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.