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 population standard deviation of 6.8 hours. The population distribution is assumed to be normal.

Construct a 90% confidence interval for the population mean time to complete the tax forms.

(iii) Calculate the error bound. (Round your answer to two decimal places.)

I don't understand the error bound. I keep getting lost when I work through it.

Answer 1

Solution:

since population standard deviation is given ,it is z critical value we have to use in calculations

Givne data

n=100

xbar=23.6

sigma=6.8

z critical for 90%=

=NORMINV(0.95,0,1)

=1.644854

the error bound is margin of error denoted by

=Zcrit*sigma/sqrt(n)

=(1.644854*6.8)/sqrt(100)

=1.118501

=1.12

the error bound=1.12

90% confidence interval for mean is

sample mean-error bound,sample mean+error bound

23.6-1.118501,23.6+1.118501,

22.4815, 24.7185

90% lower limit mean=22.4815

90% upper limit mean=24.7185