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**

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