Question

In order to estimate the average electric usage per month, a sample of 196 houses was selected, and their electric usage determined. a) Assume a population standard deviation of 350 kilowatt hours. Determine the standard error of the mean. b) With a 0.95 probability, determine the margin of error. c) If the sample mean is 2,000 KWH, what is the 95% confidence interval estimate of the population mean

Answer 1

a) SE = 25

b) MOE = 41

c) CI = 1951 ; 2049

Explanation:

Normal distribution

Population mean unknown

Population standard deviation σ = 350 Kwh

a) The standard error of the mean SE is

SE = σ/√

SE = 350 /√196

SE = 350/14

SE = 25

b) If confidence nterval is 95% or 0,95 then

α = 0,05

And from z table we get z(c) = 1,64

MOE = z(c) * SE

MOE = 1,64 * 350/√196

SE = 1,64 * (350)/14

SE = 41

MOE = And from z tabl we get z(c) = 1,64

MOE = 1,64 * 350/√196

MOE = 1,64 * (350)/14

MOE = 1,64 * 25

MOE = 41

c) The confidence interval is:

Z = 2000

α = 1- 0,95

α = 0,05 ⇒ α/2 = 0,025

CI = Z - z(α/2) * σ/√n ; Z + z(α/2) * σ/√n

z(α/2) from z-table is: z(0,025) = 1,96

CI = 2000 - 1,96* 350/√196 ; 2000 + 1,96* 350/√196

CI = 2000 - 1,96*25 ; 2000 + 1,96*25

CI = 2000 - 49 ; 2000 + 49

CI = 1951 ; 2049