SOLUTION: 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

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?
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 the hospital's success rate is different from the national average:
**1. State the Hypotheses:**
* **Null Hypothesis (H0):** The hospital's success rate is equal to the national average (p = 0.70).
* **Alternative Hypothesis (H1):** The hospital's success rate is different from the national average (p ≠ 0.70). This is a two-tailed test.
**2. Significance Level:** α = 0.05 (Assuming this common value if not provided)
**3. Calculate the Sample Proportion (p̂):**
* p̂ = (Number of successful operations) / (Total number of operations)
* p̂ = 32 / 42 ≈ 0.7619
**4. Calculate the Test Statistic (z-score):**
z = (p̂ - p) / √(p(1 - p) / n)
Where:
* p̂ = sample proportion (0.7619)
* p = hypothesized population proportion (0.70)
* n = sample size (42)
z = (0.7619 - 0.70) / √(0.70 * (1 - 0.70) / 42)
z = 0.0619 / √(0.70 * 0.30 / 42)
z = 0.0619 / √(0.21 / 42)
z = 0.0619 / √0.005
z = 0.0619 / 0.0707
z ≈ 0.875
**5. Determine the P-value:**
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:
* P(z < -0.875) ≈ 0.1909
* P(z > 0.875) ≈ 0.1909
* P-value = 2 * 0.1909 ≈ 0.3818
**6. Make a Decision:**
Compare the p-value to the significance level (α):
* p-value (0.3818) > α (0.05)
Since the p-value is *greater than* the significance level, we *fail to reject* the null hypothesis.
**7. Conclusion:**
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.

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