Answer to Question #186155 in Statistics and Probability for Ghulamdastgeer

Question

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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

$n= 30 \\ \mu = 375 \\ \sigma = 81$

z = 2.57 for 99% confidence interval

$CI = \mu ±z \times \frac{\sigma}{\sqrt{n}} \\ = 375 ± 2.57 \times \frac{81}{\sqrt{30}} \\ = 375 ± 38.0$

(337, 413)