# Z-tests for the Results of Prostate Cancer Study

• Mar 14th 2013, 02:18 PM
oldwarplanes
Z-tests for the Results of Prostate Cancer Study
Hello! I was wondering if I could get some help with this statistics problem!

In short, there was a prostate cancer study that took place from 1989 to 2000. There were 2 groups created out of 695 men diagnosed with prostate cancer. 348 men were in the control group, "watch and wait" while the other 347 men had "radical prostatecomies". The results of the study were as follows in the table.

 died of prostate cancer died of all causes combined watch and wait 31 62 radical prostatecomy 16 53

Using the Z-test, test the null hypothesis of no difference between the proportions for death due to prostate cancer.

I have the equation for the Z-test, but I'm a little confused as to how to apply it to this problem. Thanks in advance for the help!
• Mar 14th 2013, 03:18 PM
Shakarri
Re: Z-tests for the Results of Prostate Cancer Study
Your table has 162 samples in it, what happened to the other 533?
• Mar 14th 2013, 03:37 PM
oldwarplanes
Re: Z-tests for the Results of Prostate Cancer Study
The 533 are alive and not a part of the problem.
• Mar 14th 2013, 05:41 PM
Shakarri
Re: Z-tests for the Results of Prostate Cancer Study
Quote:

Originally Posted by oldwarplanes
The 533 are alive and not a part of the problem.

Oh thats good :)

Keep in mind that it is a binomial distribution. Let dying of combined causes be a success and dying of prostate cancer be a fail.
In 'wait and watch' the mean is 62/93=0.667

The standard deviation is $\displaystyle \sqrt{0.667\cdot(1-0.667)}=0.471$
The sample size is 93

In 'radical' the mean is 53/69=0.768

The standard deviation is $\displaystyle \sqrt{0.768\cdot(1-0.768)}=0.422$
The sample size is 69

The confidence interval for 'wait and watch' is $\displaystyle 0.667\pm\frac{0.471z}{\sqrt{93}}$

The confidence interval for 'radical' is $\displaystyle 0.768\pm\frac{0.422z}{\sqrt{69}}$
• Mar 14th 2013, 11:03 PM
oldwarplanes
Re: Z-tests for the Results of Prostate Cancer Study
Thank you so much for the help! I was getting intimidated by sigma and x-bar and the sqaure roots in the equation, but it's really not that difficult now that you wrote it out like this!

• Mar 15th 2013, 07:06 AM
oldwarplanes
Re: Z-tests for the Results of Prostate Cancer Study
Quote:

Originally Posted by Shakarri
Oh thats good :)

Keep in mind that it is a binomial distribution. Let dying of combined causes be a success and dying of prostate cancer be a fail.
In 'wait and watch' the mean is 62/93=0.667

The standard deviation is $\displaystyle \sqrt{0.667\cdot(1-0.667)}=0.471$
The sample size is 93

In 'radical' the mean is 53/69=0.768

The standard deviation is $\displaystyle \sqrt{0.768\cdot(1-0.768)}=0.422$
The sample size is 69

The confidence interval for 'wait and watch' is $\displaystyle 0.667\pm\frac{0.471z}{\sqrt{93}}$

The confidence interval for 'radical' is $\displaystyle 0.768\pm\frac{0.422z}{\sqrt{69}}$

I lied. I have some questions still.

So I don't understand why you used 62 instead of 31 if the question is asking for a Z-test of the null hypothesis of no difference between proportions of death due to prostate cancer. Wouldn't you need to use both 31/93 and 16/69 to test the difference between the two proportions?

Since there is another similar question with the only change being test the null hypothesis of no difference between the proportions for death from all causes. In this case wouldn't you use 62/93 and 16/69?

But then the question is how to you incorporate that into Z = xbar - µ0 / σ - √n if there are two separate proportions? Am I using the wrong equation or something?
• Mar 15th 2013, 07:30 AM
Shakarri
Re: Z-tests for the Results of Prostate Cancer Study
Quote:

Originally Posted by oldwarplanes
I lied. I have some questions still.

So I don't understand why you used 62 instead of 31 if the question is asking for a Z-test of the null hypothesis of no difference between proportions of death due to prostate cancer. Wouldn't you need to use both 31/93 and 16/69 to test the difference between the two proportions?

Since there is another similar question with the only change being test the null hypothesis of no difference between the proportions for death from all causes. In this case wouldn't you use 62/93 and 16/69?

But then the question is how to you incorporate that into Z = xbar - µ0 / σ - √n if there are two separate proportions? Am I using the wrong equation or something?

If there is no difference in the deaths due to prostate cancer then their is no difference in the deaths due to all causes. Since it is a binomial distribution the standard deviation of both types of deaths is the same so the calculation will be the same.

I'm not sure what you mean by Z = xbar - µ0 / σ - √n you shouldn't have a σ - √n anywhere
µ0 is the exact mean but you don't have an exact mean just two sample means.

I was confusing a z test with something else, ignore what I said about confidence intervals you only need the standard deviations.
Is this the equation you were given?

$\displaystyle z=\frac{x_1-x_2}{\sqrt{\frac{(\sigma_1)^2}{n_1}+\frac{(\sigma_ 2)^2}{n_2}}}$

If so then you have all the information you need to find the z value. Do you know what to do with the z value once you have found it?
• Mar 15th 2013, 07:47 AM
oldwarplanes
Re: Z-tests for the Results of Prostate Cancer Study
Ah. A fellow student pointed me in the wrong direction with that other equation. No wonder I was having such an issue. Thanks for the clarification.

Honestly, I don't know what to do with z. Haha, I'm a mess.

Also, what would x1 and x2​ be? The sample means?
• Mar 15th 2013, 08:05 AM
Shakarri
Re: Z-tests for the Results of Prostate Cancer Study
When you have the value of z use this calculator to find the percentage associated with that value of z.
Standard Normal Distribution Table
Make sure the calculator is set to '0 up to z'
When you have this percentage multiply it by 2.

So for example, if z=1 the the calcualtor will tell you that at z=1 the percentage is 34.1%
2(34.1)= 68.2%

This means that if you choose to reject the null hypothesis (reject that the means are the same) there is a 68.2% chance that you are wrong.