Question

The electric cooperative needs to know the mean household usage of
electricity by its non-commercial customers in kWh per day. They
would like the estimate to have a maximum error of 0.13 kWh. A
previous study found that for an average family the standard
deviation is 2.1 kWh and the mean is 15.8 kWh per day. If they are
using a 99% level of confidence, how large of a sample is required
to estimate the mean usage of electricity? Round your answer up to
the next integer.

Answer #1

Solution :

Given that,

standard deviation = =2.1

Margin of error = E = 0.13

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_{/2}
= 2.58 ( Using z table ( see the 0.005 value in standard normal (z)
table corresponding z value is 2.58 )

sample size = n = [Z_{/2}*
/ E] ^{2}

n = ( 2.58* 2.1/ 0.13 )^{2}

n =1737

Sample size = n =1737

( when you have z value 3 decimal than Z_{/2}
= 2.576 n=1732)

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