Results 1 to 5 of 5

Math Help - Hypothesis Test

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    37

    Hypothesis Test

    I was wondering if I am even answering these questions right.

    The Better Business Bureau has several complaints that a flour company is under-filling their 5 pound flour bags. The Bureau randomly selects 750 bags of flour and determines the weight of each bag. The sample average of the bags is 4.80 pounds with a standard deviation of .15 pounds. Is there overwhelming evidence at the .01 level that the bags are under-filled?

    M = 5
    x = 4.8
    n = 750
    s = .15

    z = (x-M)/(s/n^(1/2))
    z = (4.8-5)/(.15/750^(1/2))
    = -36.5

    The next step I am unsure of.

    P = P(z>-36.5) < .005?
    How do I get the z value? I have a table in the back of the book that only shows up to -3.4 but that is .0003 so -36.5 has to be much less then that so it is true? Not even sure I am doing this right.

    A fuel oil company claims that one-fifth of the homes in a certain city are heated by oil. Do we have reason to doubt this claim if, in a random sample of 1000 homes in this city, it is found that 136 are heated by oil? Use a .01 level of significance.

    p = .20
    q = .8
    n = 1000
    x = 136

    z = (x-np)/(npq)^(1/2)
    z = (136-1000(.2))/(1000*.2*.8)^(1/2)
    = -5.06

    P = P(z > -5.06) < .005

    The oil company is right?

    Again not sure I am using right formula or doing it right. Also q is just 1-p, correct?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kenshinofkin View Post
    I was wondering if I am even answering these questions right.

    The Better Business Bureau has several complaints that a flour company is under-filling their 5 pound flour bags. The Bureau randomly selects 750 bags of flour and determines the weight of each bag. The sample average of the bags is 4.80 pounds with a standard deviation of .15 pounds. Is there overwhelming evidence at the .01 level that the bags are under-filled?

    M = 5
    x = 4.8
    n = 750
    s = .15

    z = (x-M)/(s/n^(1/2))
    z = (4.8-5)/(.15/750^(1/2))
    = -36.5

    The next step I am unsure of.

    P = P(z>-36.5) < .005?
    How do I get the z value? I have a table in the back of the book that only shows up to -3.4 but that is .0003 so -36.5 has to be much less then that so it is true? Not even sure I am doing this right.

    [snip]
    Since the population sd is unknown you're using the sample sd. This means that you have to use the t-distribution, NOT the z-distribution. BUT ..... since n is really big there's no real difference between using z-distribution and t-distribution for 750 - 1 = 749 degrees of freedom. So you're OK this time.

    You should realise that Pr(z < -36.5) is for all practical purposes equal to zero. And zero is certainly less than 0.01 ..... Overwhelming is certainly a good adjective to use here ......

    Note: It's Pr(z < - 36.5) NOT Pr(z > - 36.5) ....... Capisce?
    Last edited by mr fantastic; August 3rd 2008 at 10:31 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2008
    Posts
    37
    Quote Originally Posted by mr fantastic View Post
    Since the population sd is unknown you're using the sample sd. This means that you have to use the t-distribution, NOT the z-distribution. BUT ..... since n is really big there's no real difference between using z-distribution and t-distribution for 750 - 1 - 749 degrees of freedom. So you're OK this time.

    You should realise that Pr(z < -36.5) is for all practical purposes equal to zero. And zero is certainly less than 0.01 ..... Overwhelming is certainly a good adjective to use here ......

    Note: It's Pr(z < - 36.5) NOT Pr(z > - 36.5) ....... Capisce?
    Ok so I guess I got the second one right?

    So the first question should be t-distributions? So

    t = (x-M)/(s/n^(1/2))
    t = (4.8-5)/(.15/750^(1/2))
    = -36.5

    So the v = 749 and a(forgot what the symbol is called) = .01 the critical value equals 2.326 or so. So I should reject the value?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kenshinofkin View Post
    [snip]
    A fuel oil company claims that one-fifth of the homes in a certain city are heated by oil. Do we have reason to doubt this claim if, in a random sample of 1000 homes in this city, it is found that 136 are heated by oil? Use a .01 level of significance.

    p = .20
    q = .8
    n = 1000
    x = 136

    z = (x-np)/(npq)^(1/2)
    z = (136-1000(.2))/(1000*.2*.8)^(1/2)
    = -5.06

    P = P(z < -5.06) < .005

    The oil company is right?

    Again not sure I am using right formula or doing it right. Also q is just 1-p, correct? Mr F says: Yes.

    Thanks
    You seem to have a general problem getting the correct direction for your inequality signs. Note my correction in red.

    The P-value is 2 P(z < -5.06) < 0.01 (since it's a two-tailed test).

    Do you understand that the P-value is the probability that you get the data that you got given that the null hypothesis is true. So the null hypothesis is rejected.

    The null hypothesis is that the proportion is equal to 0.2 .......
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    Since the population sd is unknown you're using the sample sd. This means that you have to use the t-distribution, NOT the z-distribution. BUT ..... since n is really big there's no real difference between using z-distribution and t-distribution for 750 - 1 = 749 degrees of freedom. So you're OK this time.

    You should realise that Pr(z < -36.5) is for all practical purposes equal to zero. And zero is certainly less than 0.01 ..... Overwhelming is certainly a good adjective to use here ......

    Note: It's Pr(z < - 36.5) NOT Pr(z > - 36.5) ....... Capisce?
    Quote Originally Posted by kenshinofkin View Post
    Ok so I guess I got the second one right?

    So the first question should be t-distributions? So

    t = (x-M)/(s/n^(1/2))
    t = (4.8-5)/(.15/750^(1/2))
    = -36.5

    So the v = 749 and a(forgot what the symbol is called) = .01 the critical value equals 2.326 or so. So I should reject the value?
    Look again at what I said and you will see that I said that because n was so large there is no real difference between using z-distribution and t-distribution for 750 - 1 = 749 degrees of freedom. So it's OK not to use the t-distribution. Notice that z-critical = 2.33 and t-critical = 2.33 ......
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Test the hypothesis
    Posted in the Advanced Statistics Forum
    Replies: 7
    Last Post: December 19th 2009, 03:13 AM
  2. Hypothesis test for
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 17th 2009, 07:40 PM
  3. hypothesis test help
    Posted in the Statistics Forum
    Replies: 1
    Last Post: May 5th 2009, 11:46 AM
  4. hypothesis test ,test statistic wrong!
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: March 28th 2009, 04:59 AM
  5. Test the hypothesis
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 17th 2007, 11:48 AM

Search Tags


/mathhelpforum @mathhelpforum