With 98% confidence interval and n = 25. Find left critical value for Zinterval.
0 -2.492
0 -2.05
0 -2.797
-2.326

Respuesta :

Answer:

[tex]\alpha=1-0.98 =0.02[/tex]

And [tex]\alpha/2 0.01[/tex], the degrees of freedom are given by:

[tex] df= n-1= 25-1=14[/tex]

Then the critical value using the t distribution with 24 degrees of freedom is:

[tex] t_{\alpha/2}= \pm 2.492[/tex]

And the best solution would be:

0 -2.492

Step-by-step explanation:

For this problem we know that the sample size is n = 25. The confidence level is 98% or 0.98 then the significance would be:

[tex]\alpha=1-0.98 =0.02[/tex]

And [tex]\alpha/2 0.01[/tex], the degrees of freedom are given by:

[tex] df= n-1= 25-1=14[/tex]

Then the critical value using the t distribution with 24 degrees of freedom is:

[tex] t_{\alpha/2}= \pm 2.492[/tex]

And the best solution would be:

0 -2.492