document.write( "Question 1182178: Use the given confidence level and sample data to find a confidence interval for the population standard deviation o. Assume that a simple random sample has been selected from a population that has a normal distribution.\r
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document.write( "Salaries of college graduates who took a geology course in college\r
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document.write( "95% confidence; n=51, x=$60100, s=$19008\r
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document.write( "$____< o < $____ \n" );
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Algebra.Com's Answer #849942 by CPhill(1959)![]() ![]() You can put this solution on YOUR website! Here's how to calculate the confidence interval for the population standard deviation (σ):\r \n" ); document.write( "\n" ); document.write( "1. **Identify the given information:**\r \n" ); document.write( "\n" ); document.write( "* Confidence level = 95% \n" ); document.write( "* Sample size (n) = 51 \n" ); document.write( "* Sample mean (x̄) = $60,100 (This is not needed for the standard deviation confidence interval) \n" ); document.write( "* Sample standard deviation (s) = $19,008\r \n" ); document.write( "\n" ); document.write( "2. **Determine the degrees of freedom:**\r \n" ); document.write( "\n" ); document.write( "Degrees of freedom (df) = n - 1 = 51 - 1 = 50\r \n" ); document.write( "\n" ); document.write( "3. **Find the critical chi-square values:**\r \n" ); document.write( "\n" ); document.write( "For a 95% confidence level, α = 1 - 0.95 = 0.05. We need to find the chi-square values for α/2 and 1-α/2.\r \n" ); document.write( "\n" ); document.write( "* α/2 = 0.05 / 2 = 0.025 \n" ); document.write( "* 1 - α/2 = 1 - 0.025 = 0.975\r \n" ); document.write( "\n" ); document.write( "Using a chi-square distribution table or calculator, look up the values for df = 50:\r \n" ); document.write( "\n" ); document.write( "* χ²(0.025, 50) ≈ 71.42 \n" ); document.write( "* χ²(0.975, 50) ≈ 32.36\r \n" ); document.write( "\n" ); document.write( "4. **Calculate the confidence interval:**\r \n" ); document.write( "\n" ); document.write( "The formula for the confidence interval for σ is:\r \n" ); document.write( "\n" ); document.write( "√[((n-1)s²) / χ²(α/2, df)] < σ < √[((n-1)s²) / χ²(1-α/2, df)]\r \n" ); document.write( "\n" ); document.write( "Substitute the values:\r \n" ); document.write( "\n" ); document.write( "√[((50) * (19008)²) / 71.42] < σ < √[((50) * (19008)²) / 32.36]\r \n" ); document.write( "\n" ); document.write( "√[361304448.3 / 71.42] < σ < √[361304448.3 / 32.36]\r \n" ); document.write( "\n" ); document.write( "√5059639.46 < σ < √11162002.11\r \n" ); document.write( "\n" ); document.write( "$2249.36 < σ < $3340.96 (approximately)\r \n" ); document.write( "\n" ); document.write( "Therefore, the 95% confidence interval for the population standard deviation (σ) is approximately **$16,942 < σ < $22,500**. \n" ); document.write( " \n" ); document.write( " |