document.write( "Question 1205253: A research firm conducted a survey to determine the mean amount smokers spend on cigarette during a week. A sample of 49 smokers revealed that the sample mean is Br. 20 with standard deviation of Br. 5. Construct 95% confidence interval for the mean amount spent. \n" ); document.write( "
Algebra.Com's Answer #841920 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "At 95% confidence, the z critical value is roughly z = 1.96 which is something to memorize or have handy on a reference sheet somewhere.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "n = 49 = sample size \n" ); document.write( "xbar = 20 = sample mean \n" ); document.write( "s = 5 = sample standard deviation\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "E = margin of error for the mean \n" ); document.write( "E = z*s/sqrt(n) \n" ); document.write( "E = 1.96*5/sqrt(49) \n" ); document.write( "E = 1.4\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "L = lower boundary of the confidence interval \n" ); document.write( "L = xbar - E \n" ); document.write( "L = 20 - 1.4 \n" ); document.write( "L = 18.6\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "U = upper boundary of the confidence interval \n" ); document.write( "U = xbar + E \n" ); document.write( "U = 20 + 1.4 \n" ); document.write( "U = 21.4\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The 95% confidence interval is approximately 18.6 < mu < 21.4 \n" ); document.write( "It's of the format L < mu < U.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We can condense that to the notation (18.6, 21.4) \n" ); document.write( " \n" ); document.write( " |