Question

With 98% confidence interval and n = 25. Find left critical value for Zinterval.

0 -2.492
0 -2.05
0 -2.797
-2.326

Answer 1

`\alpha=1-0.98 =0.02`

And
`\alpha/2 0.01`, the degrees of freedom are given by:

`df= n-1= 25-1=14`

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

`t_(\alpha/2)= \pm 2.492`

And the best solution would be:

0 -2.492

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:

`\alpha=1-0.98 =0.02`

And
`\alpha/2 0.01`, the degrees of freedom are given by:

`df= n-1= 25-1=14`

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

`t_(\alpha/2)= \pm 2.492`

And the best solution would be:

0 -2.492