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Math Help - random sample-help!

  1. #1
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    Post random sample-help!

    Hi all,
    Someone please help me with the number below.thanks

    In a city A, 20% of a random sample of 900 school boys had a certain slight physical defect. In another city B, 18.5% of a random sample of 1600 school boys had the same defect.
    Is the difference between the proportions significant ?
    Last edited by okello; February 3rd 2009 at 10:00 AM. Reason: missing word
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  2. #2
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    Quote Originally Posted by okello View Post
    Hi all,
    Someone please help me with the number below.thanks

    In a city A, 20% of a random sample of 900 school boys had a certain slight physical defect. In another city B, 18.5% of a random sample of 1600 school boys had the same defect.
    Is the difference between the proportions significant ?
    Suppose that X_1 ~ Binomial (n_1, \theta_1) and X_2 ~ Binomial (n_2, \theta_2) and you want to test \epsilon = \theta_1 - \theta_2.


    If n_1 and n_2 are large then \frac{X_1}{n_1} \approx N \left(\theta_1, \frac{\theta_1 (1 - \theta_1)}{n_1} \right) and \frac{X_2}{n_2} \approx N \left(\theta_2, \frac{\theta_2 (1 - \theta_2)}{n_2} \right)


    and so \frac{\frac{X_1}{n_1} - \frac{X_2}{n_2} - (\theta_1 - \theta_2)}{\sqrt{\frac{\theta_1 (1 - \theta_1)}{n_1} + \frac{\theta_2 (1 - \theta_2)}{n_2} }} \approx N (0, 1).

    With \hat{\epsilon} = \frac{X_1}{n_1} - \frac{X_2}{n_2} then se(\hat{\epsilon}) = \sqrt{\frac{\hat{\theta_1} (1 - \hat{\theta_1})}{n_1} + \frac{\hat{\theta_2} (1 - \hat{\theta_2})}{n_2} } and so the test is based on \frac{\hat{\epsilon} - \epsilon}{se(\hat{\epsilon})} \approx N(0, 1).


    You are testing the hypothesis H0: \theta_1 = \theta_2, that is, \epsilon = 0 against (I suppose) a two-sided alternative.

    Decide on a level of significance, substitute the data and crunch the numbers.
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