Computing Power

[amanning]

I know that power is one minus beta...but how do you calculate beta?

[quark]

Hi amanning,

Beta is the type II error. It's the probability of not rejecting H0 when H0 is false.

To calculate beta, first you need to get the rejection/acceptance region based on your H0, H1 and significance level, then you can get the conditional probability of the acceptance region given H0 is false.

Let me know if you have further questions.

[amanning]

Ok. The problem I am working on is Let X:N(?, 64)
n=16
H1: m=50
H2: m>50
alpha= >.05
Compute Power for m=56
(m is mew as I don't know how to type in greek)

[quark]

You want to find the critical value first, then the rejection region, ie reject H0 when sample mean xbar is....

[amanning]

would this make sense:
-
X= 59.29
Z(59.29)=1.645
Beta=.05
1-Beta=.95

[quark]

would this make sense:
-
X= 59.29
Z(59.29)=1.645
Beta=.05
1-Beta=.95

It's more complex than that. :)

You have alpha=.05, critical value of 1.645 is correct. Calculate the standard error of the mean, then the critical value is the z-value for the boundary of the rejection region. Rejection region should be Xbar>...

[amanning]

standard error of the mean:
2
found by taking the square root of the variance (variance 64) which is 8 and dividing it by the square root of the sample (square root of 16) which is 4.

[quark]

So you would reject H0 if

(Xbar-mu)/se_Xbar > 1.645
(Xbar-50)/2 > 1.645
Xbar >53.29

Next we get power directly by calculating P(Xbar >53.29 | mu=56)

P(Xbar >53.29 | mu=56)
=P[(Xbar-mu)/se_Xbar > (53.29-mu)/se_Xbar | mu=56]
=P[Z > (53.29-56)/2]
=...

You can do the rest. :)

[amanning]

OK, I guess a better question would be how do I figure out the z score?
I have a 'formula' written down that looks like Z(x)= x-mx/standard error, I don't know what to do with it.

[quark]

Substitute the values.

P(Xbar >53.29 | mu=56)
=P[(Xbar-mu)/se_Xbar > (53.29-mu)/se_Xbar | mu=56]
=P[Z > (53.29-56)/2]
=P(Z > -1.36)
=P(Z<1.36)
=...

The final answer can be found in the normal table.

[amanning]

Thank you so much for all of your help. Sometimes I need to see things written differently for them to make sense...

[quark]

You are welcome. Please pass a good word for us. Thanks.

[amanning]

In reviewing this I do have one more, most likely annoying question, why in post 8 is it x bar -50/2>1.645 and not x bar-56/2>1.645

[quark]

Very good question. When you try to find the rejection region, you use the mu in your H0. After you get the region, you use mu=56, notice I've always included the condition mu=56 afterwards. This is a KEY concept in power calculation.

[amanning]

Oooh. I hadn't noticed, or at least hadn't understood. But that makes sense.
Thanks again...
I'm sure I'll be back...

[quark]

Great! I would be glad to help you again.