If aâ rifleman's gunsight is adjusted correctly but he has shakyâ arms, he might shoot bullets scattered widely around theâ bull
's-eye target. Draw a sketch of the target with the bullet holes. Does this show variationâ (lack ofâ precision) orâ bias?
1 answer:
Answer:
Lack of precision is the cause
Step-by-step explanation:
Based on the condition that the riffle man has a shaky arm, in my honest opinion I would say it's due to lack of precision caused by his shaky arm
Since the riffle man might shoot bullets scattered widely around the bulls eye he lacks precision. Precision has to do with shooting a particular spot consistently
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Answer:
A. E ( U ) = 21.5454 , E ( F ) = 8.39333
B. M ( U ) = 17.0 , M ( F ) = 18.0
C. E ( U' ) = 17.0 , E ( F' ) = 7.95384
D. T ( U ) = 9.091% , T ( F ) = 6.667%
Step-by-step explanation:
Solution:-
- Two sample sets ( U ) and ( F ) that define the concentration ( EU/mg ) of endotoxin found in urban and farm homes as follows:
U: 6.0 5.0 11.0 33.0 4.0 5.0 80.0 18.0 35.0 17.0 23.0
F: 2.0 15.0 12.0 8.0 8.0 7.0 6.0 19.0 3.0 9.8 22.0 9.6 2.0 2.0 0.5
- To determine the mean of a sample E ( U ) or E ( F ) the following formula from descriptive statistics is used:
Where,
Xi : Data iteration
n: Sample size
Therefore,
- To determine the sample median we need to arrange the data for both samples ( U ) and ( F ) in ascending order as follows:
U: 4.0 5.0 5.0 6.0 11.0 17.0 18.0 23.0 33.0 35.0 80.0
F: 0.5 2.0 2.0 2.0 3.0 6.0 7.0 8.0 8.0 9.6 9.8 12.0 15.0 19.0 22.0
- Now find the mid value for both sets:
Median term ( U ) = ( n + 1 ) / 2
= ( 11 + 1 ) / 2 = 12/2 = 6th term
Median ( U ), 6th term = 17.0
Median term ( F ) = ( n + 1 ) / 2
= ( 15 + 1 ) / 2 = 16/2 = 8th term
Median ( F ), 8th term = 8.0
- We will now trim the smallest and largest observation from each set.
- In set ( U ) we see that smallest data corresponds to ( 4.0 ) while the largest data corresponds to ( 80.0 ). We will exclude these two terms and the trimmed set is defined as:
U': 5.0 5.0 6.0 11.0 17.0 18.0 23.0 33.0 35.0
- In set ( F ) we see that the smallest data corresponds to ( 0.5 ) while the largest data corresponds to ( 22.0 ). We will exclude these two terms and the trimmed set is defined as:
F': 2.0 15.0 12.0 8.0 8.0 7.0 6.0 19.0 3.0 9.8 9.6 2.0 2.0
- We will again use the previous formula to calculate means of trimmed samples ( U' ) and ( F' ) as follows:
- The trimming percentage is known as the amount of data removed from the original sample from top and bottom of sample size of 11 and 15, respectively.
- We removed the smallest and largest value from each set. Hence, a single value was removed from both top and bottom of each data set. We can express the trimming percentage for each set as follows:
%
- The trimming pecentages for each data set are 9.091% and 6.667% respectively.