Question

The level of lead in the blood was determined for a sample of \(152\) male hazardous-waste workers ages \(20 - 30\) and also for a sample of \(86\) female workers, resulting in a mean \(6\) standard error of \(5.5 \pm 0.3\)for the men and \(3.8 \pm 0.2\) for the women (“Temporal Changes in Blood Lead Levels of Hazardous Waste Workers in New Jersey, 1984–1987,” Environ. Monitoring and Assessment, 1993: 99–107). Calculate an estimate of the difference between true average blood lead levels for male and female workers in a way that provides information about reliability and precision.

Short answer

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

Step-by-step solution

calculate the difference between average blood level male and female workers

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.

conclusion

the final solution is

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