To calculate the required sample size for estimating the mean with a specified maximum error, we use the following formula for sample size $n$:

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

Where:

• $Z$ is the z-score corresponding to the desired confidence level,
• $\sigma$ is the population standard deviation,
• $E$ is the maximum error (or margin of error) allowed.

Step-by-Step Calculation:

1. Identify the values:

   • Confidence level = 98% → corresponding z-score ($Z$) = 2.33
   • Standard deviation ($\sigma$) = 1.9 kWh
   • Maximum error ($E$) = 0.15 kWh

2. Substitute the values into the formula:

$n = \left( \frac{2.33 \cdot 1.9}{0.15} \right)^2$

3. Simplify the expression:

$n = \left( \frac{4.427}{0.15} \right)^2 = \left( 29.5133 \right)^2$

4. Square the result:

$n = 870.90$

Since the sample size must be a whole number, and we round up, the required sample size is:

$n = 871$

Final Answer:

The required sample size is 871.

Would you like more details or have any other questions?

Here are 5 related questions to consider:

1. How does changing the confidence level affect the required sample size?
2. What happens if the standard deviation is different in future studies?
3. How is the z-score determined for different confidence levels?
4. What if a smaller margin of error is desired?
5. How can the cooperative ensure the sample is representative of the population?

Tip: When estimating means, a higher confidence level or a smaller margin of error will always increase the required sample size.