Question

Consider the population of electric usage per month for houses. The standard deviation of this population is 128 kilowatt-hours. What is the smallest sample size to provide a 90% confidence interval for the population mean with a margin of error of 30 or less

Answer 1

Answer: 50

Step-by-step explanation:

Formula for sample size(n) if population standard deviation $(\sigma)$ is known:

$n=(\dfrac{\sigma\times z_c}{E})^2$ , E = Margin of error , $z^c$ =  Critical z-value

As per given,

$\sigma= 128$ kilowatt-hours.

Critical z-value for 90% confidence = 1.645

E =  30

Required sample size : $n=(\dfrac{128\times1.645}{30})^2=(7.01867)^2$

$=49.2617285689\approx50$

Hence,  the smallest sample size to provide a 90% confidence interval for the population mean with a margin of error of 30 or less = 50