Question

Agricultural Exports. The U.S. Department of Agriculture collects data pertaining to the value of agricultural exports and publishes its findings in the U.S. An agricultural Trade Update. For one year, the values of these exports, by state, are provided on the WeissStats site. Data are in millions of dollars.

• a. Obtain the mean, median, and mode(s) of the data. Determine which of these measures of the center is best, and explain your answer
• b. Determine the range and sample standard deviation of the data.
• c. Fond the five-number summary and interquartile range of the data.
• d. Identify potential outliers, if any:
• e. Obtain and interpret a boxplot.

Short answer

Ans:

Part (a). The Mean of the data is $1563$and the median is $995$.

Part (b). The Range of the data is $11298$and the sample standard deviation is $1983$

Part (c).$3.70,234,995,2118,11302,1883$

Part (d). There are 3 potential outliers in the given data : $(5198.6,5246.8,and11301.7)$

Part (e). The boxplot is given below with 3 potential outliers .

Step-by-step solution

Step 1. Given information:

• For one year, the values of these exports, by state, are provided on the WeissStats site.
• Data are in millions of dollars

Step 2. Solving Part (a):

Using MINITAB,

Step1: Select Stat>Basic statistics>Display descriptive statistics

Step2: In the dialogue box, select Value in variables

Step3: Select Statistics. Check Mean, Median.

Step4: Click OK and OK

Following the above steps, we get the output as follows:

Descriptive Statistics: VALUE

Variable	Total Count	Mean	Median
VALUE	50	1563	995

We can observe the Mean of the data is 1563 and the median is 995.

As all observations are unique and non-repeating, there is no mode for the given data.

As the observations are numerical, the Mean is assumed to be the best measure of center.

Step 3. Solving Part (b):

Using MINITAB,

Step1: Select Stat>Basic statistics>Display descriptive statistics

Step2: In the dialogue box, select Value in variables

Step3: Select Statistics. Check range, and standard deviation.

Step4: Click OK and OK

Following the above steps, we get the output as follows:

Descriptive Statistics: VALUE

Variable	StDev	Range
VALUE	1983	11298

From the above output, we can see the Range of the data is $11298$and the sample standard deviation is $1983$

Step 4. Solving Part (c):

Using MINITAB,

Step1: Select Stat>Basic statistics>Display descriptive statistics

Step2: In the dialogue box, select Value in variables

Step3: Select Statistics. Check Minimum, First quartile, median, third quartile,

Maximum and Inter-Quartile range

Step4: Click OK and OK

Following the above steps, we get the output as follows:

Descriptive Statistics: VALUE

Variable	Minimum	Q1	Median	Q3	Maximum	IQR
VALUE	3.70	234	995	2118	11302	1883

Minimum value $=3.70$

First quartile $=234$

Median $=995$

Third quartile $=2118$

Maximum =$11302$

Also, the inter-quartile range of the data is$1883$

Step 5. Solving Part (d):

An observation is known to be a potential outlier if the value of the observation is beyond the limits.

The limits of the data are calculated as follows:

Lower Limit $=Q_1-1.5\times IQR$

$$\begin{gathered} =234-1.5\times 1883 \\ =-2590.5 \end{gathered}$$

Upper Limit$=Q_3+1.5\times IQR$

$=Q_3+1.5\times IQR$

The observations which are less than the lower limit or greater than the upper limit are assumed to be potential outliers. By observing the given data, we can see that there are three observations $(5198.6,5246.8,and11301.7)$that are greater than the upper limit. In other words, there exist 3 potential outliers in the given data.

Step 6. Solving Part (e):

From the above box plot, we can see that there exist three outliers in the given data. The area below the median is smaller than the area above the median line. Hence, we can say that the distribution is right-skewed.