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10. Comparing Two Population Means

10.5 Confidence Intervals for the Difference of Two Means

The form of the confidence interval is

    \[(\bar{x}_1 - \bar{x}_2) - E < (\mu_{1} - \mu_{2}) < (\bar{x}_1 - \bar{x}_2) + E\]

but, as with hypothesis testing, we have two cases to choose from to get the formula for E :

Case 1 : Variances of the 2 populations unequal}

    \[E = t_{\cal{C}} \sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}\]

where the degrees of freedom to use when looking up t_{\cal{C}} in the t Distribution Table is

    \[ \nu = \min[(n_{1}-1), (n_{2}-1)] \]

Case 2 : Variances of the 2 populations equal

    \[E = t_{\cal{C}} \sqrt{\frac{(n_1-1)s^2_1 + (n_2-1)s^2_2}{n_1+n_2-2}} \sqrt{\frac{1}{n_1}+ \frac{1}{n_2}}\]

where we use

    \[ \nu= n_1 + n_2 - 2 \]

when looking up t_{\cal{C}}.

To select the appropriate formula for E we need to do a preliminary hypothesis test on H_{0} : \sigma_{1}^{2} = \sigma_{2}^{2}. An odd combination of hypothesis test followed by confidence interval calculation.

Insight! By now you should have noticed that the formulae for E are just t times standard error of the mean. This whole z-transformation thing should be becoming somewhat transparent.

Example 10.6 : Find the 95\% confidence interval for \mu_{1} - \mu_{2} for the data of Example 10.4 :

s_1=38 \overline{x}_1 = 191 n_1 = 8
s_2 = 12 \overline{x}_2 = 199 n_2 = 100

Solution :

First use F-test to see which formula to use.  We did this already in Example 10.4 (the data come from that question) and found that we believed \sigma_{1}^{2} \neq \sigma_{2}^{2} with \alpha=0.05.

Next, look up t_{\cal{C}} in the t Distribution Table for 95\% confidence interval for \nu = 7:

    \[ t_{95\%} = 2.365\]

Compute

    \begin{eqnarray*} E &=& t_{95\%} \sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}} \\ E &=& 2.365\sqrt{\frac{38^2}{8} + \frac{12^2}{10}} = 33.01 \end{eqnarray*}

So

    \begin{eqnarray*} (\bar{x}_1 - \bar{x}_2) - E &< \mu_1 - \mu_2 &< (\bar{x}_1 - \bar{x}_2) +E \\ (191-199) - 33.02 &< \mu_1 - \mu_2 &< (191-199) + 33.02 \\ -8 - 33.02 &< \mu_1 - \mu_2 &< -8 + 33.02 \\ -41.02 &< \mu_1 - \mu_2 &< 25.02 \end{eqnarray*}

be careful of the order!

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Introduction to Applied Statistics for Psychology Students Copyright © 2022 by Gordon E. Sarty is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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