Question

The electric cooperative wishes to know the mean household usage of electricity by its non-commercial customers. They believe that the mean is 16.3 kWh per day for each family. Assume a previous study found the standard deviation to be 2.4. How large of a sample would be required in order to estimate the mean number of kWh of electricity used daily per family at the 90% confidence level with an error of at most 0.14 kWh?

Answer 1

Solution :

Given that,

standard deviation = = 2.4

margin of error = E = 0.14

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Sample size = n = ((Z_/2 * ) / E)²

= ((1.645 * 2.4) / 0.14)²

= 795.24 = 796

Sample size = 796