Question

A random sample of 30 households was selected as part of a study on electricity usage, and
the number of kilowatt-hours (kWh) was recorded for each household in the sample for the
March quarter of 2006. The average usage was found to be 375kWh. In a very large study in the
March quarter of the previous year it was found that the standard deviation of the usage was
81kWh. Assuming the standard deviation is unchanged and that the usage is normally
distributed, provide an expression for calculating a 99% confidence interval for the mean usage
in the March quarter of 2006.

Answer 1

n = Sample Size = 30

= Sample Mean = 375

= Population SD = 81

SE = /

= 81/

= 14.7885

=0.01

From Table, critical values of Z = 2.576

Confidence Interval:

375 (2.576 X 14.7885)

= 375 38.0952

= (336.9048 ,413.0952)

Confidence Interval:

336.9048 <   <   413.0952