In: Statistics and Probability
3. Determine the critical value of t in each of the following circumstances: a. 1 − α = 0.95, n=15
b. 1 − α = 0.99, n=20
c. 1 − α = 0.90, n=40
n = 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 ( using student t table)
b.
Degrees of freedom = df = n - 1 = 20 - 1 = 19
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2 df = t0.005,19 = 2.861 ( using student t table)
c.
Degrees of freedom = df = n - 1 =40 - 1 = 39
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
t /2,df = t0.05,39 = 1.685 ( using student t table)