Question

Determine the t critical value for a two-sided confidence interval in each of the following situations. (Round your answers to three decimal places.) (a) Confidence level = 95%, df = 10 (b) Confidence level = 95%, df = 15 (c) Confidence level = 99%, df = 15 (d) Confidence level = 99%, n = 5 (e) Confidence level = 98%, df = 22 (f) Confidence level = 99%, n = 38 You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Solution :

Given that,

(a)

Degrees of freedom = df = 10

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,10 = 2.228

critical value = 2.228

(b)

Degrees of freedom = df = 15

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,15 = 2.131

critical value = 2.131

(c)

Degrees of freedom = df =15

At 99% confidence level the t is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

t_{$\alpha$ /2,df} = t_0.005,15 = 2.947

critical value = 2.947

(d)

sample size = n = 5

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

At 99% confidence level the t is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

t_{$\alpha$ /2,df} = t_0.005,4 = 4.604

critical value = 4.604

(e)

Degrees of freedom = df = 22

At 98% confidence level the t is ,

$\alpha$ = 1 - 98% = 1 - 0.98 = 0.02

$\alpha$ / 2 = 0.02 / 2 = 0.01

t_{$\alpha$ /2,df} = t_0.01,22 = 2.508

critical value = 2.508

(f)

sample size = n = 38

Degrees of freedom = df = n - 1 = 38 - 1 = 37

At 99% confidence level the t is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

t_{$\alpha$ /2,df} = t_0.005,37 = 2.715

critical value = 2.715