document.write( "Question 1185356: An academic medical center surveyed all of its patients in 2002 to assess their satisfaction with medical care. Satisfaction was measured on a scale of 0 to 100, with higher scores indicative of more satisfaction. The mean satisfaction score in 2002 was 84.5. Several quality-improvement initiatives were implemented in 2003 and the medical center is wondering whether the initiatives increased patient satisfaction. A random sample of 125 patients seeking medical care in 2003 was surveyed using the same satisfaction measure. Their mean satisfaction score was 89.2 with a standard deviation of 17.4. Is there evidence of a significant improvement in satisfaction? Run the appropriate test at the 1% level of significance. \n" ); document.write( "
Algebra.Com's Answer #849757 by CPhill(1959)\"\" \"About 
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Here's how to conduct a hypothesis test to determine if there's a significant improvement in patient satisfaction:\r
\n" ); document.write( "\n" ); document.write( "**1. State the Hypotheses:**\r
\n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H₀):** Patient satisfaction has not increased (or has decreased). The mean satisfaction score in 2003 is less than or equal to the mean satisfaction score in 2002. μ ≤ 84.5
\n" ); document.write( "* **Alternative Hypothesis (H₁):** Patient satisfaction has increased. The mean satisfaction score in 2003 is greater than the mean satisfaction score in 2002. μ > 84.5 (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 = 0.01\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 large (n = 125 > 30) and the population standard deviation is unknown, we can use a one-sample z-test. While a t-test would also be appropriate, with such a large sample, the results will be very similar.\r
\n" ); document.write( "\n" ); document.write( "**4. Calculate the test statistic:**\r
\n" ); document.write( "\n" ); document.write( "The z-statistic is calculated as:\r
\n" ); document.write( "\n" ); document.write( "z = (sample mean - population mean) / (sample standard deviation / √sample size)
\n" ); document.write( "z = (89.2 - 84.5) / (17.4 / √125)
\n" ); document.write( "z = 4.7 / (17.4 / 11.18)
\n" ); document.write( "z = 4.7 / 1.556
\n" ); document.write( "z ≈ 3.02\r
\n" ); document.write( "\n" ); document.write( "**5. Determine the critical value (or p-value):**\r
\n" ); document.write( "\n" ); document.write( "* **Using a z-table:** For a one-tailed test with α = 0.01, the critical z-value is approximately 2.33.
\n" ); document.write( "* **Using a calculator or statistical software:** A calculator or statistical software can provide a more precise p-value.\r
\n" ); document.write( "\n" ); document.write( "**6. Calculate the p-value:**\r
\n" ); document.write( "\n" ); document.write( "Using statistical software or a z-table, with z ≈ 3.02, the p-value is very small, much less than 0.01 (approximately 0.0013).\r
\n" ); document.write( "\n" ); document.write( "**7. Make a decision:**\r
\n" ); document.write( "\n" ); document.write( "* **Using the critical value:** Our calculated z-statistic (3.02) is greater than the critical z-value (2.33). Therefore, we reject the null hypothesis.
\n" ); document.write( "* **Using the p-value:** The p-value (≈ 0.0013) is less than the significance level (0.01). Therefore, we reject the null hypothesis.\r
\n" ); document.write( "\n" ); document.write( "**8. Conclusion:**\r
\n" ); document.write( "\n" ); document.write( "There is very strong evidence at the α = 0.01 level of significance to conclude that the quality-improvement initiatives have increased patient satisfaction.
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