Question

Consider a population with a known standard deviation of 13.6. In order to compute an interval estimate for the population mean, a sample of 40 observations is drawn. [You may find it useful to reference the z table.]

a. Is the condition that X− is normally distributed satisfied? Yes No

b. Compute the margin of error at a 99% confidence level. (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places and final answer to 2 decimal places.)

c. Compute the margin of error at a 99% confidence level based on a larger sample of 240 observations. (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places and final answer to 2 decimal places.)

d. Which of the two margins of error will lead to a wider confidence interval?

99% confidence with n = 40.

99% confidence with n = 240.

Answer 1

Solution :

Given that,

Population standard deviation = = 13.6

a) Yes, A population with a known standard deviation of 13.6.

b)

Sample size = n = 40

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 * (13.6 / 40 )

= 5.54

c)

Sample size = n = 240

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 * (13.6 / 240 )

= 2.26

d)

99% confidence with n = 40.