Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. What level of significance are you using? P = .05?
95% = mean ± 1.96 SEm
Many dairy cows now receive injections of BST, a hormone intended to spur greater milk production. A group of 53 Jersey cows increased average milk production from 43 to 52 pounds per day, with a standard deviation of 4.8 lbs.
Is this evidence that the hormone maybe effective in this breed of cattle? Assume the assumptions/conditions have been met.
The 95% confidence interval is (, ). Show one decimal places in your answer
2 answers
paired test hard