Question

A study has indicated that the sample size necessary to estimate the average electricity use by residential customers of a large western utility company is 900 customers. Assuming that the margin of error associated with the estimate will be ±30 watts and the confidence level is stated to be 90 percent, what was the value for the population standard deviation?

Answer 1

Answer:

$\sigma=547.1125$

Step-by-step explanation:

-The margin of error is calculated using the formula:

$ME=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

-We substitute the values, n=900 and ME=30 in the formula to solve for standard deviation:

$ME=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}\\\\\\30=1.645\times\frac{\sigma}{\sqrt{900}}\\\\\sigma=\frac{30^2}{1.645}\\\\=547.1125$

Hence, the population standard deviation is 547.1125