document.write( "Question 1179922: Doctors nationally believe that 70% of a certain type of operation are successful. In a particular hospital, 42 of these operations were observed and 32 of them were successful. At is this hospital's success rate different from the national average? \n" ); document.write( "
Algebra.Com's Answer #850158 by CPhill(1959)![]() ![]() You can put this solution on YOUR website! Here's how to conduct a hypothesis test to determine if the hospital's success rate is different from the national average:\r \n" ); document.write( "\n" ); document.write( "**1. State the Hypotheses:**\r \n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H0):** The hospital's success rate is equal to the national average (p = 0.70). \n" ); document.write( "* **Alternative Hypothesis (H1):** The hospital's success rate is different from the national average (p ≠ 0.70). This is a two-tailed test.\r \n" ); document.write( "\n" ); document.write( "**2. Significance Level:** α = 0.05 (Assuming this common value if not provided)\r \n" ); document.write( "\n" ); document.write( "**3. Calculate the Sample Proportion (p̂):**\r \n" ); document.write( "\n" ); document.write( "* p̂ = (Number of successful operations) / (Total number of operations) \n" ); document.write( "* p̂ = 32 / 42 ≈ 0.7619\r \n" ); document.write( "\n" ); document.write( "**4. Calculate the Test Statistic (z-score):**\r \n" ); document.write( "\n" ); document.write( "z = (p̂ - p) / √(p(1 - p) / n)\r \n" ); document.write( "\n" ); document.write( "Where:\r \n" ); document.write( "\n" ); document.write( "* p̂ = sample proportion (0.7619) \n" ); document.write( "* p = hypothesized population proportion (0.70) \n" ); document.write( "* n = sample size (42)\r \n" ); document.write( "\n" ); document.write( "z = (0.7619 - 0.70) / √(0.70 * (1 - 0.70) / 42) \n" ); document.write( "z = 0.0619 / √(0.70 * 0.30 / 42) \n" ); document.write( "z = 0.0619 / √(0.21 / 42) \n" ); document.write( "z = 0.0619 / √0.005 \n" ); document.write( "z = 0.0619 / 0.0707 \n" ); document.write( "z ≈ 0.875\r \n" ); document.write( "\n" ); document.write( "**5. Determine the P-value:**\r \n" ); document.write( "\n" ); document.write( "Since this is a two-tailed test, we need to find the probability of getting a z-score as extreme as 0.875 or -0.875. Using a z-table or calculator:\r \n" ); document.write( "\n" ); document.write( "* P(z < -0.875) ≈ 0.1909 \n" ); document.write( "* P(z > 0.875) ≈ 0.1909 \n" ); document.write( "* P-value = 2 * 0.1909 ≈ 0.3818\r \n" ); document.write( "\n" ); document.write( "**6. Make a Decision:**\r \n" ); document.write( "\n" ); document.write( "Compare the p-value to the significance level (α):\r \n" ); document.write( "\n" ); document.write( "* p-value (0.3818) > α (0.05)\r \n" ); document.write( "\n" ); document.write( "Since the p-value is *greater than* the significance level, we *fail to reject* the null hypothesis.\r \n" ); document.write( "\n" ); document.write( "**7. Conclusion:**\r \n" ); document.write( "\n" ); document.write( "There is not sufficient evidence at the α = 0.05 level of significance to conclude that the hospital's success rate is different from the national average. Therefore, the hospital's success rate is not significantly different from the national average. \n" ); document.write( " \n" ); document.write( " |