In: Statistics and Probability
a) Use the Student's t distribution to find tc for a 0.99 confidence level when the sample is 25. (Round your answer to three decimal places.)
b) Use the Student's t distribution to find tc for a 0.90 confidence level when the sample is 29. (Round your answer to three decimal places.)
c) Use the Student's t distribution to find tc for a 0.95 confidence level when the sample is 14. (Round your answer to three decimal places.)
solution
(a)n = 25
Degrees of freedom = df = n - 1 =25 - 1 =24
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2 df = t0.005,24 = 2.797 ( using student t table)
tc = 2.797
(B)
n = 29
Degrees of freedom = df = n - 1 = 29- 1 = 28
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
t /2,df = t0.05,28 = 1.701 ( using student t table)
tc=1.701
(c)
n =14
Degrees of freedom = df = n - 1 = 14- 1 =13
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2= 0.05 / 2 = 0.025
t /2,df = t0.025,13 = 2.160 ( using student t table)