Question

Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal.

a. What is a point estimate for the true average time needed to complete a person’s tax

forms?

23.6

b. What is the margin of error?

1.519

c. What is a 97% confidence interval for the true average time needed to complete a

person’s tax forms?

[22.081, 25.119]

Can you explain those answers?

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = = 23.6

Population standard deviation =    = 7.0

Sample size = n = 100

b) At 97% confidence level

= 1 - 97%  

= 1 - 0.97 =0.03

/2 = 0.015

Z/2 = Z_{0.015 = 2.17}

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

E = 2.17 * ( 7.0 /  100 )

E = 1.519

At 97% confidence interval estimate of the population mean is,

  ± E

23.6 ± 1.519

( 22.081, 25.119)