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

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