a.Use the Student's t distribution to find tc for a 0.95 confidence level when the sample...

a.Use the Student's t distribution to find t_c for a 0.95 confidence level when the sample is 9. (Round your answer to three decimal）

b.Use the Student's t distribution to find t_c for a 0.99 confidence level when the sample is 31. (Round your answer to three decimal places.)

Solutions

Expert Solution

Degrees of freedom = df = n - 1 =9 - 1 = 8

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2= 0.05 / 2 = 0.025

t /2,df = t0.025,8 = 2.306 ( using student t table)

Degrees of freedom = df = n - 1 =31 - 1 = 30

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2 df = t0.005,30 =2.750 ( using student t table)

orchestra answered 2 years ago

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