document.write( "Question 1192314: with sample size of 900 the standard error is 3 what should be sample size so that we would be 95$ confident that population mean is within 4 sample of mean?
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #848969 by CPhill(2189)\"\" \"About 
You can put this solution on YOUR website!
**1. Determine the Z-score for 95% Confidence:**\r
\n" ); document.write( "\n" ); document.write( "* For a 95% confidence level, the Z-score is 1.96.\r
\n" ); document.write( "\n" ); document.write( "**2. Calculate the Standard Deviation:**\r
\n" ); document.write( "\n" ); document.write( "* Standard Error (SE) = Standard Deviation (σ) / √(sample size)
\n" ); document.write( "* 3 = σ / √900
\n" ); document.write( "* σ = 3 * √900
\n" ); document.write( "* σ = 3 * 30
\n" ); document.write( "* σ = 90\r
\n" ); document.write( "\n" ); document.write( "**3. Determine the Required Sample Size:**\r
\n" ); document.write( "\n" ); document.write( "* Margin of Error (E) = 4
\n" ); document.write( "* Z-score = 1.96
\n" ); document.write( "* Standard Deviation (σ) = 90\r
\n" ); document.write( "\n" ); document.write( "* Formula: n = (Z-score * σ / E)²
\n" ); document.write( "* n = (1.96 * 90 / 4)²
\n" ); document.write( "* n = (44.1)²
\n" ); document.write( "* n = 1944.81\r
\n" ); document.write( "\n" ); document.write( "* **Round up to the nearest whole number:** n = 1945\r
\n" ); document.write( "\n" ); document.write( "**Therefore, the required sample size to be 95% confident that the population mean is within 4 units of the sample mean is 1945.**
\n" ); document.write( " \n" ); document.write( "
\n" );