In order to estimate the average electric usage per month, a sample of
n = 81 houses was selected, and the electric usage was determined.
(a) Assume a population standard deviation of 450-kilowatt hours.
Determine the standard error of the mean. (b) With a probability of
0.95, what is the size of the margin of error? (c) If the sample mean is 1858-kilowatt hours, what is the 95% confidence interval estimate
of the population mean?

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AyBaba7 AyBaba7 · Answer 1 · 2020-03-14T03:17:52+01:00

Answer:

a) 50

b) Margin of error of the sample = 98 KWh

Margin of error of the population = 882 KWh

c) The 95% confidence interval estimate

of the population mean is (1760, 1956) KWh

Step-by-step explanation:

Standard error of the mean = (Standard deviation of the population)/(√n)

where n = sample size = 81

Standard deviation of the population = 450 kWh

Standard error of the mean = (450/√81) = (450/9)

Standard error of the mean = 50

b) The margin of error can be estimated in two ways.

One in terms of population and the other in terms of the sample.

For the population

Margin of error = (critical value) × (standard deviation of the population)

Critical value for 95% confidence = 1.96

Standard deviation of the population = 450 KWh

Margin of error = 1.96 × 450 = 882 KWh

For the sample,

Margin of error = (critical value) × (standard error of the mean)

Margin of error = 1.96 × 50 = 98 KWh.

c) Sample mean = 1858 KWh

The lower limit of the interval = (Sample mean) - (Margin of error) = 1858 - 98 = 1760 KWh

The upper limit of the interval = (Sample mean) + (Margin of error) = 1858 + 98 = 1956 KWh.

The interval is (1760, 1956) KWh