Question 1179488: A group of 10-foot surgery patients had a mean weight of 240 pounds. The sample
standard deviation was found to be pounds. Assuming that the foot surgery
patients’ weights are normally distributed. Construct a 95% confidence interval for a
sample for the true mean weight of all foot surgery patients.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! It seems like there's a typo in the problem statement, and the sample standard deviation is missing. I'll assume the sample standard deviation is **20 pounds** to illustrate the process. If you have the correct value, you can plug it in and follow the same steps.
**Here's how to construct the 95% confidence interval:**
**1. Identify the given information:**
* Sample size (n) = 10
* Sample mean (x̄) = 240 pounds
* Sample standard deviation (s) = 20 pounds (assumed)
* Confidence level = 95%
**2. Determine the degrees of freedom (df):**
df = n - 1 = 10 - 1 = 9
**3. Find the t-value:**
For a 95% confidence interval and 9 degrees of freedom, we need the t-value that leaves 2.5% in each tail (α/2 = 0.05/2 = 0.025).
Using a t-table or calculator, the t-value for df = 9 and α/2 = 0.025 is approximately 2.262.
**4. Calculate the margin of error (E):**
E = t * (s / √n) = 2.262 * (20 / √10) ≈ 14.28 pounds
**5. Construct the confidence interval:**
Confidence interval = x̄ ± E = 240 ± 14.28
Lower bound = 240 - 14.28 = 225.72 pounds
Upper bound = 240 + 14.28 = 254.28 pounds
**Therefore, the 95% confidence interval for the true mean weight of all foot surgery patients is approximately (225.72 pounds, 254.28 pounds).**
**If you provide the correct sample standard deviation, I can recalculate the confidence interval with the accurate value.**
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