document.write( "Question 1174988: The high temperatures (in degrees Fahrenheit) of a random sample of 10 small towns are:
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document.write( "98.8
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document.write( "97
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document.write( "97.1
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document.write( "98.3
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document.write( "97.5
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document.write( "98.5
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document.write( "99.1
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document.write( "96.3
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document.write( "97.9
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document.write( "96.7\r
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document.write( "Assume high temperatures are normally distributed. Based on this data, find the 95% confidence interval of the mean high temperature of towns. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).\r
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document.write( "95% C.I. = \n" );
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Algebra.Com's Answer #800510 by math_tutor2020(3817) ![]() You can put this solution on YOUR website! \n" ); document.write( "Answer: (97.04, 98.40)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "=================================================== \n" ); document.write( "Work Shown:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "xbar = sample mean \n" ); document.write( "xbar = 97.72 \n" ); document.write( "This is found by adding up all of the numbers given in the table, then dividing by 10 since there are 10 values in this data table. \n" ); document.write( "So this means n = 10 is the sample size.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "s = sample standard deviation \n" ); document.write( "s = 0.946103 which is approximate \n" ); document.write( "While it is possible to find this by hand, it's a bit of a pain with this many numbers. I recommend using either a spreadsheet or some other similar form of technology to quickly compute this value.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Because n is not greater than 30, and because we don't know the population standard deviation (sigma), this means we must use a T distribution (instead of a Z distribution)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "At 95% confidence and with degrees of freedom = n-1 = 10-1 = 9, the critical t value is roughly t = 2.262 \n" ); document.write( "Use a table or calculator to determine this. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The lower bound (L) of the confidence interval is \n" ); document.write( "L = xbar - t*s/sqrt(n) \n" ); document.write( "L = 97.72 - 2.262*0.946103/sqrt(10) \n" ); document.write( "L = 97.72 - 0.676754 \n" ); document.write( "L = 97.043246 \n" ); document.write( "L = 97.04\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The upper bound (U) of the confidence interval is \n" ); document.write( "U = xbar + t*s/sqrt(n) \n" ); document.write( "U = 97.72 + 2.262*0.946103/sqrt(10) \n" ); document.write( "U = 97.72 + 0.676754 \n" ); document.write( "U = 98.396754 \n" ); document.write( "U = 98.40\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The 95% confidence interval that estimates the population mean mu is approximately (97.04, 98.40) \n" ); document.write( "This is the same as writing 97.04 < mu < 98.40 \n" ); document.write( "We're 95% confident that mu is somewhere between 97.04 and 98.40\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "mu = population mean of high temperatures (in degrees Fahrenheit) of towns. \n" ); document.write( " \n" ); document.write( " |