Question

1. A simple random sample of 50 customers is selected from an account receivable portfolio and the sample mean account balance is $1000. The population standard deviation σ is known to be $200.

(12 marks)

a. Construct a 95% confidence interval for the mean account balance of the population.

b. What is the margin of error for estimating the mean account balance at the 95% confidence level?

c. Construct a 95% confidence interval for the mean account balance of the population if the sample mean account balance is obtained from a random sample of 100 instead of 50.

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = = 1000

Population standard deviation =    = 200

Sample size = n = 50

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z_{0.25 = 1.96}

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

= 1.96 * (200 /   50)

= 55.44

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

  ± E

1000 ± 55.44

(944.56 , 1055.44)

b)  Margin of error = E = Z/2 * ( /n)

= 1.96 * (200 /   50)

= 55.44

c) n = 100

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

= 1.96 * (200 /   100)

= 39.2

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

  ± E

1000 ± 39.2

(960.8 , 1039.2)