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)\"\" \"About 
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\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
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\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
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\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
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\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
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\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
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\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
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\n" ); document.write( "\n" ); document.write( "We can condense that to the notation (18.6, 21.4)
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