An article considered various issues associated with determining endotoxin concentration. The following data on concentration (EU/mg) in settled dust for one sample of urban homes and another of farm homes was kindly supplied by the authors of the article. U: 6.0 5.0 11.0 33.0 4.0 5.0 89.0 18.0 35.0 17.0 23.0
F: 4.0 14.0 11.0 9.0 9.0 8.0 4.0 20.0 5.0 8.9 21.0 9.2 3.0 2.0 0.3
(a) Determine the value of the sample standard deviation for each sample. [Hint: sum. gifxi = 246.0 for the urban sample and 128.4 for the farm sample, and sum. gifxi2 = 11,600 for the urban sample and 1617.94 for the farm sample.](Round your answers to two decimal places.)
urban-settlings=?
farm-settlings=?
Interpret these values.
Contrast variability in the two samples:
*The variability in endotoxin concentration is far less in urban homes than in farm homes.
*The variability in endotoxin concentration is far greater in urban homes than in farm homes.
*The variability in endotoxin concentration is similar in urban homes than in farm homes.
(b) Compute the fourth spread for each sample and compare. Do the fourth spreads convey the same message about variability that the standard deviations do? Explain.
*Yes, The fourth spreads do communicate the same message as the standard deviations did.
*No, The fourth spreads do not communicate the same message as the standard deviations did.

The variability in endotoxin concentration is far greater in urban homes than in farm homes.

Yes, The fourth spreads do communicate the same message as the standard deviations did.

Step-by-step explanation:

Given the data:

U: 6.0 5.0 11.0 33.0 4.0 5.0 89.0 18.0 35.0 17.0 23.0

F: 4.0 14.0 11.0 9.0 9.0 8.0 4.0 20.0 5.0 8.9 21.0 9.2 3.0 2.0 0.3

Farm sample:

Σx = 128.4

Σx² = 1617.94

n = 15

Sample mean, m = Σx / n ; 128.4 / 15 = 8.56

Standard deviation :

sqrt[(Σx²/n-1) - (m²)]

sqrt[(1617.94/14) - (8.56^2)]

sqrt(42.293542) = 6.50

Fourth spread = 3rd quartile - 1st quartile

3rd quartile = 3/4(n+1)th term

1st quartile = 1/4(n+1)th term

Using calculator :

Q3 = 11 ; Q1 = 4

Fourth spread = 11 - 4 = 7

Urban sample:

Σx = 246.0

Σx² = 11600

n = 11

Sample mean, m = Σx / n ; 246 / 11 = 22.36

Standard deviation :

sqrt[(Σx²/n-1) - (m²)]

sqrt[(11600/10) - (22.36²)]

sqrt(660.0304) = 25.69

Fourth spread = 3rd quartile (Q3) - 1st quartile(Q1)

3rd quartile = 3/4(n+1)th term

1st quartile = 1/4(n+1)th term

Q3 = 33 ; Q1 = 5

Fourth spread = 33 - 5 = 28

Standard deviation of farm sample = 6.50

Standard deviation of urban sample = 25.69

From the standard deviation values obtained, it could be observed that the urban sample has greater Variability than the farm sample, this is due to the higher standard deviation value of the urban sample and this shows that variation about the mean value is higher for urban sample.

Looking at the fourth spread value for both urban and farm samples ; the fourth spread value for urban is greater Than for farm sample.

answer:   y=2x-3

step-by-step explanation:

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step-by-step explanation: