Results 1 to 12 of 12

Math Help - Hypothesis testing for pairs of means

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    Hypothesis testing for pairs of means

    1a) The strength of concrete depends to some extent on the method used for drying it. Two different drying methods were tested independently on specimens. The strength using each of the methods follow a normal distribution with mean μ_x and μ_y respectively and the same variance. The results are:
    method 1: n1=7, x bar=3250, s1=210
    method 2: n2=10, y bar=3240, s2=190
    Do the methods appear (use alpha=0.05) to produce concrete with different mean strength ?

    For this one, the test is H_o: μ_x=μ_y v.s. H_a: μ_xμ_y. I computed a p-value of >0.2, so the p-value is greater than alpha(which is 0.05), and so we fail to reject H_o and the answer is "no".

    Now I am stuck with part b...
    1b) Suppose σ_x=210, σ_y=190 (n1, n2, x bar, y bar same as part a). Find the probability of deciding that the methods are not different when the true difference in means is 2.

    I am having trouble understanding what the question is asking for. Is it asking for P(type II error)? If so, how can I find it in this case?

    Any help is greatly appreciated!


    P.S. By the way, this topic is also discussed in SOS mathematics cyberboard forum.
    Last edited by kingwinner; April 1st 2009 at 06:52 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    They want you to compute \beta under the SIMPLE hypothesis of H_a:\mu_1-\mu_2=2
    Last edited by matheagle; March 31st 2009 at 01:10 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Now I am a bit confused...

    (i) P(fail to reject H_o | H_o is false) = β

    (ii) P(fail to reject H_o | μ_x-μ_y=2)

    Are the two probabilities above equal? In (i), we are given (conditional on) that H_o is false, so μ_x≠μ_y. In (ii), we are given that μ_x-μ_y=2.

    But μ_x≠μ_y and μ_x-μ_y=2 aren't equivalent (μ_x≠μ_y does not imply that μ_x-μ_y=2, it could be that μ_x-μ_y=3, μ_x-μ_y=4, etc.), so I think (i) and (ii) are not equal. And it looks like the question is asking for (ii), right?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by kingwinner View Post
    Now I am a bit confused...

    (i) P(fail to reject H_o | H_o is false) = β

    (ii) P(fail to reject H_o | μ_x-μ_y=2)

    Are the two probabilities above equal? In (i), we are given (conditional on) that H_o is false, so μ_x≠μ_y. In (ii), we are given that μ_x-μ_y=2.

    But μ_x≠μ_y and μ_x-μ_y=2 aren't equivalent (μ_x≠μ_y does not imply that μ_x-μ_y=2, it could be that μ_x-μ_y=3, μ_x-μ_y=4, etc.), so I think (i) and (ii) are not equal. And it looks like the question is asking for (ii), right?

    Thanks!

    There's a different value for beta for every single possibility under the alternative hypothesis.
    There's \beta (\mu_1-\mu_2 =1), \beta (\mu_1-\mu_2 =2)....

    You are asked to find \beta (\mu_1-\mu_2 =2).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    In hypothesis testing, if H_o is false, does this imply that H_a is true?
    e.g. H_o: μ_x=μ_y is false => H_a: μ_x-μ_y=2 is true ?
    Last edited by kingwinner; March 31st 2009 at 09:42 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    But let's rephrase my question in this way: from a purely mathematical point of view, μ_x≠μ_y does not GUARANTEE that μ_x-μ_y=2, no arguments about that, right? So is there something special about hypothesis testing?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    one set is contained in the other
    μ_x-μ_y=2 GUARANTEEs μ_x≠μ_y
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Quote Originally Posted by matheagle View Post
    one set is contained in the other
    μ_x-μ_y=2 GUARANTEEs μ_x≠μ_y
    Yes, but for P(fail to reject H_o | H_o is false) and P(fail to reject H_o | μ_x-μ_y=2) to be equal, the condition must be perfectly equivalent, i.e. must have H_o is false if and only if μ_x-μ_y=2

    μ_x-μ_y=2 guarantees μ_x≠μ_y, but the converse is false, hence not if and only if...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    As I wrote yesterday. There is a different beta for every different possible outcome under the alternative hypothesis. And one minus beta is the power FUNCTION. There are a lot of beta's. YOU were asked to obtain just one, when the difference of the two populations means were equal to 2.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    I've been seeing two versions of the definition for β...

    Definition 1: β = P(fail to reject H_o | H_o is false)

    Definition 2: β = P(fail to reject H_o | H_a is true)

    Are these two definitions consistent?


    So in hypothesis testing, saying H_o is false is EXACTLY THE SAME thing as saying that H_a is true, am I right?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    yes, we end up deciding on one or the other
    But most statisticians either say we accept H_a
    or we fail to accept H_a.
    Some won't say we accept H_0, because this really is like a proof by contradiction.
    We assume H_0 to be correct, only to hope that it isn't and thus deciding on H_a.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    So we have:
    Ho: μ_x = μ_y
    Ha: μ_x - μ_y = 2
    Need to find: P(type II error) = P(fail to reject H_o | μ_x-μ_y=2)

    Now I am having some trouble with the red part. How can I find the "rejection region" in this case??

    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Testing for Differences between Independent Means
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 19th 2011, 09:08 PM
  2. Another bonus question - testing the means?
    Posted in the Statistics Forum
    Replies: 2
    Last Post: December 27th 2009, 11:07 PM
  3. Hypothesis Testing with 2 means or proportions
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 19th 2008, 04:38 PM
  4. Testing two variances is the same as testing two means?
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 13th 2007, 11:38 PM
  5. testing the difference between two means
    Posted in the Statistics Forum
    Replies: 1
    Last Post: August 25th 2007, 05:26 PM

Search Tags


/mathhelpforum @mathhelpforum