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3.171

Expert-verifiedFound in: Page 137

Book edition
9th

Author(s)
Weiss, Neil

Pages
590 pages

ISBN
9780321989505

Hurricanes. An article by D. Schaefer et al. (journal of Tropical Ecology, Vol. 16, pp. 189-207) reported on a long-term study of the effects of hurricanes on tropical streams of the Luquillo Experimental Forest in Puerto Rico. The study shows that Hurricane Hugo had a significant impact on stream water chemistry. The following table shows a sample of 10 ammonia fluxes in the first year after Hugo. Data are in kilograms per hectare per year.

a. obtain and interpret the quartiles.

b. determine and interpret the interquartile range.

c. find and interpret the five-number summary:

d. identify potential outliers, if am:

e. construct and interpret a boxplot.

a)Interpretation: Based on the foregoing findings, ammonia fluxes in the first year following Hugo are less than 154 kg per hectare per year for$25\%$ of the time.

b)Interpretation: In the first year following Hugo, the middle 50% of ammonia fluxes is over 66 kilograms per hectare per year.

c)Interpretation: It can be seen from the above results that the variation between the quartiles is substantial, with the second quartile having the greatest variation in the data set.

d)There are no potential outliers because all of the data are included in the lower and upper limits.

e)From the box plot, there have no outlier

Given in the question that, a table of data, which represents the effects of hurricanes on tropical streams of the Luquillo Experimental Forest in Puerto Rico.

The quartiles should be calculated as follows:

To start, arrange the data in ascending order:

$\begin{array}{|lllllllllll|}\hline \text{FLUX}& 57& 66& 88& 96& 116& 147& 147& 154& 154& 175\\ \hline\end{array}$

The Number of observations is 10 .

So the median is at position

localid="1653981003840" $=\frac{(n+1)}{2}$

localid="1653981008948" $=\frac{11}{2}$

= 5.5

The median is average of localid="1653981013325" ${5}^{th}$ and localid="1653981017117" ${6}^{th}$ position values, it can be shown in boldface in ordered data set. Thus, the median of the entire data set is

localid="1653981032071" ${Q}_{2}=\frac{116+147}{2}$

= 131.5

As a result, the dataset's second quartile is 131.5.

$\begin{array}{|lllll|}\hline 57& 66& 88& 96& 116\\ \hline\end{array}$

The median is the middle term of the supplied data because there are 5 observations.

The data set's first quartile islocalid="1653981035594" ${Q}_{1}=88$

As a result, the data set's first quartile is 88.

Consider the second part of the complete data set, which is at or below the data set's median

$\begin{array}{|lllll|}\hline 147& 147& 154& 154& 175\\ \hline\end{array}$

There are a total of 5 observations. As a result, the median is the data's middle term.

The data set's third quartile islocalid="1653981041653" $Q3=154.$

As a result, the data set's third quartile is154

Interpretation: Based on the foregoing findings, ammonia fluxes in the first year following Hugo are less than 154 kg per hectare per year forlocalid="1653981051325" $25\%$ of the time.

Given in the question that,

Calculate the range between the quartiles.

The difference between the first and third quartiles is the inter quartile range (IQR).

$IQR={Q}_{3}-{Q}_{1}$

= 154 -88

= 66

Interpretation: In the first year following Hugo, the middle 50% of ammonia fluxes is over 66 kilograms per hectare per year.

We have to determine and interpret the five number summary.

Calculate the five numbers summery

The data set's minimum value is 57.

The data set's bottom quartile is $Q1=88.$

The data set's median is $Q2=131.5.$

The data set's top quartile is$Q3=154.$

The given data set's maximum value is 175

The measure of variation of the middle quarter is

$={Q}_{2}-{Q}_{1}$

= 131.5 -88

= 43.5

The measure of variation of the third quarter is

$={Q}_{3}-{Q}_{2}$

= 154 - 131.5

= 22.5

The Variation of the first quarter is

$={Q}_{1}-\mathrm{min}$

= 88 -57

= 31

The Variation of the fourth quarter is

$=\mathrm{max}-{Q}_{3}$

= 175 -154

= 21

Interpretation: It can be seen from the above results that the variation between the quartiles is substantial, with the second quartile having the greatest variation in the data set.

Given in the question that,

Calculate the lower and upper limit of data set.

Lower limit

$={Q}_{1}-1.5\left(IQR\right)$

= 88- 1.5(66)

= -11

Upper limit

$={Q}_{3}+1.5\left(IQR\right)$

= 154 + 1.5(66)

= 253

Potential outliers are observations that fall below or over the lower or higher limits.

There are no potential outliers based on the provided data.

As a result, there are no possible outliers.

There are no potential outliers because all of the data are included in the lower and upper limits.

We have to construct and interpret a boxplot

We have to use MINITAB, for the box plot

From the above box plot, there have no outlier

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