The air in poultry-processing plants often contains fungus spores. Inadequate ve
ID: 3159143 • Letter: T
Question
The air in poultry-processing plants often contains fungus spores. Inadequate ventilation can affect the health of the workers. The problem is most serious during the summer and least serious during the winter. To measure the presence of spores, air samples are pumped to an agar plate and "colony forming units (CFUs)" are counted after an incubation period. Here are data from the "kill room" of a plant that processes 37,000 turkeys per day, taken on four separate days in the summer and in the winter. The units are CFUs per cubic meter of air.
Summer 3080 2691 1581 1202
Winter 384 119 267 102
The counts are clearly much higher in the summer. The mean (±0.01) and the standard deviation (±0.001) for the summer are: x¯ = s = . The mean (±0.01) and the standard deviation (±0.001) for the winter are: x¯ = s = . The standard error of the difference in sample means is (±0.001) The critical value (±0.001) from the t distribution with the conservative degrees of freedom for a confidence interval of 80% is: t* = . Give a 80% confidence interval to estimate how much higher (±0.1) the mean count is during the summer:-------------- to---------------
Explanation / Answer
The mean (±0.01) and the standard deviation (±0.001) for the summer are: x¯ = s = .
By technology,
X1 = 2138.5 [ANSWER]
s1 = 890.6022307 [ANSWER]
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The mean (±0.01) and the standard deviation (±0.001) for the winter are: x¯ = s = .
By technology,
X2 = 218 [ANSWER]
s2 = 133.1840831 [ANSWER]
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The standard error of the difference in sample means is (±0.001)
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 4
n2 = sample size of group 2 = 4
hence,
sD = 450.2527994 [ANSWER]
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The critical value (±0.001) from the t distribution with the conservative degrees of freedom for a confidence interval of 80% is: t* =
For the 0.8 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.1
Thus, at df = n1 - 1 = 3,
t(alpha/2) = 1.637744354 [ANSWER]
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Give a 80% confidence interval to estimate how much higher (±0.1) the mean count is during the summer:-------------- to---------------
As
lower bound = [X1 - X2] - t(alpha/2) * sD = 1183.10102
upper bound = [X1 - X2] + t(alpha/2) * sD = 2657.89898
Thus, the confidence interval is
( 1183.10102 , 2657.89898 ) [ANSWER]
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