A herd of 1,500 steer was fed a special high‐protein grain for a month. A random sample of 29 were weighed and had gained an average of 6.7 pounds. If the standard deviation of weight gain for the entire herd is 7.1, test the hypothesis that the average weight gain per steer for the month was more than 5 pounds.

By formula we know that:

z (x) = (x - m) / [sd / sqrt (n)]

where x is the value we want to know (6.7), m is the mean (5), sd is the standard deviation (7.1) and n is the sample size (29).

Replacing we have:

z (6.7) = (6.7 - 5) / [7.1 / sqrt (29)]

z = 1.289

If we look in the normal distribution table (attached), we have that the probability is 0.8997, therefore:

1 - 0.8897 = 0.1003

So the conditional probability of a herd sample earning at least 6.7 pounds per steer is 10.03%.

Now the hypothesis tells me:

m> 5

The probability is somewhat low, therefore, the most correct thing is to reject the hypothesis even though it is a fact that can occur.

step-by-step explanation:

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this