91影视

The confidence interval for \({\mu _1} - {\mu _2}\) with a confidence level of around \(100(1 - \alpha )\% \) percent is for \(m\), \(n\)large enough.

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

where \( + \) and \( - \) denote the interval's appropriate upper and lower limits. An upper/lower bound is obtained by substituting \({z_{\alpha /2}}\) with \({z_\alpha }\) and \( \pm \)with only \( + \) and \( - \).

Because both sample sizes are large enough, the confidence interval specified before can be employed. The stated information regarding the first sample is in the exercise.

\(\begin{array}{l}\bar x = 5.5\\S{E_1} = \frac{{s_1^2}}{m} = 0.{3^2}\end{array}\)

And about the second sample are

\(\begin{array}{l}\bar y = 3.8\\S{E_2} = \frac{{s_2^2}}{n} = 0.{2^2}\end{array}\)

Using confidence level of \(\alpha = 0.05\), the \( z\) value is \({z_{\alpha /2}} = {z_{0.025}} = 1.96\). The \( 95\% \)confidence interval becomes

\(\begin{array}{l}\bar x - \bar y \pm {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} = 5.5 - 3.8 \pm 1.96 \times \sqrt {0.{3^2} + 0.{2^2}} \\ = (0.99,2.41).\end{array}\)

The genuine average blood lead level for male workers is somewhere between \({\rm{0}}{\rm{.99}}\) and greater than the average for female workers, according to \(95\) confidence.

the final solution is

\({\rm{(0}}{\rm{.99,2}}{\rm{.41)}}\)