Score: 0 of 3 points
You are interested in estimating the average monthly electricity consumption (in kilowatt-hours) of households in your neighborhood. To do this, you plan on constructing a
confidence interval; however, you’re unsure about how many households should be included in the sample. If you want your sample estimate to be within 6 kilowatt-hours of the
actual mean with a confidence level of , how many households should be included in your sample? Assume that the standard deviation of the monthly electricity
consumption for all households in this neighborhood is 39 kilowatt-hours.
Sample Size

Question

Score: 0 of 3 points
You are interested in estimating the average monthly electricity consumption (in kilowatt-hours) of households in your neighborhood. To do this, you plan on constructing a
confidence interval; however, you’re unsure about how many households should be included in the sample. If you want your sample estimate to be within 6 kilowatt-hours of the
actual mean with a confidence level of

, how many households should be included in your sample? Assume that the standard deviation of the monthly electricity
consumption for all households in this neighborhood is 39 kilowatt-hours.
Sample Size

UpStudy AI · Accepted Answer

The margin of error equation for estimating a population mean is given by

$$
E = z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}},
$$

where  
$$ E = 6 $$ kilowatt-hours (the desired margin of error),  
$$ \sigma = 39 $$ kilowatt-hours (the population standard deviation), and  
$$ z_{\frac{\alpha}{2}} $$ is the critical value corresponding to a $$96\%$$ confidence level.

For a confidence level of $$96\%$$, we have

$$
\alpha = 1 - 0.96 = 0.04 \quad \text{and} \quad \frac{\alpha}{2} = 0.02.
$$

The critical value $$ z_{0.02} $$ is approximately $$2.054$$.

Setting up the equation, we have

$$
6 = 2.054 \frac{39}{\sqrt{n}}.
$$

Solve for $$ \sqrt{n} $$ by rearranging:

$$
\sqrt{n} = 2.054 \frac{39}{6}.
$$

Calculating the numerator,

$$
2.054 \times 39 \approx 80.126,
$$

so

$$
\sqrt{n} \approx \frac{80.126}{6} \approx 13.354.
$$

Squaring both sides,

$$
n \approx (13.354)^2 \approx 178.3.
$$

Since the sample size must be a whole number and to ensure the margin of error is not exceeded, round up to the next integer:

$$
n = 179.
$$
To estimate the average monthly electricity consumption within 6 kilowatt-hours with a 96% confidence level, you need to include **179 households** in your sample.

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