Question

The electric cooperative needs to know the main household usage of electricity by it’s non commercial customers in KWH per day. They would like the estimate to have a maximum error of 0.1 3 KWH. A previous study found that for an average Family the standard deviation is 2.1 KWH and the mean is 15.8 KWH per day. If they’re using a 99% level of confidence how large of a sample is required to estimate the mean usage of electricity. Round your answer to the next integer

Answer 1

1737

Explanation:

Given:

standard deviation (σ) = 2.1

Margin of error = E = 0.13

At 99% confidence level the z is:

`\begin{gathered} \alpha=1-99\% \\ \\ \alpha=1-0.99=0.01 \\ \\ \alpha\text{/2}=(0.01)/(2)=0.005 \\ \\ \text{ The corresponding value of z is the following:} \\ \\ Z_{\alpha\text{/2}}=2.58 \end{gathered}`

We can calculate the sample size using the following formula:

`\begin{gathered} n=\left(\frac{Z_{\alpha \text{/2}}\cdot \sigma }{E}\right)^2 \\ \\ \text{ We replacing:} \\ \\ n=\:\left((2.58\cdot2.1)/(0.13)\right)^2 \\ \\ n=1736.97\approx1737 \end{gathered}`

The sample size is 1737