document.write( "Question 1185360: A journal article reported that the mean hospital stay following a particular surgical procedure
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document.write( "in 2001 was 7.1 days. A researcher feels that the mean hospital stay in 2002 should be less due
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document.write( "to initiatives aimed at reducing health care costs. A random sample of 40 patients undergoing
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document.write( "the same surgical procedure in 2002 had a mean length of stay of 6.85 days with a standard
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document.write( "deviation of 7.01 days. Run the appropriate statistical test at 𝛼 = 0.05. \n" );
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Algebra.Com's Answer #849755 by CPhill(1959)![]() ![]() You can put this solution on YOUR website! Here's how to conduct a hypothesis test to see if the mean hospital stay in 2002 was less than in 2001:\r \n" ); document.write( "\n" ); document.write( "**1. State the Hypotheses:**\r \n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H₀):** The mean hospital stay in 2002 is the same as or greater than the mean hospital stay in 2001. μ ≥ 7.1 days \n" ); document.write( "* **Alternative Hypothesis (H₁):** The mean hospital stay in 2002 is less than the mean hospital stay in 2001. μ < 7.1 days (This is a one-tailed test).\r \n" ); document.write( "\n" ); document.write( "**2. Determine the Significance Level:**\r \n" ); document.write( "\n" ); document.write( "α = 0.05\r \n" ); document.write( "\n" ); document.write( "**3. Choose the appropriate test statistic:**\r \n" ); document.write( "\n" ); document.write( "Since the sample size is greater than 30 and the population standard deviation is unknown, we will use a one-sample t-test.\r \n" ); document.write( "\n" ); document.write( "**4. Calculate the test statistic:**\r \n" ); document.write( "\n" ); document.write( "The t-statistic is calculated as:\r \n" ); document.write( "\n" ); document.write( "t = (sample mean - population mean) / (sample standard deviation / √sample size) \n" ); document.write( "t = (6.85 - 7.1) / (7.01 / √40) \n" ); document.write( "t = -0.25 / (7.01 / 6.324) \n" ); document.write( "t = -0.25 / 1.11 \n" ); document.write( "t ≈ -0.225\r \n" ); document.write( "\n" ); document.write( "**5. Determine the degrees of freedom:**\r \n" ); document.write( "\n" ); document.write( "Degrees of freedom (df) = sample size - 1 = 40 - 1 = 39\r \n" ); document.write( "\n" ); document.write( "**6. Find the critical value (or p-value):**\r \n" ); document.write( "\n" ); document.write( "* **Using a t-table:** For a one-tailed test with α = 0.05 and df = 39, the critical t-value is approximately -1.685. \n" ); document.write( "* **Using a calculator or statistical software:** A calculator or statistical software can give a more precise p-value.\r \n" ); document.write( "\n" ); document.write( "**7. Calculate the p-value:**\r \n" ); document.write( "\n" ); document.write( "Using statistical software or a t-table with interpolation, with t ≈ -0.225 and df = 39, the p-value is approximately 0.412.\r \n" ); document.write( "\n" ); document.write( "**8. Make a decision:**\r \n" ); document.write( "\n" ); document.write( "* **Using the critical value:** Our calculated t-statistic (-0.225) is greater than the critical t-value (-1.685). Therefore, we fail to reject the null hypothesis. \n" ); document.write( "* **Using the p-value:** The p-value (0.412) is greater than the significance level (0.05). Therefore, we fail to reject the null hypothesis.\r \n" ); document.write( "\n" ); document.write( "**9. Conclusion:**\r \n" ); document.write( "\n" ); document.write( "There is not sufficient evidence at the α = 0.05 level to conclude that the mean hospital stay in 2002 is less than 7.1 days. \n" ); document.write( " \n" ); document.write( " |