document.write( "i believe that rounding up to the next higher integer is what is normally recommended.\r<BR>\n" );
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document.write( "this insures your margin of error is less than or equal to the desired margin of error.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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