document.write( "Question 1175141: You intend to estimate a population mean μ from the following sample.
\n" ); document.write( "51.7 41.4 56.8 74.3
\n" ); document.write( "82 101.6 78 104.9
\n" ); document.write( "55.5 77.3 72.6 72.2
\n" ); document.write( "63.5 82.6 66.2 68.2
\n" ); document.write( "79.5 64.5 56.4 55.8
\n" ); document.write( "64.7 63 43.6 35.4
\n" ); document.write( "82.7 59.8 89.7 83.5
\n" ); document.write( "71.2 48.8 64 62.4
\n" ); document.write( "83.7 77 64.7 60.6
\n" ); document.write( "52.7 78.8 59.5 74.2
\n" ); document.write( "65.5 83.8 43.7 63.4
\n" ); document.write( "74 32 93.1 57.8
\n" ); document.write( "86.6 \r
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\n" ); document.write( "\n" ); document.write( "Find the 90% confidence interval. Enter your answer as a tri-linear inequality accurate to two decimal place (because the sample data are reported accurate to one decimal place) ... <μ< ...\r
\n" ); document.write( "\n" ); document.write( "2. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ = 36.7. You would like to be 90% confident that your esimate is within 0.1 of the true population mean. How large of a sample size is required? n = Do not round mid-calculation. However, use a critical value accurate to three decimal places — this is important for the system to be able to give hints for incorrect answers.
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Algebra.Com's Answer #800743 by math_tutor2020(3817)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Problem 1
\n" ); document.write( "Use a spreadsheet or calculator to compute the following
\n" ); document.write( "xbar = 67.8551020408163
\n" ); document.write( "s = 15.8895151649404\r
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\n" ); document.write( "\n" ); document.write( "xbar is the sample mean, while s is the sample standard deviation
\n" ); document.write( "Because we have n = 49 items here, this means we have enough to use the Z distribution. We can use this if we know sigma, or if n > 30. When n > 30, the student T distribution is approximately the same as the standard normal distribution.\r
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\n" ); document.write( "\n" ); document.write( "At 90% confidence, the z critical value is roughly z = 1.645; use a table or calculator to compute this\r
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\n" ); document.write( "\n" ); document.write( "L = lower bound of confidence interval
\n" ); document.write( "L = xbar - z*s/sqrt(n)
\n" ); document.write( "L = 67.8551020408163 - 1.645*15.8895151649404/sqrt(49)
\n" ); document.write( "L = 67.8551020408163 - 3.734036063761
\n" ); document.write( "L = 64.1210659770552
\n" ); document.write( "L = 64.12\r
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\n" ); document.write( "\n" ); document.write( "U = upper bound of confidence interval
\n" ); document.write( "U = xbar + z*s/sqrt(n)
\n" ); document.write( "U = 67.8551020408163 + 1.645*15.8895151649404/sqrt(49)
\n" ); document.write( "U = 67.8551020408163 + 3.734036063761
\n" ); document.write( "U = 71.5891381045772
\n" ); document.write( "U = 71.59\r
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\n" ); document.write( "\n" ); document.write( "The 90% confidence interval of the form L < mu < U is therefore 64.12 < mu < 71.59\r
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\n" ); document.write( "Problem 2\r
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\n" ); document.write( "\n" ); document.write( "z = 1.645 from earlier (critical value for 90% confidence interval)
\n" ); document.write( "sigma = 36.7 = given population standard deviation
\n" ); document.write( "E = 0.1 = desired error\r
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\n" ); document.write( "\n" ); document.write( "n = minimum sample size needed
\n" ); document.write( "n = (z*sigma/E)^2
\n" ); document.write( "n = (1.645*36.7/0.1)^2
\n" ); document.write( "n = 364,471.801225
\n" ); document.write( "n = 364,472
\n" ); document.write( "Always round up to the nearest whole number for this type of problem.\r
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\n" ); document.write( "\n" ); document.write( "The min sample size needed is 364,472\r
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\n" ); document.write( "\n" ); document.write( "Side note: you may need to ignore the comma if you are typing this answer into a computer system
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