Question

The table t-value associated with 8 degrees of freedom and used to calculate a 99% confidence interval is _______.

Select one:

a. 3.355

b. 1.860

c. 1.397

d. 2.896

Cameron Sinclair, Information Services Manager with Global Financial Service (GFS), is studying employee use of GFS email for non-business communications. He plans to use a 95% confidence interval estimate of the proportion of email messages that are non-business; he will accept a 0.05 error. Previous studies indicate that approximately 30% of employee email is not business related. Cameron should sample _______ email messages.

Select one:

a. 14

b. 323

c. 457

d. 12

Answer 1

Solution:

a ) Given thta,

Degrees of freedom = 8

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _{/2,df f} = t_0.005,8 = 3.355

A 99% confidence interval is =3.355

Answer a ) is correct.

b ) Given that,

= 30% = 0.30

1 - = 1 - 0.30 = 0.70

margin of error = E = 0.05

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 = 1.960

Sample size = n = ((Z _{/ 2}) / E)² * * (1 - )

= (1.960 / 0.05)² * 0.30 * 0.70

= 322.6944

= 323

n = sample size = 323

Answer b ) is correct.