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 to the power of 2 cm . What is the smallest sample size needed? [8]

To find the smallest sample size required for a 91.98% confidence interval for the mean, we need to follow these steps:

### Given Data:
- Confidence level \( (1-\alpha) = 91.98\% \)
- Population standard deviation \( \sigma = 25 \, \text{cm} \)
- The desired margin of error \( E \leq 2 \, \text{cm} \)

### 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 need to 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 \( E = 2 \, \text{cm} \), \( \sigma = 25 \, \text{cm} \), and \( z_{\alpha/2} = 1.75 \), we can set up the equation:
\[
2 = 1.75 \cdot \frac{25}{\sqrt{n}}
\]

### Step 3: Solve for \( n \) (sample size).
Rearrange the equation to solve for \( n \):
\[
\sqrt{n} = \frac{1.75 \cdot 25}{2}
\]

\[
\sqrt{n} = \frac{43.75}{2} = 21.875
\]

\[
n = 21.875^2
\]

\[
n \approx 478.52
\]

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 \( \boxed{479} \).

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 do you determine the z-value for different confidence levels?
2. What happens if the desired margin of error is reduced to 1 cm?
3. How does the sample size change if the population variance is larger?
4. What is the impact of a lower confidence level on the required sample size?
5. How would you calculate the sample size if the population standard deviation was unknown?

**Tip:** Always round up the sample size when dealing with confidence intervals to ensure the margin of error is not exceeded.

Learn how to calculate the smallest sample size needed for a 91.98% confidence interval with a margin of error no longer than 2 cm. This problem involves using statistical concepts such as confidence intervals, normal distribution, and the margin of error formula. Suitable for college-level students and professionals.

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