14. Golf Averages A study of 35 golfers showed that their average score on a particular course was 92 . The standard deviation of the population is 5 . a. Find the best point estimate of the mean. b. Find the $$ 95 \% $$ confidence interval of the mean score for all golfers. c. Find the $$ 95 \% $$ confidence interval of the mean score if a sample of 60 golfers is used instead of a sample of 35 . d. Which interval is smaller? Explain why.

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Let's tackle each part of the problem step by step.

### **Given:**
- **Sample Size (n):** 35 (and 60 for part c)
- **Sample Mean ($$\bar{x}$$):** 92
- **Population Standard Deviation ($$\sigma$$):** 5
- **Confidence Level:** 95%

---

### **a. Best Point Estimate of the Mean**

**Point Estimate:** The best point estimate for the population mean ($$\mu$$) is the **sample mean**.

$$
\text{Best Point Estimate} = \bar{x} = 92
$$

---

### **b. 95% Confidence Interval for the Mean Score (n = 35)**

Since the population standard deviation ($$\sigma$$) is known and the sample size is greater than 30, we can use the **Z-distribution** to construct the confidence interval.

**Formula for Confidence Interval:**

$$
\bar{x} \pm Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right)
$$

Where:
- $$Z_{\alpha/2}$$ is the Z-score corresponding to the desired confidence level. For 95%, $$Z_{0.025} = 1.96$$.
- $$\sigma$$ is the population standard deviation.
- $$n$$ is the sample size.

**Calculations:**

1. **Standard Error (SE):**

$$
SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{35}} \approx \frac{5}{5.916} \approx 0.845
$$

2. **Margin of Error (ME):**

$$
ME = Z_{\alpha/2} \times SE = 1.96 \times 0.845 \approx 1.655
$$

3. **Confidence Interval:**

$$
\bar{x} \pm ME = 92 \pm 1.655
$$

$$
\Rightarrow (92 - 1.655, \ 92 + 1.655) = (90.345, \ 93.655)
$$

**Rounded to Two Decimal Places:**

$$
\boxed{(90.35,\ 93.65)}
$$

---

### **c. 95% Confidence Interval for the Mean Score (n = 60)**

Using the same approach as in part b, but with $$n = 60$$.

**Calculations:**

1. **Standard Error (SE):**

$$
SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{60}} \approx \frac{5}{7.746} \approx 0.645
$$

2. **Margin of Error (ME):**

$$
ME = Z_{\alpha/2} \times SE = 1.96 \times 0.645 \approx 1.265
$$

3. **Confidence Interval:**

$$
\bar{x} \pm ME = 92 \pm 1.265
$$

$$
\Rightarrow (92 - 1.265, \ 92 + 1.265) = (90.735, \ 93.265)
$$

**Rounded to Three Decimal Places:**

$$
\boxed{(90.735,\ 93.265)}
$$

---

### **d. Comparison of the Two Confidence Intervals**

**Which Interval is Smaller?**

The **confidence interval with $$n = 60$$** is **smaller** than the one with $$n = 35$$.

**Explanation:**

- **Standard Error (SE) Decreases with Increasing Sample Size:** As the sample size increases, the standard error ($$SE = \frac{\sigma}{\sqrt{n}}$$) decreases because the denominator becomes larger.
  
- **Smaller SE Leads to a Narrower Confidence Interval:** A smaller standard error reduces the margin of error, resulting in a more precise (narrower) confidence interval.

**Conclusion:** Increasing the sample size from 35 to 60 decreases the standard error, which in turn reduces the width of the 95% confidence interval.

---

**Summary of Confidence Intervals:**
- **n = 35:** (90.35, 93.65)
- **n = 60:** (90.735, 93.265)

The interval for $$n = 60$$ is narrower than that for $$n = 35$$, providing a more precise estimate of the mean score.
**a. Best Point Estimate of the Mean:** 92 **b. 95% Confidence Interval (n=35):** (90.35, 93.65) **c. 95% Confidence Interval (n=60):** (90.735, 93.265) **d. The interval with n=60 is smaller because a larger sample size reduces the standard error, leading to a narrower confidence interval.**

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