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.(adsbygoogle = window.adsbygoogle || []).push({});

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, Lets 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# 95% Confidence Intervals

**Physics Forums | Science Articles, Homework Help, Discussion**