SOLUTION: 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

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.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to conduct a hypothesis test to determine if there's a significant improvement in patient satisfaction:
**1. State the Hypotheses:**
* **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
* **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).
**2. Determine the Significance Level:**
α = 0.01
**3. Choose the appropriate test statistic:**
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.
**4. Calculate the test statistic:**
The z-statistic is calculated as:
z = (sample mean - population mean) / (sample standard deviation / √sample size)
z = (89.2 - 84.5) / (17.4 / √125)
z = 4.7 / (17.4 / 11.18)
z = 4.7 / 1.556
z ≈ 3.02
**5. Determine the critical value (or p-value):**
* **Using a z-table:** For a one-tailed test with α = 0.01, the critical z-value is approximately 2.33.
* **Using a calculator or statistical software:** A calculator or statistical software can provide a more precise p-value.
**6. Calculate the p-value:**
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).
**7. Make a decision:**
* **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.
* **Using the p-value:** The p-value (≈ 0.0013) is less than the significance level (0.01). Therefore, we reject the null hypothesis.
**8. Conclusion:**
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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