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Math Help - In a sample of 600 men

  1. #1
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    In a sample of 600 men

    Question :

    In a sample of 600 men from a certain large city , 450 are found to be smokers. In another sample 900 from other large city, 650 are smokers. Do the data indicate that the cities are significantly different with respect to the prevalence of smoking among men?
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  2. #2
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    Quote Originally Posted by zorro View Post
    Question :

    In a sample of 600 men from a certain large city , 450 are found to be smokers. In another sample 900 from other large city, 650 are smokers. Do the data indicate that the cities are significantly different with respect to the prevalence of smoking among men?
    Well what have you done? What tests do you know for comparing proportions?

    CB
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  3. #3
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    Maybe I am wrong, but shouldn't just get the ratios and compare them?
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  4. #4
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    I dont know from where to start

    Quote Originally Posted by CaptainBlack View Post
    Well what have you done? What tests do you know for comparing proportions?

    CB

    I dont know from where to start....
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    Quote Originally Posted by zorro View Post
    I dont know from where to start....
    Look at your notes and/or text for testing the difference of two proportions.

    Or google finds:

    http://stattrek.com/AP-Statistics-4/...px?Tutorial=AP

    despite what looks like a type in this it still looks a nice explanation:
    http://www.pinkmonkey.com/studyguide...8/s0808f01.asp

    CB
    Last edited by CaptainBlack; December 8th 2009 at 11:47 PM.
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  6. #6
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    Quote Originally Posted by zorro View Post
    Question :

    In a sample of 600 men from a certain large city , 450 are found to be smokers. In another sample 900 from other large city, 650 are smokers. Do the data indicate that the cities are significantly different with respect to the prevalence of smoking among men?
    is the answer as obvious as i think or is it some kind of trick question?

    ratio = (650/900)/(450/600)
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  7. #7
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    Quote Originally Posted by coder View Post
    is the answer as obvious as i think or is it some kind of trick question?

    ratio = (650/900)/(450/600)
    If it is obvious it is not obvious to you. Why don't you look at the links on the previous post to yours.

    CB
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  8. #8
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    The question does not provide you with a level of significance or a level of confidence. My intro textbook always provides this information in such a question. Since "significantly different" is used in the question, you probably should try the significance level approach. There are two significance levels used in my textbook, 1% and 5%. Choose the level which results in the most powerful test. Do you know which percentage gives your test the most power? Read your textbook or go to CaptainBlack's links. Spoonfeeding you the answer is not good for your learning curve.
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    Is this correct?

    Quote Originally Posted by CaptainBlack View Post
    Look at your notes and/or text for testing the difference of two proportions.

    Or google finds:

    Hypothesis Test for Difference Between Proportions

    despite what looks like a type in this it still looks a nice explanation:
    PinkMonkey.com Statistics Study Guide 8.15 Test for difference between proportions

    CB

    Is this correct >


    P_1 = \frac{x_1}{n_1} = \frac{450}{600} = 0.75

    P_2 = \frac{x_2}{n_2}= \frac{650}{900} = 0.72

    CaptainBlack in the link provided there is a mention of the term z
    It didnt understand what z stands for ......
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  10. #10
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    You now have both p-hats. All of the information about z and everything else you need to know is clearly explained in any intro to statistics textbook. Which textbook are you using?
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  11. #11
    Grand Panjandrum
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    Quote Originally Posted by zorro View Post
    Is this correct >


    P_1 = \frac{x_1}{n_1} = \frac{450}{600} = 0.75

    P_2 = \frac{x_2}{n_2}= \frac{650}{900} = 0.72

    CaptainBlack in the link provided there is a mention of the term z
    It didnt understand what z stands for ......
    The basic idea is that as the sample sizes are large and the probabilities are in the suitable range we can assume that the sampling distributions of the two proportions are normal with standard deviations \sqrt{p_i(1-p_i)/n}. Now the z-score arises when we construct a test under these assumptions as a test statistic with a standard normal distribution that we can use to look up in a table.

    CB
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  12. #12
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    Is this correct?

    Quote Originally Posted by CaptainBlack View Post
    The basic idea is that as the sample sizes are large and the probabilities are in the suitable range we can assume that the sampling distributions of the two proportions are normal with standard deviations \sqrt{p_i(1-p_i)/n}. Now the z-score arises when we construct a test under these assumptions as a test statistic with a standard normal distribution that we can use to look up in a table.

    CB


    My work Please let me know if it is correct or no

    P_1 : 0.75

    P_2 : 0.72

    H_0 : P_1 = P_2

    H_a : P_1 \neq P_2

    Formulate the Analysis Plan :

    -Significant level : 0.05
    -Test Method
    . \quad Analysis of Sampled data
    . \qquadPooled Sampled proportion:  p=\frac{((p_1 * n_1) + (p_2 * n_2))}{n_1 + n_2}=0.7
    p_1 : sampled proportion
    p_2 : sampled proportion
    n_1 : size of 1st sample
    n_1 : size of 1st sample

    Standard Error (SE) = \sqrt{p*(1-p) \left[ \frac{1}{n_1} + \frac{1}{n_2} \right]} = 0.025

    Test Statistic (z) = \frac{(P_1-P_2)}{SE} = 1.2

    z<-1.2=: 0.11507


    Is this correct??????
    Last edited by zorro; December 12th 2009 at 01:45 AM.
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