Question

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.)

Answer 1

= ( 1.39, 2.80) blood level for male is greater than female worker by a figure within confidence interval.

Explanation:

From the data given:

`n_1 =152`

`\bar{x_1} = 5.4 se_1 = 0.3`

`n_2 = 86 \bar{x_2} = 3.1 se_2 = 0.2`

Difference between average blood lead levels for male and female is calculated as

`\bar{x_1} -\bar{x_2} = 5.4 - 3.1 = 2.1`

standard deviation for men is

`s_1 = se_1 √(n_1)`

`= 0.3* √(152) = 3.70`

standard deviation for women is

`s_2 = se_2 √(n_2)`

`= 0.2* √(86) = 1.85`

The degree of freedom calculated as

`df = n_1 +n_2 -2`

= 152+86-2 = 236

The t critical value for alpha 0.05 and 236 degree of freedom is 1.97

`CI = \bar{x_1} -\bar{x_2} \pm t_(cr) \sqrt{(s_1^2)/(n_1)+ (s_2^2)/(n_2)}`

`= 2.1 \pm 1.97 \sqrt{(3.70^2)/(152)+ (1.58^2)/(86)}`

= ( 1.39, 2.80)