Question

Part 1. For a 95% confidence interval for u when df = 19, the t-critical value is _______

Part 2: For a 90% confidence interval for u when df = 19, the t-critical value is _______

Part 3: For a 99% confidence interval for u when df = 19, the t-critical value is _______

Part 4: For a 95% confidence interval for u when df = 42, the t-critical value is _______

Part 5: For a 99% confidence interval for u when df = 1000, the t-critical value is _______

Answer 1

Solution :

Given that,

a ) Degrees of freedom = df = 19

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,19=2.101

The critical value =2.101

b ) Degrees of freedom = df = 19

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,19 =1.734

  The critical value =1.734

c ) Degrees of freedom = df = 19

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 = 2.878

The critical value =2.878

d ) Degrees of freedom = df = 42

At 95% confidence lea dvel the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,42 = 2.019

The critical value =2.019

e ) Degrees of freedom = df = 1000

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,1000 =2.581

The critical value =2.581