MATH SOLVE

2 months ago

Q:
# The level of lead in the blood was determined for a sample of 152 male hazardous-waste workers age 20-30 and also for a sample of 86 female workers, resulting in a mean ± standard error of 5.4 ± 0.3 for the men and 3.1 ± 0.2 for the women. 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. (Use a 95% confidence interval. Round your answers to two decimal places.)

Accepted Solution

A:

Answer: = ( 1.39, 2.80) blood level for male is greater than female worker by a figure within confidence interval.Step-by-step explanation:From the data given:[tex]n_1 =152[/tex][tex]\bar{x_1} = 5.4 se_1 = 0.3[/tex][tex]n_2 = 86 \bar{x_2} = 3.1 se_2 = 0.2[/tex]Difference between average blood lead levels for male and female is calculated as[tex]\bar{x_1} -\bar{x_2} = 5.4 - 3.1 = 2.1[/tex] standard deviation for men is [tex]s_1 = se_1 \sqrt{n_1}[/tex] [tex]= 0.3\times \sqrt{152} = 3.70[/tex]standard deviation for women is [tex]s_2 = se_2 \sqrt{n_2}[/tex] [tex]= 0.2\times \sqrt{86} = 1.85[/tex]The degree of freedom calculated as[tex]df = n_1 +n_2 -2[/tex] = 152+86-2 = 236The t critical value for alpha 0.05 and 236 degree of freedom is 1.97[tex]CI = \bar{x_1} -\bar{x_2} \pm t_{cr} \sqrt{\frac{s_1^2}{n_1}+ \frac{s_2^2}{n_2}}[/tex][tex]= 2.1 \pm 1.97 \sqrt{\frac{3.70^2}{152}+ \frac{1.58^2}{86}}[/tex] = ( 1.39, 2.80)