If Y' = a +bX for all i and a=the mean of Y - b*the mean of X, prove that SigmaY' = SigmaY (subscript i).

Hint: replace a with mean of Y minus b*mean of X and take the sum of both sides of the equation--that is, SigmaY'(subscript i)=Sigma(mean of Y-b*Mean of X + bX(subscript i).

I just don't even know where to begin with this problem. Please Help!!

To prove that ΣY' = ΣY, we need to substitute the given equation Y' = a + bX into the sum and show that it simplifies to ΣY.

Step 1: Start with the equation Y' = a + bX.

Step 2: Substitute a with the given expression for a - the mean of Y minus b times the mean of X:
Y' = mean of Y - b * mean of X + bX

Step 3: Next, take the sum of both sides of the equation:
ΣY' = Σ(mean of Y - b * mean of X + bX)

Step 4: We can split the sum into three parts and apply the properties of sums separately:
ΣY' = Σ(mean of Y) - Σ(b * mean of X) + Σ(bX)

Step 5: Recall that the mean of Y is a constant, so the sum of a constant times the number of terms is just the constant times the number of terms:
Σ(mean of Y) = (mean of Y) * n, where n is the number of terms.

Step 6: Similarly, the mean of X is also a constant, so the sum of a constant times the number of terms is just the constant times the number of terms:
Σ(b * mean of X) = b * (mean of X) * n

Step 7: The sum Σ(bX) is the same as b times the sum ΣX:
Σ(bX) = b * ΣX

Step 8: Now we can substitute the simplified sums back into the equation:
ΣY' = (mean of Y) * n - b * (mean of X) * n + b * ΣX

Step 9: Notice that n is a factor common to all terms, so we can factor it out:
ΣY' = n * [(mean of Y) - b * (mean of X) + b * ΣX]

Step 10: Finally, recall that (mean of Y) - b * (mean of X) is equal to a (given in the problem statement):
ΣY' = n * (a + b * ΣX)

Step 11: We know that ΣY = n * (mean of Y), so we substitute that in:
ΣY' = ΣY

Hence, we have proven that ΣY' = ΣY using the given equation and the properties of sums.