Results 1 to 9 of 9
Like Tree2Thanks
  • 1 Post By Shakarri
  • 1 Post By Shakarri

Math Help - Z-tests for the Results of Prostate Cancer Study

  1. #1
    Newbie oldwarplanes's Avatar
    Joined
    Mar 2013
    From
    United States
    Posts
    7

    Unhappy 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!
    Last edited by oldwarplanes; March 14th 2013 at 02:28 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    586
    Thanks
    155

    Re: Z-tests for the Results of Prostate Cancer Study

    Your table has 162 samples in it, what happened to the other 533?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie oldwarplanes's Avatar
    Joined
    Mar 2013
    From
    United States
    Posts
    7

    Re: Z-tests for the Results of Prostate Cancer Study

    The 533 are alive and not a part of the problem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    586
    Thanks
    155

    Re: Z-tests for the Results of Prostate Cancer Study

    Quote Originally Posted by oldwarplanes View Post
    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 \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 \sqrt{0.768\cdot(1-0.768)}=0.422
    The sample size is 69

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

    The confidence interval for 'radical' is 0.768\pm\frac{0.422z}{\sqrt{69}}
    Thanks from oldwarplanes
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie oldwarplanes's Avatar
    Joined
    Mar 2013
    From
    United States
    Posts
    7

    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!

    Last edited by oldwarplanes; March 15th 2013 at 07:05 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie oldwarplanes's Avatar
    Joined
    Mar 2013
    From
    United States
    Posts
    7

    Re: Z-tests for the Results of Prostate Cancer Study

    Quote Originally Posted by Shakarri View Post
    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 \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 \sqrt{0.768\cdot(1-0.768)}=0.422
    The sample size is 69

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

    The confidence interval for 'radical' is 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?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    586
    Thanks
    155

    Re: Z-tests for the Results of Prostate Cancer Study

    Quote Originally Posted by oldwarplanes View Post
    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?

    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie oldwarplanes's Avatar
    Joined
    Mar 2013
    From
    United States
    Posts
    7

    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?
    Last edited by oldwarplanes; March 15th 2013 at 07:55 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    586
    Thanks
    155

    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.
    Last edited by Shakarri; March 15th 2013 at 08:07 AM.
    Thanks from oldwarplanes
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cancer Diet
    Posted in the New Users Forum
    Replies: 0
    Last Post: January 1st 2013, 09:02 PM
  2. causes for lung cancer
    Posted in the New Users Forum
    Replies: 0
    Last Post: November 22nd 2012, 12:04 AM
  3. Comparing the results of two McNemar's tests
    Posted in the Statistics Forum
    Replies: 1
    Last Post: August 2nd 2012, 10:41 AM
  4. probability of cancer
    Posted in the Statistics Forum
    Replies: 1
    Last Post: May 12th 2011, 10:33 AM
  5. Replies: 1
    Last Post: March 13th 2011, 03:39 AM

Search Tags


/mathhelpforum @mathhelpforum