91影视

For m, n large enough, confidence interval for\({\mu _1} - {\mu _2}\) with confidence level of approximately\(100(1 - \alpha )\% \) is

\(\bar x - \bar y \pm {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \)

where + and - give the adequate upper/lower limit of the interval. By replacing\({z_{\alpha /2}} by {z_\alpha }\) and pm with only + and - an upper/lower bound is obtained.

Value\(\bar x - \bar y\) is in the middle of the interval (the center), thus

\(\bar x - \bar y = \frac{{ - 473.3 + 1691.9}}{2} = 609.3\)

Value for

\(\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \)

can be obtained from

\({z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} = \frac{w}{2}\)

where\(w\)is the width of interval

\(w = 1691.9 - ( - 473.3) = 2165.2\)

Also, \({z_{\alpha /2}} for 95\% is {z_{\alpha /2}} = {z_{0.025}} = 1.96,\) which was obtained using the table in the appendix. Therefore,

\(\begin{array}{l}{z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} = \frac{w}{2}\\1.96 \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} = 1082.6 \\\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} = 552.35\end{array}\)

The only missing value to compute 90 % confidence interval is \({z_{\alpha /2}} = {z_{0.1/2}} = 1.645\) which was obtained using the table in the appendix. Thus, the 90% \(Cl\ ) for this difference is

\(\bar x - \bar y \pm {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} = 609.3 \pm 1.645 \times 552.35\ )

\( = ( - 299.3,1517.9)\)