Weaning weights are used to evaluate di erences in growth potential of calves. A random sample of 41 steer calves is taken from a large ranch. The sample yields a mean weaning weight of 578 with a standard deviation of 87 lbs. This particular ranch is targeting an average weaning weight of 610 lbs. Perform a hypothesis test using = 0:01 to determine if the average weaning weight of cows at this ranch is di erent than 610 lbs.

We accept H₀  the weaning weights of cows at this ranch is 610 lbs

Step-by-step 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

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation:

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.