Questions: Hip surgery: In a sample of 126 hip surgeries of a certain type, the average surgery time was 137.5 minutes with a standard deviation of 22.9 minutes. Part: 0 / 2 Part 1 of 2 (a) Construct a 90% confidence interval for the mean surgery time for this procedure. Round the answers to at least one decimal place. A 90% confidence interval for the mean surgery time for this procedure is <μ< .

Solution Steps

We have a sample of hip surgeries with the following statistics:

• Sample size (\(n\)) = 126
• Sample mean (\(\bar{x}\)) = 137.5 minutes
• Sample standard deviation (\(s\)) = 22.9 minutes
• Confidence level = 90%

For a 90% confidence level, the significance level (\(\alpha\)) is: \[ \alpha = 1 - 0.90 = 0.10 \] The critical value \(z\) for a 90% confidence level can be found using the standard normal distribution. For a two-tailed test, we look for \(z\) such that: \[ P(Z < z) = 1 - \frac{\alpha}{2} = 0.95 \] This gives us \(z \approx 1.645\).

The margin of error (\(E\)) is calculated using the formula: \[ E = z \cdot \frac{s}{\sqrt{n}} \] Substituting the values: \[ E = 1.645 \cdot \frac{22.9}{\sqrt{126}} \approx 1.645 \cdot 2.037 \approx 3.344 \]

The confidence interval for the mean is given by: \[ \bar{x} \pm E \] Calculating the lower and upper bounds: \[ \text{Lower bound} = 137.5 - 3.344 \approx 134.156 \] \[ \text{Upper bound} = 137.5 + 3.344 \approx 140.844 \] Rounding to one decimal place, we have: \[ \text{Lower bound} \approx 134.1, \quad \text{Upper bound} \approx 140.9 \]

Final Answer