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 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 99% confidence interval to estimate how much higher the spore count is in the kill room?

-611.031 < μd < 4260.031

Step-by-step explanation:

This is a matched pair design :

Day1 Day 2 Day 3 Day 4

x1 Kill room 3175 2526 1763 1090

x2 Processing 529 141 362 224

First we obtain the difference, d ;

d = (x1 - x2)

d = 2646, 2385, 1401, 866

The mean difference, dbar = Σd/ n = 7298 / 4 = 1824.5

The standard deviation of difference,

Sd = √(Σx-dbar)²

Using calculator, Sd = 834.086

The 99% confidence interval :

Mean ± margin of error

dbar ± margin of error

Margin of Error = Tcritical * Sd/√n

Tcritical at 99%, df = n-1 ; df = 4-1 = 3

Tcritical = 5.84

Margin of Error = 5.84 * 834.086/√4 = 2435.53112

Confidence interval = 1824.5 ± 2435.53112

(-611.031 ; 4260.031)

it has the same slope and a different y-intercept

hope this : )

answer: the answer is 30 miles per hour.

ft/s -> mphmph = ft/s 1.466667mph = 44 divided by 1.466667mph = 30.00000

- r3ktforgood ☕