Question 13 of 14, Step 1 of 1 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 2.3 kWh . The mean electricity usage per family was found to be 15.3 kWh per day for a sample of 3611 families. Construct the $$ 99 \% $$ confidence interval for the mean usage of electricity. Round your answers to one decimal place.

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The confidence interval for the mean is given by

$$
\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}
$$

where  
$$\bar{x} = 15.3$$ kWh,  
$$\sigma = 2.3$$ kWh, and  
$$n = 3611$$.

For a $$ 99\% $$ confidence interval, the critical value is approximately

$$
z_{0.005} \approx 2.576.
$$

First, compute the standard error:

$$
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{2.3}{\sqrt{3611}}.
$$

Calculating the square root:

$$
\sqrt{3611} \approx 60.1.
$$

Thus,

$$
\text{SE} \approx \frac{2.3}{60.1} \approx 0.0383.
$$

Now, compute the margin of error:

$$
\text{ME} = z_{0.005} \cdot \text{SE} \approx 2.576 \times 0.0383 \approx 0.0987.
$$

Rounded to one decimal place, the margin of error is approximately $$0.1$$ kWh.

Finally, the confidence interval is

$$
15.3 \pm 0.1,
$$

which gives us

$$
(15.2, 15.4) \text{ kWh per day}.
$$
The 99% confidence interval for the mean electricity usage is 15.2 to 15.4 kWh per day.

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