A sample of size 36 was taken out of a population with population standard deviation of  = 3. The sample
mean was calculated to be 20.
a) Construct a 95% confidence interval for the population mean. Do not round.
b) Construct a 98% confidence interval for the above problem. Do not round.
c) Explain how and why the two intervals are different, what happens to the confidence interval if the
confidence level is increased?
d) Increase the sample size from n = 36 to n = 56, and construct 95% confidence interval for population
mean. Round to the nearest thousandth.
e) Explain the resulting difference, what happens to the confidence interval when the sample size is
increased?

### Step-by-Step Solutions:

---

#### **(a) Construct a 95% Confidence Interval for the Population Mean:**
For a confidence interval:
\[
\text{CI} = \bar{x} \pm Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
\]
Where:
- \(\bar{x} = 20\) (sample mean),
- \(\sigma = 3\) (population standard deviation),
- \(n = 36\) (sample size),
- \(Z_{\alpha/2}\) for 95% confidence level = 1.96.

Substitute the values:
\[
\text{Margin of Error} = 1.96 \cdot \frac{3}{\sqrt{36}} = 1.96 \cdot \frac{3}{6} = 1.96 \cdot 0.5 = 0.98
\]

Thus, the confidence interval:
\[
\text{CI} = 20 \pm 0.98 \quad \Rightarrow \quad (19.02, 20.98)
\]

---

#### **(b) Construct a 98% Confidence Interval for the Population Mean:**
For 98% confidence level, \(Z_{\alpha/2} = 2.33\).

Substitute the values:
\[
\text{Margin of Error} = 2.33 \cdot \frac{3}{\sqrt{36}} = 2.33 \cdot 0.5 = 1.165
\]

Thus, the confidence interval:
\[
\text{CI} = 20 \pm 1.165 \quad \Rightarrow \quad (18.835, 21.165)
\]

---

#### **(c) Explain the Difference Between the Two Intervals:**
- **Difference**: The 98% confidence interval \((18.835, 21.165)\) is wider than the 95% confidence interval \((19.02, 20.98)\). 
- **Reason**: As the confidence level increases, the interval becomes wider to ensure greater certainty that the population mean falls within the interval.

**Effect of Increasing Confidence Level:**
- Increasing the confidence level increases the width of the interval because a higher level of certainty requires accounting for more variability.

---

#### **(d) Increase Sample Size to \(n = 56\) and Construct a 95% Confidence Interval:**
For \(n = 56\), with 95% confidence (\(Z_{\alpha/2} = 1.96\)):

Substitute the values:
\[
\text{Margin of Error} = 1.96 \cdot \frac{3}{\sqrt{56}} = 1.96 \cdot \frac{3}{7.4833} = 1.96 \cdot 0.4009 = 0.7858
\]

Thus, the confidence interval:
\[
\text{CI} = 20 \pm 0.7858 \quad \Rightarrow \quad (19.214, 20.786)
\]

---

#### **(e) Explain the Effect of Increasing Sample Size:**
- **Resulting Difference**: The 95% confidence interval with a larger sample size \((19.214, 20.786)\) is narrower than the interval with the smaller sample size \((19.02, 20.98)\).
- **Reason**: As the sample size increases, the standard error (\(\sigma / \sqrt{n}\)) decreases, leading to a smaller margin of error and a narrower interval.

**Effect of Increasing Sample Size:**
- Increasing the sample size decreases variability and narrows the confidence interval, making the estimate of the population mean more precise.

---

### **Summary Results:**
1. **95% CI (n = 36):** \((19.02, 20.98)\)
2. **98% CI (n = 36):** \((18.835, 21.165)\)
3. **95% CI (n = 56):** \((19.214, 20.786)\)

---

Do you want further clarification or additional examples?  
Here are five related questions to explore:
1. How is \(Z_{\alpha/2}\) determined for different confidence levels?
2. What happens to the confidence interval if the population standard deviation changes?
3. Can we use a \(t\)-distribution instead of \(Z\)-distribution here? Why or why not?
4. How does increasing the variability of the data affect the confidence interval?
5. What is the relationship between margin of error and sample size?

**Tip:** Always ensure the sample size is large enough to reduce the margin of error for better population parameter estimates.

A sample of size 36 was taken out of a population with population standard deviation of σ = 3. The sample mean was calculated to be 20. a) Construct a 95% confidence interval for the population mean. Do not round. b) Construct a 98% confidence interval for the above problem. Do not round. c) Explain how and why the two intervals are different, what happens to the confidence interval if the confidence level is increased? d) Increase the sample size from n = 36 to n = 56, and construct a 95% confidence interval for population mean. Round to the nearest thousandth. e) Explain the resulting difference, what happens to the confidence interval when the sample size is increased?

Learn how to construct 95% and 98% confidence intervals for a population mean with a sample size of 36 and a population standard deviation of σ = 3. Explore the effect of increasing the confidence level and sample size on the interval width. Includes step-by-step calculations and explanations.

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