An organizational psychologist measures levels of job satisfaction in a sample of 30 participants. To measure the variance of job satisfaction, it is calculated that the SS = 120 for this sample.
What are the degrees of freedom for the variance?
Compute the variance and standard deviation?
The formula for a single sample t-Test is N-1 or in this case N (number of samples) minus 1. It is then 30 - 1 = 29.
Dont post responses if you dont know what youre talking about
Variance =30-1=29
Variance 120/29= 4.138
Standard =2.034
SS/n = variance
SD = √variance
I don't know a whole lot about adding and subtracting liner expressions
i go to middle school but i need help with a math problem or maybe 2 so if you have time people then i need you to help me out thank you
Thank you so much PsyDAG,
So given your formula can you tell me if I'm correct.
SS/N =
120/30 = 4
Degrees of freedom for the variance is 4
SD=√Variance
Compute variance and SD = 2
I don't think I really understand SD=√Variance. May you provide me more help with it if I have it wrong...and for the Degrees of freedom too. Thank you kindly!
√ = square root of...
The variance is 4, but I don't know about the degrees of freedom.