Prolonged exposure to silica dust generated during crushing, blasting, drilling,

Question

Prolonged exposure to silica dust generated during crushing, blasting, drilling, and grinding of silica-containing materials such as granite, sandstone, coal, and some metallic ores causes silicosis. This disease is caused by the deposition of silica particles in the alveoli sacs and ducts of the lungs, which are unable to remove the silica particles by coughing or expelling of mucous. The deposited silica particles obstruct gaseous exchange between the lung and the blood stream. This sparks an inflammatory response which recruits large numbers of leukocytes (neutrophils) to the affected region. Roursgaard et al. (2008) studied quartz-induced lung inflam- mation in mice. The increase in the number of neutrophils is associated with higher levels of MIP-2 (anti-macrophage inflammatory protein 2). Bronchoalveolar fluid in the lungs of 23 mice exposed to 50 g quartz was found to have a mean MIP-2 expression level of 11.4 pg/ml of bronchoalveolar fluid with a standard deviation of 5.75. (a) What would you predict is the standard error of the mean of the MIP-2 expression levels? (b) The expression levels may not be normally distributed, but since n = 23 we assume that the central-limit theorem holds. Construct a 95% confidence inter- val for the population mean.

Explanation / Answer

TRADITIONAL METHOD
given that,
standard deviation, ? =5.75
sample mean, x =11.4
population size (n)=23
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 5.75/ sqrt ( 23) )
= 1.199
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, ? = 0.05
from standard normal table,right tailed z ?/2 =1.645
since our test is right-tailed
value of z table is 1.645
margin of error = 1.645 * 1.199
= 1.972
III.
CI = x ± margin of error
confidence interval = [ 11.4 ± 1.972 ]
= [ 9.428,13.372 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, ? =5.75
sample mean, x =11.4
population size (n)=23
level of significance, ? = 0.05
from standard normal table,right tailed z ?/2 =1.645
since our test is right-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 11.4 ± Z a/2 ( 5.75/ Sqrt ( 23) ) ]
= [ 11.4 - 1.645 * (1.199) , 11.4 + 1.645 * (1.199) ]
= [ 9.428,13.372 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [9.428 , 13.372 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 11.4
standard error =1.199
z table value = 1.645
margin of error = 1.972
confidence interval = [ 9.428 , 13.372 ]

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