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Test for proportions

  1. Dec 16, 2008 #1
    Test and CI for Two Proportions

    Sample X N Sample p
    1 3 839 0.003576
    2 20 1662 0.012034


    Difference = p (1) - p (2)
    Estimate for difference: -0.00845801
    95% upper bound for difference: -0.00290435
    Test for difference = 0 (vs < 0): Z = -2.51 P-Value = 0.006

    Fisher's exact test: P-Value = 0.025



    Cannot understand how the value for upperbound and z, p value were obtained....Fisher's exact test value too? Was checking answer for a problem...and this is given with no explanatations...could anyone please ellaborate? am trying meanwhile

    Thanks
     
  2. jcsd
  3. Dec 16, 2008 #2
    So far I only looked at the upper bound

    The following formulas were used

    [tex] \hat{p_{1}} = 0.003575685, \hat{p_{2}} = 0.012033694 \Rightarrow \hat{p_{1}} - \hat{p_{2}} = -0.008458009 [/tex]

    [tex] SE_{1} = \sqrt{\frac{\hat{p_{1}} (1 - \hat{p_{1}})}{N_{1}}} = 0.002060729 [/tex]

    and

    [tex] SE_{2} = \sqrt{\frac{\hat{p_{2}} (1 - \hat{p_{2}})}{N_{2}}} = 0.002674577
    [/tex]

    Therefore

    [tex] SE_{Total} = \sqrt{SE^{2}_{1} + SE^{2}_{2}} = 0.003376383 [/tex]

    And finally upper CI is

    [tex] \hat{p_{1}} - \hat{p_{2}} + 1.645 \cdot SE_{Total} = -0.002903859 [/tex]
     
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