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)
solution
At 82% confidence level the z is ,
= 1 - 82% = 1 - 0.82 = 0.18
/ 2 = 0.18 / 2 = 0.09
Z/2 = Z0.09 = 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 = t0.025,14 = 2.145
Get Answers For Free
Most questions answered within 1 hours.