Question

The average monthly electric bill of a random sample of 256 residents of a city is $90 with a standard deviation of $24.

Construct a 95% confidence interval for the mean monthly electric bills of all residents. (Round to two decimal places) [Answer, Answer]

Construct a 99% confidence interval for the mean monthly electric bills of all residents. (Round to two decimal places) [Answer, Answer]

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 90

Population standard deviation = = 24

Sample size = n = 256

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (24 / 256)

= 2.94

At 95% confidence interval estimate of the population mean is,

- E < < + E

90 - 2.94 < < 90 + 2.94

87.06 < < 92.94

(87.06 , 92.94)

(b)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (24 / 256)

= 3.86

At 99% confidence interval estimate of the population mean is,

- E < < + E

90 - 3.86 < < 90 + 3.86

86.14 < < 93.86

(86.14 , 93.86)