2.3 You are required to provide a 91.98% confidence interval for a mean, it is important that the interval be no longer than 2 cm . Assume that the population is normally distributed with the variance of 2 25 cm . What is the smallest sample size needed? [8]

Let's carefully analyze the problem and find the smallest sample size required.

### Given Data:
- **Confidence level:** \( 91.98\% \)
- **Desired margin of error \( E \):** \( 2 \, \text{cm} \)
- **Population variance \( \sigma^2 \):** \( 2 \times 25 \, \text{cm} = 50 \, \text{cm}^2 \)
- **Population standard deviation \( \sigma \):** \( \sqrt{50} \, \text{cm} \)

### Step 1: Determine the z-value corresponding to the confidence level.
The confidence level is 91.98%, so the significance level \( \alpha = 1 - 0.9198 = 0.0802 \).

Since the confidence interval is symmetric, \( \alpha/2 = 0.0401 \).

Now, we find the z-value corresponding to \( 1 - \alpha/2 = 0.9599 \).

This z-value is approximately:
\[
z_{\alpha/2} \approx 1.75
\]

### Step 2: Use the margin of error formula.
The margin of error \( E \) for the confidence interval is given by:
\[
E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
\]
Given that \( E = 2 \, \text{cm} \), \( \sigma = \sqrt{50} \, \text{cm} \), and \( z_{\alpha/2} = 1.75 \), we set up the equation:
\[
2 = 1.75 \cdot \frac{\sqrt{50}}{\sqrt{n}}
\]

### Step 3: Solve for \( n \) (sample size).
Rearrange the equation to solve for \( n \):
\[
\sqrt{n} = 1.75 \cdot \frac{\sqrt{50}}{2}
\]
\[
\sqrt{n} = \frac{1.75 \cdot 7.0711}{2} \approx \frac{12.3744}{2} \approx 6.1872
\]

\[
n \approx (6.1872)^2 \approx 38.28
\]

Since the sample size \( n \) must be a whole number, we round up to the next whole number.

### Final Answer:
The smallest sample size needed is \( \boxed{39} \).

Would you like any additional details or have any questions?

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Here are five related questions you might find interesting:
1. How does the z-value change for different confidence levels?
2. How does increasing the desired margin of error affect the sample size?
3. What if the variance was \( 4 \times 25 \, \text{cm}^2 \)? How would that affect the sample size?
4. What happens to the sample size if the confidence level is increased to 95%?
5. How do we determine the sample size if the population variance is unknown?

**Tip:** When calculating sample size, always ensure you round up to the next whole number to maintain the desired confidence level and margin of error.

Learn how to calculate the smallest sample size needed to achieve a 91.98% confidence interval with a margin of error no larger than 2 cm. Understand the concepts of confidence intervals, normal distribution, and sample size determination using specific formulas and calculations.

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