document.write( "Question 1028909: how can i get standard deviation and sample size from the following data \r
\n" );
document.write( "\n" );
document.write( "Data from five paediatric centres were compared in children
\n" );
document.write( "who received 0.05 (41 episodes) or 0.1 units/kg/h (52 episodes)
\n" );
document.write( " In the low vs. standard dose group, at 6 h following admission,
\n" );
document.write( "the fall in blood glucose levels [11.3 (95% confidence interval 8.6 to 13.9) vs. 11.8 (8.4 to 15.2) mmol/L, p = 0.86] and rise in pH [0.13 (0.09 to 0.18) vs.
\n" );
document.write( "0.11 (0.07 to 0.15), p = 0.78] were similar\r
\n" );
document.write( "\n" );
document.write( "it's not home work it's a research
\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #644070 by rothauserc(4718)![]() ![]() You can put this solution on YOUR website! Is p = 0.86 a statistics p-value? If not, what is it? \n" ); document.write( ": \n" ); document.write( "if we assume a standard normal distribution \n" ); document.write( ": \n" ); document.write( "************************** \n" ); document.write( "consider the first two confidence intervals with p = 0.86 \n" ); document.write( ": \n" ); document.write( "upper 95% limit = sample mean + (standard error for sample mean(SEM) * 1.96) \n" ); document.write( ": \n" ); document.write( "13.9 = 11.3 + (SEM * 1.96) \n" ); document.write( ": \n" ); document.write( "SEM = (13.9 - 11.3) / 1.96 = 1.326530612 \n" ); document.write( ": \n" ); document.write( "sample size = (critical value * SEM / margin of error (ME))^2 \n" ); document.write( ": \n" ); document.write( "note that ME for a 95% confidence interval is 0.05 \n" ); document.write( ": \n" ); document.write( "sample size = (1.96 * 1.326530612 / 0.05)^2 is approx 2704 \n" ); document.write( " |