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 = [tex]\dfrac{1-0.99}{2}=0.005[/tex]
Using student's t-value table, we have
[tex]t_{0.005,\ 21}=2.831[/tex]
Hence, the t score for a 99% confidence interval if we take a sample of size 22= 2.831
We have that for the Question "What is the value of the t score for a 99% confidence interval if we take a sample of size 22" it can be said that
Hence,the t-score for a 99% confidence interval is [tex]t_{0.005,21}=2.831[/tex]
From the question we are told
What is the value of the t score for a 99% confidence interval if we take a sample of size 22?
Generally the equation for the confidence level is mathematically given as
Where
[tex]Degree\ of\ freedom = n-1 df= 22-1df=21[/tex]
Therefore
[tex]CI=\frac{1-0.99}{2}\\\\CI=0.005[/tex]
Using t table
[tex]t_{0.005,21}=2.831[/tex]
Hence,the t-score for a 99% confidence interval is [tex]t_{0.005,21}=2.831[/tex]
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