Question

Consider a normal population distribution with the value of o known. (a) What is the confidence level for the interval x 2.810/ n? (Round your answer to one decimal place.) % (b) What is the confidence level for the interval + 1.450/ n? (Round your answer to one decimal place.) % (c) What value of zul in the CI formula below results in a confidence level of 99.7%? (Round your answer to two decimal places.) (8–2012. o. X + 22 Za/2 = (d) Answer the question posed in part (c) for a confidence level of 78%. (Round your answer to two decimal places.) Za/2 = You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

a.)

Here given critical value of z = $\pm$ 2.81

Now,

P(z > 2.81) = 1−P ( Z < 2.81 )=1−0.9975 = 0.0025

P(z < -2.81) = 1 - P(Z > -2.81) = 0.0025

Hence, P(-2.81 < Z < 2.81) = 0.005

Hence Confidence level = 1 - 0.005 = 0.995 = 99.5%

b.)

Here given critical value of z = $\pm$ 1.45

Now,

P(z > 1.45) = 1−P ( Z < 1.45 )=1−0.9265 = 0.0735

P(z < -1.45) = 1 - P(Z > -1.45) = 0.0735

Hence, P(-1.45 < Z < 1.45) = 0.147

Hence Confidence level = 1 - 0.147 = 0.853 = 85.3%

c.)

Here given confidence level = 99.7

Hence, $\alpha$ = 1 - 0.997 = 0.003

For this level critical value is given as 2.97

d.)

Here confidence level = 78%

Hence, $\alpha$ = 1 - 0.78 = 0.22

For this critical value = 1.23