To determine the minimum sample size required for the estimate, we will use the formula for the sample size in estimating a population mean when the population variance is known:

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

Where:

• $Z_{\alpha/2}$ is the z-score corresponding to the desired confidence level.
• $\sigma$ is the standard deviation of the population.
• $E$ is the margin of error.
• $n$ is the required sample size.

Step 1: Identify the values

• Confidence level = 90%, so $\alpha = 1 - 0.90 = 0.10$, which gives $\alpha/2 = 0.05$.
• From the z-score table, the z-score for $Z_{0.05}$ (i.e., the critical value for a 90% confidence interval) is approximately 1.645.
• Population variance $\sigma^2 = 1.69$, so $\sigma = \sqrt{1.69} = 1.3$.
• Margin of error $E = 0.13$.

Step 2: Plug values into the formula

$n = \left( \frac{1.645 \cdot 1.3}{0.13} \right)^2$ $n = \left( \frac{2.1385}{0.13} \right)^2$ $n = (16.45)^2 = 270.7025$

Step 3: Round up

Since the sample size must be a whole number, round 270.7025 up to the next integer.

$n = 271$

Conclusion:

The minimum sample size required to ensure the estimate has an error of at most 0.13 at the 90% confidence level is 271.

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Would you like more details on any step? Feel free to ask!

Here are some related questions:

1. What is the relationship between confidence level and margin of error?
2. How does increasing the confidence level affect the sample size?
3. Why do we use a z-score for calculating sample size?
4. Can you explain why the sample size formula involves squaring?
5. How does population variance affect the required sample size?

Tip: Higher confidence levels result in larger sample sizes, as they require a more precise estimate of the population parameter.