- #1
bhoover05
- 20
- 0
I was under the impression that 95% C.I requires that the critical value in the error term comes from a t-distribution with 25 degrees of freedom.
I was taught in class today that a 25% CI requires that the critical value in the error term comes from a t-distribution with 23 degrees of freedom? I am unsure as to why this is?
~For example, Let's use this. . .
Average length of workweeks of 15 randomly selected employees in the mining industry and 10 randomly selected employees in the manufacturing industry were obtained.
Miners(x)=15 observation, Mean of 47.5, and Std. Dev. of 5.5
Manufacturers= 10 observations, Mean of 42.5, and Std. Dev of 4.9.
With this data I obtained that Sp 5.27, t=2.069, and a 95% CI =(0.55,9.45)
~I never used 23 degrees of freedom to obtain those answers, were are correct. Why should the 95% CI for the difference between x-y require the critical value in the error term comes from a t-distribtuion with 23 degrees of freedom?
I was taught in class today that a 25% CI requires that the critical value in the error term comes from a t-distribution with 23 degrees of freedom? I am unsure as to why this is?
~For example, Let's use this. . .
Average length of workweeks of 15 randomly selected employees in the mining industry and 10 randomly selected employees in the manufacturing industry were obtained.
Miners(x)=15 observation, Mean of 47.5, and Std. Dev. of 5.5
Manufacturers= 10 observations, Mean of 42.5, and Std. Dev of 4.9.
With this data I obtained that Sp 5.27, t=2.069, and a 95% CI =(0.55,9.45)
~I never used 23 degrees of freedom to obtain those answers, were are correct. Why should the 95% CI for the difference between x-y require the critical value in the error term comes from a t-distribtuion with 23 degrees of freedom?