To determine the required sample size for constructing a 99% confidence interval for the population mean, and the estimate must be within [tex]$10 of the population mean, we can use the following formula:
\[ n = \left(\frac{Z \cdot \sigma}{E}\right)^2 \]
where:
- \( n \) is the sample size,
- \( Z \) is the z-score corresponding to the desired confidence level,
- \( \sigma \) is the population standard deviation, and
- \( E \) is the margin of error.
Given:
- Confidence level (CL) = 99%
- \( \sigma \) (population standard deviation) = $[/tex]22.50
- [tex]\( E \)[/tex] (Margin of error) = [tex]$10
To find the z-score that corresponds to a 99% confidence level, we look up the z-value in a standard normal distribution table that accumulates a probability of 0.995 (since 99% confidence interval leaves 0.5% on each side of the distribution).
The z-score for a 99% confidence interval is approximately 2.576.
Now, we can substitute the given values into the formula:
\[ n = \left(\frac{2.576 \times 22.50}{10}\right)^2 \]
\[ n = \left(\frac{57.96}{10}\right)^2 \]
\[ n = \left(5.796\right)^2 \]
\[ n = 33.601 \]
Since the sample size must be a whole number and you cannot have a fraction of a sample, we round up to the next whole number.
Therefore, the required sample size is \( \lceil 33.601 \rceil = 34 \).
You would need a sample size of at least 34 dishwashers to estimate the mean repair cost within $[/tex]10 of the population mean with a confidence level of 99%.
