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 25 cm². What is the smallest sample size needed?

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To solve this problem, we need to calculate the smallest sample size \( n \) required to achieve a confidence interval for the mean with a specified confidence level (91.98%) and a maximum allowable length of 2 cm. Given that the population is normally distributed with a variance of \( \sigma^2 = 25 \, \text{cm}^2 \), we will follow these steps:

### Step-by-Step Solution

1. **Confidence Level and Z-score**:
   A 91.98% confidence level corresponds to a Z-score. Since the confidence level is not a standard value, we need to find the Z-score that leaves 4.01% in each tail (i.e., 100% - 91.98% = 8.02%, so 4.01% in each tail).

   We will denote this Z-score as \( Z_{\alpha/2} \), where \( \alpha = 1 - 0.9198 = 0.0802 \).

2. **Formula for the Confidence Interval**:
   The formula for the confidence interval for a mean \( \mu \) is:

   \[
   \text{CI} = \bar{x} \pm Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right)
   \]

   The length of the confidence interval is given by:

   \[
   \text{Length of CI} = 2 \times Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right)
   \]

   We are given that this length should be no more than 2 cm, so:

   \[
   2 \times Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right) \leq 2
   \]

3. **Rearrange to Find \( n \)**:
   Substituting the value for \( \sigma \):

   \[
   Z_{\alpha/2} \left( \frac{\sqrt{25}}{\sqrt{n}} \right) \leq 1
   \]

   Since \( \sqrt{25} = 5 \):

   \[
   Z_{\alpha/2} \left( \frac{5}{\sqrt{n}} \right) \leq 1
   \]

   Solving for \( n \):

   \[
   \frac{5 Z_{\alpha/2}}{\sqrt{n}} \leq 1 \implies 5 Z_{\alpha/2} \leq \sqrt{n}
   \]

   \[
   n \geq (5 Z_{\alpha/2})^2
   \]

4. **Find the Z-score**:
   We need the Z-score corresponding to the 91.98% confidence level, which leaves 4.01% in each tail.

Let's calculate the Z-score and the smallest sample size \( n \).The Z-score corresponding to a 91.98% confidence level is approximately \( 1.75 \).

The smallest sample size needed to ensure that the 91.98% confidence interval for the mean is no longer than 2 cm is \( \boxed{77} \).

Would you like more details on this calculation, or do you have any questions?

Here are some related questions you might find useful:
1. How would the required sample size change with a different confidence level?
2. What if the variance were different, say 16 cm² or 36 cm²?
3. How can you interpret confidence intervals in the context of hypothesis testing?
4. Why do we use the Z-score for normally distributed populations?
5. How would this calculation differ if the population were not normally distributed?

**Tip:** Confidence intervals provide a range of values within which the true population parameter is expected to lie, given a specified level of confidence.

Learn how to calculate the minimum sample size needed for a 91.98% confidence interval with a maximum length of 2 cm. This problem involves statistical concepts such as confidence intervals, normal distribution, and sample size determination based on variance. Suitable for advanced college-level statistics.

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