Question

Please explain

(1 point) 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. 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 two locations in a plant that processes 37,000 turkeys per day, taken on four days in the summer. The units are CFUs per cubic meter of air. Day 1 Day 2 Day 3 Day 4 Kill room 3175 2526 1763 1090 Processing 529 141 362 224 The spore count is clearly higher in the kill room. Give sample means and a 98% confidence interval to estimate how much higher the spore count is in the kill room. To do this problem correctly you need to think carefully about whether this is really a matched pairs design (one sample with two measurement) or a two sample design. X (kill): x (processing): 98% confidence interval: to

Answer 1

	Kill	Processing	Difference = Kill - Processing
	3175	529	2646
	2526	141	2385
	1763	362	1401
	1090	224	866
Average	2138.5	314	1824.5
St. Dev.	906.429	169.861	834.086
n	4	4	4

Xˉ Kill​=2138.5

X_ Processing= 314

98% confidence interval can be calculated as,

u Ps X 7 F P = 15

=1824.5 ± 4.541 x 834.086/√4

=1824.5 ±1893.79

= (-69.29 ,3718.29)

98% confidence interval is −69.168<μD​<3718.168.