Question

for a sample of size n=15

the critical z-score for 82% confidence is _ (two decimal accuracy, positive value only)

the critical t-score for 95% confidence is _ (three- decimal accuracy, positive value only)

Answer 1

solution

At 82% confidence level the z is ,

= 1 - 82% = 1 - 0.82 = 0.18

/ 2 = 0.18 / 2 = 0.09

Z_/2 = Z_0.09 = 1.34

(B)n = 15

Degrees of freedom = df = n - 1 = 15- 1 = 14

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,14 = 2.145