Question

Determine the t critical value for a lower or an upper confidence bound in each of the following situations. (Round your answers to three decimal places.) (a) Confidence level = 95%, df = 5 (b) Confidence level = 95%, df = 20 (c) Confidence level = 99%, df = 20 (d) Confidence level = 99%, n = 5 (e) Confidence level = 97.5%, df = 23 (f) Confidence level = 99%, n = 36 You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Solution :

Given that,

Degrees of freedom = df = 5

a) At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.01

$\alpha$/2 = 0.025

t$\alpha$/2,df = t_0.025,5 = 2.571

b) Degrees of freedom = df = 20

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.01

$\alpha$/2 = 0.025

t$\alpha$/2,df = t_0.025,20 = 2.086

c) Degrees of freedom = df = 20

At 99% confidence level

$\alpha$ = 1 - 99%

$\alpha$ =1 - 0.99 =0.01

$\alpha$/2 = 0.005

t$\alpha$/2,df = t_0.005,20 = 2.845

d) n = 5

Degrees of freedom = df = n - 1 = 5 - 1 = 4

At 99% confidence level

$\alpha$ = 1 - 99%

$\alpha$ =1 - 0.99 =0.01

$\alpha$/2 = 0.005

t$\alpha$/2,df = t_0.005,4 = 4.604

d) Degrees of freedom = df = 23

At 97.5% confidence level

$\alpha$ = 1 - 97.5%

$\alpha$ =1 - 0.975 =0.025

$\alpha$/2 = 0.0125

t$\alpha$/2,df = t_0.0125,23 = 2.398

f) n = 36

Degrees of freedom = df = n - 1 = 36 - 1 = 35

At 99% confidence level

$\alpha$ = 1 - 99%

$\alpha$ =1 - 0.99 =0.01

$\alpha$/2 = 0.005

t$\alpha$/2,df = t_0.005,35 = 2.724