Answer to Question #157603 in Statistics and Probability for umar

Question

A random sample of 20 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 sample standard deviation of the usage was 81kWh. Assuming the that the usage is normally distributed, provide an expression for calculating a 95% confidence interval for the mean usage in the March quarter of 2006. 

Answer 1

$n= 20 \\ \bar{X}=375 \;kWh \\ σ = 81 \;kWh$

Confidence interval for population mean:

$(\bar{X} -Z_{α/2} \times \frac{σ}{\sqrt{n}} , \bar{X} + Z_{α/2} \times \frac{σ}{\sqrt{n}}) \\ 1 -α = 0.95 \\ α = 0.05$

$Z_{α/2} = Z_{0.025} = 1.96$

Z is the standard normal variable.

95% confidence interval for mean usage in the March quarter of 2006:

$(375 -1.96 \times \frac{81}{\sqrt{20}} , 375 + 1.96 \times \frac{81}{\sqrt{20}}) \\ = (339.5, 410.5)$