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11.2: Confidence Interval for the Difference between Two Proportions

  • Page ID
    52007
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    The form of the confidence interval is

    \[ (\hat{p}_{1} - \hat{p}_{2}) - E < (p_{1} - p_{2}) < (\hat{p}_{1} - \hat{p}_{2}) + E \]

    with

    \[ E = z_{\cal{C}} \sqrt{\frac{\hat{p}_{1}\hat{q}_{1}}{n_{1}} + \frac{\hat{p}_{2}\hat{q}_{2}}{n_{2}} } \]

    where, as usual you can get \(z_{\cal{C}}\) from the last line of the t Distribution Table.

    Example 11.2 : Using the data from Example 11.1, find the 95\(\%\) confidence interval for \(p_{1} - p_{2}\).

    Solution : The relevant numbers from Example 11.1 are: \(n_{1} = 34\), \(\hat{p}_{1} = 0.35\), \(\hat{q}_{1} = 1 - 0.35 = 0.65\) and \(n_{2} = 24\), \(\hat{p}_{2} = 0.71\), \(\hat{q}_{1} = 1 - 0.71 = 0.29\).

    Compute (after finding \(z_{95\%} = 1.96\) from the t Distribution Table)

    \[\begin{eqnarray*} E & = & z_{95\%} \sqrt{\frac{\hat{p}_{1}\hat{q}_{1}}{n_{1}} + \frac{\hat{p}_{2}\hat{q}_{2}}{n_{2}} }\\ E & = & 1.96 \sqrt{\frac{(0.35)(0.65)}{34} + \frac{(0.71)(0.29)}{24} }\\ E & = & 0.242 \end{eqnarray*}\]

    and

    \[ \hat{p}_{1} - \hat{p}_{2} = 0.35 - 0.71 = -0.36 \]

    So

    \[\begin{eqnarray*} (\hat{p}_{1} - \hat{p}_{2}) - E & < (p_{1} - p_{2}) < & (\hat{p}_{1} - \hat{p}_{2}) + E \\ -0.36 - 0.242 & < (p_{1} - p_{2}) < & -0.36 + 0.242 \\ -0.602 & < (p_{1} - p_{2}) < & -0.118 \end{eqnarray*}\]

    with 95\(\%\) confidence. (Note that this corresponds with the rejection of \(H_{0}\) in Example 11.1 since 0 is not in the confidence interval.)


    This page titled 11.2: Confidence Interval for the Difference between Two Proportions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gordon E. Sarty via source content that was edited to the style and standards of the LibreTexts platform.

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