Questions: You work for a large farm with many fields of corn. You are investigating the mass of a sample of ears of corn. You gather the following data:
Mass(g) of ears of corn
655.4, 494.6, 736.8, 651.1, 452.6, 607.1, 786.8, 521.9, 446.7, 604.2, 647.4, 441.4, 430.3
Checksum: 7476.3
Some of the masses in the sample seem much larger than the rest. You decide to make several calculations describing the "spread" of the data set. You hope to use them to help in the search for outliers. Find the following:
a) IQR 160.
b) Apply the 1.5 IQR rule to search for outliers. Report the lower and upper cutoffs.
Lower: Upper:
Are there any outliers by the 1.5 IQR rule? (Enter "yes" or "no")
Transcript text: You work for a large farm with many fields of corn. You are investigating the mass of a sample of ears of corn. You gather the following data:
Mass(g) of ears of corn
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline 655.4 & 494.6 & 736.8 & 651.1 & 452.6 & 607.1 & 786.8 \\
\hline 521.9 & 446.7 & 604.2 & 647.4 & 441.4 & 430.3 & \\
\hline
\end{tabular}
Checksum: 7476.3
Some of the masses in the sample seem much larger than the rest. You decide to make several calculations describing the "spread" of the data set. You hope to use them to help in the search for outliers.
Find the following:
a) IQR 160. $\square$
b) Apply the 1.5 IQR rule to search for outliers. Report the lower and upper cutoffs.
Lower: $\square$ Upper: $\square$
Are there any outliers by the 1.5 IQR rule? (Enter "yes" or "no") $\square$
Solution
Solution Steps
The sorted data is [430.3 441.4 446.7 452.6 494.6 521.9 604.2 607.1 647.4 651.1 655.4 736.8
786.8].
Step 1: Calculate the Interquartile Range (IQR)
The first quartile (Q1) is 452.6, and the third quartile (Q3) is 651.1.
Therefore, the Interquartile Range (IQR) is Q3 - Q1 = 651.1 - 452.6 = 198.5.
Step 2: Calculate the sample standard deviation
The mean of the dataset is $mean = \frac{data_sorted}{n}$ = 575.1, and the sample standard deviation is $\sqrt{\frac{x-mean}{n}} = 119$.
Step 3: Apply the 1.5 IQR rule for outliers
The lower cutoff for outliers using the 1.5 IQR rule is Q1 - 1.5 * IQR = 154.85, and the upper cutoff is Q1 + 1.5 * IQR = 948.85.
Step 4: Apply the 2-standard deviation rule for outliers
The lower cutoff for outliers using the 2-standard deviation rule is mean - 2 * std_dev = 337.1, and the upper cutoff is mean + 2 * std_dev = 813.1.
Final Answer:
Outliers identified using the 1.5 IQR rule: []
Outliers identified using the 2-standard deviation rule: []