Question

Suppose that an accounting firm does a study to determine the time needed to complete one persons tax form. 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. Construct a 90% confidence interval for the population mean time to complete the tax forms.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 23.6

Population standard deviation = = 7.0

Sample size = n = 100

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z0.05 = 1.645

Z/2 = 1.645  
Margin of error = E = Z/2 * ( /n)

= 1.645 * (7 /  100 )

= 1.151

At 90 % confidence interval estimate of the population mean is,

- E < < + E

23.6 - 1.151 <   < 23.6 + 1.151

22.449 <   < 24.751

(22.449 , 24.751 )