determine the z-sgtatistic for two population samples

I'm having some trouble trying to figure out this question which deals with two population samples. The actual question is as follows:

Quote:

On Assignment 2 we had n = 342, Σx = 6552 and Σx^{2} = 127552. On Assignment 3 we had n = 311, Σx = 3667 and Σx^{2} = 53389. Suppose I want to test H_{0}: μ_{2}-μ_{3} = 0 against H_{a}: μ_{2}-μ_{3} ≠ 0 at a significance level of α=0.05. What is the z-statistic and the conclusion.

I know the z statistic is:

z = [(x̄_{1}-x̄_{2})-D_{0}] / sqrt([s_{1}^{2}/n] + [s_{2}^{2}/n])

where x̄ = Σx/n

and where s^{2} = [Σx^{2} - (Σx)^{2}/n] / n-1

but I'm not exactly sure how I'm supposed to get the correct answer. The Solutions to this are as follows:

[quote]1.4588, -3.5045, do not reject H_{0 }[/quote] |

So, I plugged in plugged in the values and I get the following:

x̄_{1 }= 6552/342 = 19.158

x̄_{2 }= 3667/311 = 11.79099

D_{0 }= 0

s_{1}^{2 }= [127552 - (6552^{2}/342)] / 341 = 5.9515

s_{2}^{2 }= [53389 - (3667^{2}/311)] / 310 = 32.7465

So then if you plug in everything, you get:

z = [(19.158-11.79099)-0] / sqrt([5.9515/342] + [32.7465/311])

z = 21.029

and since the equation implies the rejection regions lie within the tail ends of the normal distribution, a/2 = 0.025 -> this means the non-rejection zone is μ ± 1.96

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this is where I get lost...

Re: determine the z-sgtatistic for two population samples

never mind, my prof said the answer key was incorrect. What i had was the correct answer.