document.write( "Question 1200658: A food snack manufacturer samples 7 bags of pretzels off the assembly line and weighs their contents. If the sample mean is 15.2 oz and the sample standard deviation is 0.70 oz, find the 95% confidence interval of the true mean. \n" ); document.write( "
Algebra.Com's Answer #834886 by Theo(13342)\"\" \"About 
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sample size is 7.
\n" ); document.write( "sampole mean is 15.2.
\n" ); document.write( "sample standard deviation is .7
\n" ); document.write( "critical t-score at 6 degrees of freedom at 95% confidence interval is equal to plus or minus 2.446911839.
\n" ); document.write( "t-score formula is t = (x-m)/s
\n" ); document.write( "t is the t-score
\n" ); document.write( "x is the confidence limits of the true mean.
\n" ); document.write( "m is the sample mean
\n" ); document.write( "s is the standard error.
\n" ); document.write( "standard error = .7/sqrt(7) = .2645751311.
\n" ); document.write( "on the low side, you get:
\n" ); document.write( "minus 2.446911839 = (x-15.2)/.2645751322.
\n" ); document.write( "solve for x to get x = 2.446911839 * .2645751322 + 15.2 = 14.5
\n" ); document.write( "on the high side, you get:
\n" ); document.write( "2.446911839 = (x - 15.2)/.2645751322.
\n" ); document.write( "solve for x to get x = 2.446911839 * .2645751322 + 15.2 = 15.9
\n" ); document.write( "the 95% confidence interval is from 14.5 to 15.9.
\n" ); document.write( "the true population mean will be within those limits 95% of the time.
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