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## Homework Statement

"The Daily Planet ran a recent story about Kryptonite poisoning in the water supply after a recent event in Metropolis. Their usual field reporter, Clark Kent, called in sick and so Lois Lane reported the stories. Researchers plan to sample 288 individuals from Metropolis and control city Gotham and will compare mean blood Kryptonite levels (in Lex Luthors per milliliter, LL/ml). The expect to find a mean difference in LL/ml of around 2. Assoming a two sided Z test of the relevant hypothesis at 5%, what would be the power. Assume that the standard deviation is 12 for both groups.

* Around 60%

* Around 20%

* Around 80%

* Around 70%

* Around 50%

* Around 90%

* Around 40%

* Around 10%

* Around 30%"

## Homework Equations

I'm not sure, but here's what I think:

Z-statistic for a mean difference of two independent groups (I hope this is correct):

[tex]\frac{\mu_x - \mu_y}{\sqrt{\frac{{\sigma_x}^2}{n_x} + \frac{{\sigma_y}^2}{n_y}}}[/tex]

## The Attempt at a Solution

After trying and getting it wrong twice, I am on my third and last attempt to answer this question. Previously, I answered "80%" and then "90%", but they were wrong.

What I'm having problems with is constructing a two-sided power calculation. I don't know how to do that.

What I at least know, however, is how to do a one-sided power calculation:

[tex]1 - \beta = P ( Z > z_{1 - \alpha} - \frac{\mu_a - \mu_0}{\sigma / \sqrt{n}} | \mu = \mu_a)[/tex]

This is correct, right? Well, I thought maybe for a one-sided power for a mean difference of two independent groups, it would be:

[tex]1 - \beta = P ( Z > z_{1 - \alpha} - \frac{\mu_a - \mu_0}{\sqrt{\frac{{\sigma_x}^2}{n_x} + \frac{{\sigma_y}^2}{n_y}}})[/tex]

Is this correct?

I don't know how to do a two-sided power calculation though.

Can anyone help me?

Thank you.