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

a) SE = 25

b) MOE = 41

c) CI = 1951 ; 2049

Step-by-step 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

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