Question

What is the value of the t score for a 99% confidence interval if we take a sample of size 22?

Answer 1

Answer: 2.831

Step-by-step explanation:

As per given , we have

sample size : n= 22

Degree of freedom = n-1 = 22-1=21

Significance level for 99% confidence interval = $\dfrac{1-0.99}{2}=0.005$

Using student's t-value table, we have

$t_{0.005,\ 21}=2.831$

Hence, the  t score for a 99% confidence interval if we take a sample of size 22= 2.831