To determine the required sample size $n$ for estimating a population mean with a specified confidence level and margin of error, we use the following formula:

$n = \left( \frac{Z \cdot \sigma}{E} \right)^2$

where:

• $Z$ is the Z-score corresponding to the desired confidence level,
• $\sigma$ is the population standard deviation (in this case, 48.7),
• $E$ is the margin of error (in this case, 10).

Step 1: Determine the Z-score for a 98% confidence level

For a 98% confidence level, we look up the Z-score that leaves 1% in each tail (because $100\% - 98\% = 2\%$, split between two tails). The Z-score for 98% confidence is approximately $Z = 2.33$.

Step 2: Plug values into the formula

Substitute $Z = 2.33$, $\sigma = 48.7$, and $E = 10$:

$n = \left( \frac{2.33 \cdot 48.7}{10} \right)^2$

Step 3: Calculate the sample size

1. Calculate the numerator: $2.33 \cdot 48.7 = 113.471$.
2. Divide by the margin of error: $\frac{113.471}{10} = 11.3471$.
3. Square this result: $11.3471^2 \approx 128.74$.

Since the sample size $n$ must be a whole number, we round up to ensure the desired confidence level.

Final Answer: $n = 129$

This is the required sample size to be 98% confident that the estimate is within 10 units of the true population mean.

Would you like further details on this calculation?

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1. How does the Z-score change for different confidence levels?
2. What happens if the margin of error is decreased?
3. Why do we use the Z-score and not a t-score in this case?
4. How would this calculation change if the population standard deviation were unknown?
5. Can this formula be applied to other confidence intervals, like 90% or 95%?

Tip: Always round up to the next whole number for sample sizes to maintain the desired confidence level and precision.