Question

The t value for a 95% confidence interval estimation with 24 degrees of freedom
(not the sample size n) is

a. 1.317836
b. 1.710882
c. 2.063899
d. 2.492159
e. 2.796939

Answer 1

Answer:

c. 2.063899

Step-by-step explanation:

Assuming a confidence interval bilateral.

For this case the confidence is 95% so then the significance level is:

$\alpha = 1-0.95 =0.05$

And $\alpha/2 = 0.025$

The degrees of freedom are given $df = 24 =n-1$

So then we need to find a quantile on the t distribution with 24 degrees of freedom that accumulates 0.025 of the area on each tail, and we can use the following excel code:

"=T.INV(1-0.025,24)"

Or equivalently "=T.INV.2T(0.05,24)"

And we got $t_{crit}= 2.062899$

So the best option is:

c. 2.063899

Answer 2

The best estimate for the t-value from the table will be 2.063899

To get the t value for a 95% confidence interval estimation with 24 degrees of freedom, we will use the t-table.

If the confidence interval is 95%, hence the required probability value that is realistic will b 5% i.e 0.05

Given the following:

• df = 24
• probability = 0.05

We are to check the table for df at 24 under the probability value of 0.05. The best estimate from the table will be 2.063899

Learn more on t-test here: https://brainly.com/question/6589776