Create and interpret boxplots as a means of summarizing non-symmetric data.
Calculate and explain the purpose of measures of centers (mean, median), variability (standard deviation, interquartile range).
Explain the impact of outliers on summary statistics such as mean, median and standard deviation.
The three quartiles, \(Q_1\)
, \(Q_2\)
, and \(Q_3\)
are numbers in an ordered data set that divide the data set into four equal parts. The second quartile is known as the median.
Interquartile Range (IQR for short) is the measure of variation when using the median to measure center. It is defined as the difference of the third and the first quartiles: \(\text{IQR}=Q_3-Q_1\)
.
When the center and the spread are measured by the median and the IQR, a value in the data is considered an outlier if the value is
\(\text{fence}_{lower}=Q_1 − 1.5 \cdot \text{IQR}\)
or\(\text{fence}_{upper}=Q_3 + 1.5 \cdot \text{IQR}\)
.Note: An outlier in this definition is also called a mild outlier. An outlier that is less than the extreme lower fence \(\text{extreme fence}_{lower}=Q_1 - 3 \cdot \text{IQR}\)
or greater than the extreme upper fence \(\text{extreme fence}_{upper}=Q_3 + 3 \cdot \text{IQR}\)
is also called extreme outlier.
The minimum, \(Q_1\)
, \(Q_2\)
, \(Q_3\)
and maximum are known as the "five-number summary" of the data set.
The difference of maximum and minimum is called the range.
Find the median, quartiles, IQR and outliers (if they exist) of the sample height of 15 trees.
70, 65, 63, 72, 81, 83, 66, 75, 80, 75, 79, 76, 76, 69, 75
Find the median, quartiles, IQR and outliers (if they exist) of the sample height of 15 trees.
70, 65, 63, 72, 81, 83, 66, 75, 80, 75, 79, 76, 76, 69, 75
Solution:
63, 65, 66, 69, 70, 72, 75, 75, 75, 76, 76, 79, 80, 81, 83
\(Q_2\)
. The sample size is 15. The middle of the ordered data set is the \(\lceil 15/2 \rceil=8\)
-th number which is 75.\(Q_1\)
and \(Q_3\)
. \(Q_1\)
is the median of the numbers less than the median. \(Q_3\)
is the median of the number greater than the median. In this example, \(Q_1\)
is the 4-th number 69. \(Q_3\)
is the 4-th to the last, that is 79.\(\text{IQR}=Q_3-Q_1=79-69=10\)
.\(Q_1-1.5\text{IQR}=69-1.5\cdot 10=54\)
and \(Q_3+1.5\text{IQR}=79-1.5 \cdot 10=94\)
, there is no outlier in this sample.A box plot shows a "five-number summary" of the data set. It contains a box, two whiskers and dots (for outliers).
To create the boxplot for a distribution,
Draw a box from \(Q_1\)
to \(Q_3\)
.
Draw a vertical line in the box at the median.
Extend a tail from \(Q_1\)
to the smallest value that is not an outlier and from \(Q_3\)
to the largest value that is not an outlier.
Indicate outliers with a solid dot.
Create the boxplot for the ages of 32 best actor oscar winners (1970–2001).
31, 32, 32, 33, 35, 36, 37, 37, 38, 38, 39, 40, 40, 40, 42, 42, 43, 43, 45, 45, 46, 47, 48, 48, 51, 55, 55, 56, 60, 60, 61, 76
Create the boxplot for the ages of 32 best actor oscar winners (1970–2001).
31, 32, 32, 33, 35, 36, 37, 37, 38, 38, 39, 40, 40, 40, 42, 42, 43, 43, 45, 45, 46, 47, 48, 48, 51, 55, 55, 56, 60, 60, 61, 76
Solution: We may use Excel to find the five-number summary.
The quartiles are
$$Q_2=42.5,\quad Q_1=37.5,\quad Q_3=49.5.$$
The interquartile range and the bounds for mild outliers are
$$\text{IQR}=12, \quad Q_1-1.5\text{IQR}= 19.5, \quad Q_3+1.5\text{IQR}=67.5.$$
The smallest number that is not an outlier is 31. The largest number that is not an outlier is 61. Those two numbers bound the whiskers.
The number 76 is a mild outlier because
$$Q_3+1.5\text{IQR}< 76 < Q_3+3\text{IQR}.$$
Sigma notation: in math, we denote the sum of values \(x_1\)
, \(x_2\)
, \(\dots\)
, \(x_n\)
of a variable \(x\)
by \(\sum\limits_{i=1}^n x_i\)
or simply by \(\sum x\)
.
The population mean is \(\mu= \frac{\sum x}{N}\)
, where \(N\)
is the population size, i.e the number of elements in the population.
The notation \(\mu\)
reads as mu.
The sample mean is \(\bar{x}=\frac{\sum{x}}{n}\)
, where \(n\)
is the sample size. The notation \(\bar{x}\)
reads as \(x\)
--bar.
Find the mean city mpg for a sample of 10 cars.
18, 21, 20, 21, 16, 18, 18, 18, 16, 20
Find the mean city mpg for a sample of 10 cars.
18, 21, 20, 21, 16, 18, 18, 18, 16, 20
Solution: The mean is
$$\bar{x}=\frac{18+21+20+21+16+18+18+18+16+20}{10}=18.6.$$
The mean mpg of the 10 cars is 18.6 mpg.
The weighted mean of a set of numbers \(\{x_1, \dots, x_n\}\)
with weights \(w_1\)
, \(w_2\)
, ..., \(w_n\)
is defined as $$\frac{\sum w_ix_i}{\sum w_i}.$$
The mean of a frequency table is weighted mean \(\bar{x}=\frac{\sum f x}{n}\)
, where \(x\)
is an element with frequency \(f\)
and \(n\)
is the sample size.
In a course, the overall grade is determined in the following way: the homework average counts for 10%, the quiz average counts for 10%, the test average counts 50% , and the final exam counts for 30%. What's the overall grade of the student who earned 92 on homework, 95 on quizzes, 90 on tests and 93 on the final.
In a course, the overall grade is determined in the following way: the homework average counts for 10%, the quiz average counts for 10%, the test average counts 50% , and the final exam counts for 30%. What's the overall grade of the student who earned 92 on homework, 95 on quizzes, 90 on tests and 93 on the final.
Solution: The overall grade is the weighted mean
$$\frac{\sum w_ix_i}{\sum w_i}=\frac{0.1\cdot 92+0.1\cdot 95+0.5\cdot 90+0.3\cdot 93}{0.1+0.1+0.5+0.3}=91.6.$$
Show how to use Excel
Find the average petal width for a sample of 10 iris followers.
0.2, 2.1, 0.2, 1.7, 2.3, 0.3, 1.2, 0.2, 1.8, 2.3
Find the mean from the dot plot of sepal length for a sample of 10 iris flowers.
The deviation of an entry \(x\)
in a population data set is the difference \(x-\mu\)
, where \(\mu\)
is the mean of the population.
The population variance of a population of \(N\)
entries is defined as
The population standard deviation is
The deviation of an entry \(x\)
in a sample data set is the difference \(x-\bar{x}\)
, where \(\bar{x}\)
is the mean of the sample.
The sample variance and sample standard deviation are defined similarly
\(n\)
is the sample size.
Rounding rule: for mean, variance and standard deviation, we keep at least one more digit than the accuracy of the data set.
Note: To measure the spread, one may also use the mean absolute deviation
$$MAD=\dfrac{\sum |x-\bar{x}|}{n}.$$
However, the standard deviation has better properties in applications.
Show how to use Excel to find SD
Find the mean and standard deviation ages of a sample of 32 best actor oscar winners (1970–2001).
31, 32, 32, 33, 35, 36, 37, 37, 38, 38, 39, 40, 40, 40, 42, 42, 43, 43, 45, 45, 46, 47, 48, 48, 51, 55, 55, 56, 60, 60, 61, 76
Find the mean and standard deviation ages of a sample of 32 best actor oscar winners (1970–2001).
31, 32, 32, 33, 35, 36, 37, 37, 38, 38, 39, 40, 40, 40, 42, 42, 43, 43, 45, 45, 46, 47, 48, 48, 51, 55, 55, 56, 60, 60, 61, 76
Solution: We use the Excel functions AVERAGE()
and STDEV.S()
to find the mean and sample standard deviation respectively.
The mean is 44.7. The sample standard deviation is 10.3.
A sample of GPAs from ten students random chosen from a college are recorded as follows. 1.90, 3.00, 2.53, 3.71, 2.12, 1.76, 2.71, 1.39, 4.00, 3.33
Find the standard deviation of this sample.
When we increase values in a data set by a fixed number \(c\)
, the standard deviation of a data set won't change. However, the mean increases by \(c\)
too.
When we multiple values in a data set by a factor \(k\)
, the mean and the standard deviation both scale by the factor \(k\)
.
A sample of the highest temperature of 10 days has a standard deviation \(5^\circ\mathrm{C}\)
in Celsius.
If we want to know the standard deviation in Fahrenheit, do we need to recalculate using the sample?
What is the standard deviation in Fahrenheit.
If a data set has an approximately bell-shaped distribution, then
approximately 68% of the data lie within one standard deviation of the mean.
approximately 95% of the data lie within two standard deviations of the mean.
approximately 99.7% of the data lies within three standard deviations of the mean.
For any numerical data set, at least \(1−1/k^2\)
of the data lie within \(k\)
standard deviations of the mean, where \(k\)
is any positive whole number that is at least 2.
A population data set with a bell-shaped distribution has mean \(\mu = 6\)
and standard deviation \(\sigma = 2\)
. Find the approximate proportion of observations in the data set that lie:
A population data set with a bell-shaped distribution has mean \(\mu = 6\)
and standard deviation \(\sigma = 2\)
. Find the approximate proportion of observations in the data set that lie:
Solution: Apply the Empirical Rule, there are 68% of data lie between 6-2=4 and 6+2=8. Since the distribution is symmetric, then 34% of data lie between 4 and 6, and 34% of data lie between 6 and 8. Then there are only 50%-34%=26% of data lie below 4.
A sample data set has mean \(\bar{x}=6\)
and standard deviation \(s = 2\)
. Find the minimum proportion of observations in the data set that must lie
between 2 and 10.
Solution: Apply Chebyshev's theorem, there are 75% of data are between \(\bar{x}-2s=2\)
amd \(\bar{x}+2s=10\)
.
A sample data set has mean \(\bar{x}=10\)
and standard deviation \(s = 3\)
. Find the minimum proportion of observations in the data set that must lie between 1 and 19.
A teacher decide to curve the final exam by adding 10 points for each student. Which of
the following statistic will NOT change:
A. median, B. mean, C. interquartile range, D. standard deviation?
Please explain your conclusion.
Which distribution of data has the SMALLEST standard deviation? Please explain your conclusion.
To find the median, you may use the function MEDIAN()
.
To find quartiles, you may use the function QUARTILE.EXC()
.
Note: this function calculates first and third quartiles with 25% and 75% weights. The results may be different from the results calculated by hand discussed in this course.
To find the mean, you may use the function AVERAGE()
.
To find the population standard deviation, you may use the function STDEV.P()
.
To find the sample standard deviation, you may use the function STDEV.S()
.
Select your data—either a single data series, or multiple data series.
Click Insert
> Insert Statistic Chart
> Box and Whisker
to create a boxplot.
For more information, see Create a box and whisker chart in Excel 365
Consider the following sample that consists of speeds of 20 cars.
19, 4, 17, 22, 23, 8, 20, 19, 10, 10, 13, 13, 15, 12, 20, 14, 9, 20, 12, 11
Create and interpret boxplots as a means of summarizing non-symmetric data.
Calculate and explain the purpose of measures of centers (mean, median), variability (standard deviation, interquartile range).
Explain the impact of outliers on summary statistics such as mean, median and standard deviation.
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