10.5: Confidence Intervals for the Difference of Two Means
- Page ID
- 51906
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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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