Answer:
We accept H₀ the weaning weights of cows at this ranch is 610 lbs
Explanation:
We assume normal distribution
Population mean μ₀ = 610 lbs
Population standard deviation unknown
sample size n = 41
degree of fredom df = n - 1 df = 41 - 1 df = 40
Sample mean X = 578 lbs
Sample standard deviation s = 87
As we don´t know standard deviation of the population we will use t- student test, furthemore, as we are looking for any difference upper and lower we are in presence of a two tails test
Test Hypothesis Null hypothesis H₀ X = μ₀
Alternative hypothesis Hₐ X ≠ μ₀
Now at α = 0,01 , df = 40 and two tail test we find t = 2,4347
We have in t table
30 df t = 2,457 for α = 0,01
60 df t = 2,390 for α = 0,01
Δ = 30 0,067
10 x ?? x = 0,022
then 2,457 - 0,022 = 2,4347
t = 2,4347
Now we compute the interval:
X ± t * ( s/√n) ⇒ 578 ± 2,4347 * ( 87/√41)
578 ± 2,4347 * 13,59
578 ± 33,09
P [ 578 + 33,09 ; 578 - 33,09 ]
P ( 611,09 ; 544,91]
We can see that vale 610 = μ₀ is inside the iinterval , so we accept H₀ the weaning weight of cows in the ranch is 610 lbs