Bacteria in water are counted as colony-forming units (CFU’s) per milliliter. Te
ID: 3135123 • Letter: B
Question
Bacteria in water are counted as colony-forming units (CFU’s) per milliliter. Ten bottles of water are randomly selected for sampling, with the intention of testing if two different labs produce the same results. Each bottle is divided into two parts, and then given to each of the two labs:
Bottle
1
2
3
4
5
6
7
8
9
10
Lab 1
875
959
475
589
925
1200
971
421
892
728
Lab 2
910
878
480
495
1021
980
1002
410
850
620
A) What is the probability that the average sample mean from lab 1 is more than 59 CFU’s larger than that of lab 2?
B) What is the probability that the average sample mean from lab 2 is more than 42 CFU’s larger than that of lab 1?
HINT: are the ten observations from lab 1 independent of those from lab 2?
Bottle
1
2
3
4
5
6
7
8
9
10
Lab 1
875
959
475
589
925
1200
971
421
892
728
Lab 2
910
878
480
495
1021
980
1002
410
850
620
Explanation / Answer
Lab1 : {875,959,475,589,925,1200,971,421,892,728}
SD1 = 245.64
Mean1 = 803.5
Lab2 : {910,,878,480,495,1021,980,1002,410,850,620}
SD2 = 238.07
Mean2 = 764.6
Standard Error (SE) = sqrt( (SD1^2/n1) + (SD2^2/n2) )
= sqrt( ((245.64)^2/10) + ((238.07)^2/10) )
= 108.174
A)
t = [ (Mean1 - Mean2) - d ] / SE
= ((803.5-764.6)-59)/108.174
= -0.185
dF =10+10-2 = 18
P-value = 0.5723 Answer
B)
t = [ (Mean2 - Mean1) - d ] / SE
= ((764.6-803.5)-42)/108.174
= -0.747
dF =10+10-2 = 18
P-value = 0.7676 Answer
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