Question

A public utility wants to compare the consumption of electricity during the summer season for single-family houses in two counties it services. For each household sampled, the monthly electric bill is recorded and the following table summarizes the results.

Sample MeanSample St $30 County I23 County II19 $117 $97 $23

Based upon this data find a 95% confidence interval for the difference in the average bills between the two counties.

Answer 1

As, the population standard deviation of both counties are not assumed to be equal, we will use t distribution to calculate the 95% confidence interval for the difference in the average bills between the two counties.

Let  s1 is the  standard deviation of county 1, s2 is the standard deviation of county 2, n1 is the size of county 1, n2 is the size of county 2, x1 is the mean of county 1, x2 is the mean of county 2, and SE is the standard error.

SE = sqrt[(s1²/n1) + (s2²/n2)]

= sqrt[(30²/23) + (23²/19)]

= 8.1837

Degree of freedom, DF is ,

DF = (s1²/n1 + s2²/n2)² / { [ (s1² / n1)² / (n1 - 1) ] + [ (s2² / n2)² / (n2 - 1) ] }

= (30²/23 + 23²/19)² / { [ (30² / 23)² / (23 - 1) ] + [ (23² / 19)² / (19 - 1) ] }

= 40 (Rounded to nearest interger)

Critical value of t at 95% confidence and DF = 40 is 2.021

Difference in means = x1 - x2 = 117 - 97 = 20

95% confidence interval for the difference in the average bills between the two counties

(20 - 2.021 * 8.1837, 20 + 2.021 * 8.1837)

(3.4607, 36.5393)