The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day.  Assume that the population standard deviation is 1.41.4 kWh.  The mean electricity usage per family was found to be 15.515.5 kWh per day for a sample of 142142 families.  Construct the 80%80% confidence interval for the mean usage of electricity.  Round your answers to one decimal place.

To construct an 80% confidence interval for the mean electricity usage per family, we will use the formula for the confidence interval for the population mean when the population standard deviation is known:

\[
\text{Confidence Interval} = \bar{x} \pm z \times \frac{\sigma}{\sqrt{n}}
\]

Where:
- \(\bar{x}\) is the sample mean.
- \(z\) is the z-score corresponding to the confidence level.
- \(\sigma\) is the population standard deviation.
- \(n\) is the sample size.

### Given dataTo construct an 80% confidence interval for the mean electricity usage per family, we will use the formula for the confidence interval for the population mean when the population standard deviation is known:

\[
\text{Confidence Interval} = \bar{x} \pm z \times \frac{\sigma}{\sqrt{n}}
\]

Where:
- \(\bar{x}\) is the sample mean.
- \(z\) is the z-score corresponding to the confidence level.
- \(\sigma\) is the population standard deviation.
- \(n\) is the sample size.

### Given 
- Population standard deviation (\(\sigma\)) = 1.4 kWh.
- Sample mean (\(\bar{x}\)) = 15.5 kWh.
- Sample size (\(n\)) = 142.
- Confidence level = 80%, so the z-score for an 80% confidence interval (from the z-table) is approximately \(z = 1.28\).

### Step 1: Calculate the standard error of the mean (SE)

\[
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{142}} \approx \frac{1.4}{11.916} \approx 0.1175
\]

### Step 2: Calculate the margin of error (ME)

\[
\text{ME} = z \times \text{SE} = 1.28 \times 0.1175 \approx 0.1504
\]

### Step 3: Construct the confidence interval

\[
\text{Confidence Interval} = \bar{x} \pm \text{ME} = 15.5 \pm 0.1504
\]

Lower limit = \(15.5 - 0.1504 \approx 15.3\)

Upper limit = \(15.5 + 0.1504 \approx 15.7\)

### Final Answer:
The 80% confidence interval for the mean usage of electricity is approximately **15.3 kWh to 15.7 kWh** per day.

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Would you like more details or explanations on any part? 

Here are 5 related questions you could explore:
1. How is the z-score for different confidence levels determined?
2. What happens if the population standard deviation is unknown?
3. How would the confidence interval change with a larger sample size?
4. Why is the z-score used instead of the t-score in this case?
5. How does increasing the confidence level affect the width of the confidence interval?

**Tip:** As you increase the confidence level (e.g., from 80% to 95%), the confidence interval becomes wider because you are more certain that the true population mean lies within the interval.

The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. Assume that the population standard deviation is 1.4 kWh. The mean electricity usage per family was found to be 15.5 kWh per day for a sample of 142 families. Construct the 80% confidence interval for the mean usage of electricity. Round your answers to one decimal place.

Learn how to construct an 80% confidence interval for the mean household electricity usage in kWh per day using a population standard deviation of 1.4 kWh, a sample mean of 15.5 kWh, and a sample size of 142 families. This problem explains the confidence interval formula and uses a z-score of 1.28.

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