Question 1193667
**a. 95% Confidence Interval Estimate**

* **Given:**
    * Sample mean (x̄) = 267 mg/dl
    * Population standard deviation (σ) = 45 mg/dl
    * Sample size (n) = 100
    * Confidence level = 95% 
    * Z-score for 95% confidence: 1.96

* **Standard Error of the Mean:** 
    * σ/√n = 45 / √100 = 4.5 mg/dl

* **Confidence Interval:**
    * x̄ ± Z * (σ/√n)
    * 267 ± 1.96 * 4.5
    * 267 ± 8.82

* **95% Confidence Interval:** (258.18, 275.82) mg/dl

**b. Conclusion**

* **Hypotheses:**
    * H0: μ₁ = μ₂ (Mean cholesterol level of men undergoing bypass surgery is equal to the mean cholesterol level of healthy men)
    * H1: μ₁ ≠ μ₂ (Mean cholesterol level of men undergoing bypass surgery is different from the mean cholesterol level of healthy men) 
    * where:
        * μ₁: Mean cholesterol level of men undergoing bypass surgery
        * μ₂: Mean cholesterol level of healthy men (245 mg/dl)

* **Decision:**
    * Since the 95% confidence interval for the mean cholesterol level of men undergoing bypass surgery ((258.18, 275.82) mg/dl) does *not* include the mean cholesterol level of healthy men (245 mg/dl), we can **reject the null hypothesis (H0)**.

* **Conclusion:**

    * There is sufficient evidence at the 0.05 significance level to conclude that the mean serum cholesterol level of men undergoing coronary bypass surgery is significantly different from that of healthy men. 
    * Specifically, the mean cholesterol level of men undergoing bypass surgery appears to be higher.

**Important Notes:**

* This analysis assumes that the population standard deviation is known and that the sampling distribution of the mean is approximately normal.
* This is a simplified example. In real-world scenarios, researchers often use t-distributions when the population standard deviation is unknown and estimate it using the sample standard deviation.

I hope this helps! Let me know if you have any other questions.