Question

In: Statistics and Probability

a) Use the Student's t distribution to find tc for a 0.99 confidence level when the...

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.)

Solutions

Expert Solution

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)


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