Question

Problem 1:

(a) Find the critical value z α/2 of the standard Normal distribution that corresponds to the confidence level of 100(1 − α)% = 88%

(b) Find the critical value t α/2 of the Student distribution that corresponds to the confidence level of 100(1 − α)% = 95% and 12 observations.

Answer 1

solution

At 88% confidence level the z is ,

= 1 - 88% = 1 - 0.88 = 0.12

/ 2 = 0.12 / 2 = 0.06

Z/2 = Z0.06= 1.555 ( Using z table )

b

Degrees of freedom = df = n - 1 =12 - 1 = 11

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,11 = 2.201 ( using student t table)