Question

A)Use the Student's t distribution to find tc for a 0.99 confidence level when the sample is 27. (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 6. (Round your answer to three decimal places.)

Answer 1

solution:;

(a)

n = 27

Degrees of freedom = df = n - 1 = 27 - 1 = 26

so,

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

  / 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,26 =2.779.

tc  =2.779

(b)

n = 6

Degrees of freedom = df = n - 1 = 6 - 1 = 5

so,

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

  / 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,5 =4.032 ( using student t table)

tc  = 4.032