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In some data sets, there are values (observed data points) called outliers . Outliers are observed data points that are far from the least squares line. They have large "errors", where the "error" or residual is the vertical distance from the line to the point.

Outliers need to be examined closely. Sometimes, for some reason or another, they should not be included in the analysis of the data. It is possible that an outlier is a result of erroneous data. Other times, an outlier may hold valuable information about the population under study and should remain included in the data. The key is to examine carefully what causes a data point to be an outlier.

Besides outliers, a sample may contain one or a few points that are called influential points . Influential points are observed data points that are far from the other observed data points in the horizontal direction. These points may have a big effect on the slope of the regression line. To begin to identify an influential point, you can remove it from the data set and see if the slope of the regression line is changed significantly.

Computers and many calculators can be used to identify outliers from the data. Computer output for regression analysis will often identify both outliers and influential points so that you can examine them.

Identifying outliers

We could guess at outliers by looking at a graph of the scatterplot and best fit-line. However, we would like some guideline as to how far away a point needs to be in order to be considered an outlier. As a rough rule of thumb, we can flag any point that is located further than two standard deviations above or below the best-fit line as an outlier . The standard deviation used is the standard deviation of the residuals or errors.

We can do this visually in the scatter plot by drawing an extra pair of lines that are two standard deviations above and below the best-fit line. Any data points that are outside this extra pair of lines are flagged as potential outliers. Or we can do this numerically by calculating each residual and comparing it to twice the standard deviation. On the TI-83, 83+, or 84+, the graphical approach is easier. The graphical procedure is shown first, followed by the numerical calculations. You would generally need to use only one of these methods.

In the third exam/final exam example , you can determine if there is an outlier or not. If there is an outlier, as an exercise, delete it and fit the remaining data to a new line. For this example, the new line ought to fit the remaining data better. This means the SSE should be smaller and the correlation coefficient ought to be closer to 1 or –1.

Graphical identification of outliers

With the TI-83, 83+, 84+ graphing calculators, it is easy to identify the outliers graphically and visually. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance were equal to 2 s or more, then we would consider the data point to be "too far" from the line of best fit. We need to find and graph the lines that are two standard deviations below and above the regression line. Any points that are outside these two lines are outliers. We will call these lines Y2 and Y3:

As we did with the equation of the regression line and the correlation coefficient, we will use technology to calculate this standard deviation for us. Using the LinRegTTest with this data, scroll down through the output screens to find s = 16.412 .

Line Y2 = –173.5 + 4.83 x –2(16.4) and line Y3 = –173.5 + 4.83 x + 2(16.4)

where ŷ = –173.5 + 4.83 x is the line of best fit. Y2 and Y3 have the same slope as the line of best fit.

Graph the scatterplot with the best fit line in equation Y1, then enter the two extra lines as Y2 and Y3 in the "Y="equation editor and press ZOOM 9. You will find that the only data point that is not between lines Y2 and Y3 is the point x = 65, y = 175. On the calculator screen it is just barely outside these lines. The outlier is the student who had a grade of 65 on the third exam and 175 on the final exam; this point is further than two standard deviations away from the best-fit line.

Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell if the point is between or outside the lines. On a computer, enlarging the graph may help; on a small calculator screen, zooming in may make the graph clearer. Note that when the graph does not give a clear enough picture, you can use the numerical comparisons to identify outliers.

The scatter plot of exam scores with a line of best fit.Two yellow dashed lines run parallel to the line of best fit. The dashed lines run above and below the best fit line at equal distances. One data point falls outside the boundary created by the dashed lines—it is an outlier.
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Questions & Answers

How do you get log of normal population
Shan Reply
what is the probability of getting no head face up in three tosses of a fair coin?
Epara Reply
In how many ways can probability be assigned to an event of interest?
Hey guys can someone help me with combinations and permutations
Phone lines on an airline system are occupied 50% of time assume that 10 calls are placed to the airline. What is the probability at least 1 call the lines are occupied?
Highness Reply
Phone lines on an airline system are occupied 50% of time assume that 10 calls are placed to the airline. What is the probability at least 1 call the lines are occupied?
Highness Reply
Why is the method of selecting the sample even more important than the size of the sample?
Nana Reply
formulaas for gruoped and ungrouped of quartiles
chatered Reply
Why is the method of selecting the sample even more important than the size of the sample?
una empresa productora que participa en el mercado de lapices tiene la siguente funcion de demanda Qd=a la raiz de 9 -9p si la elasticidad precio de demanda de la empresa es epd=0,5 determinar a) hallar el precio y cantidad
Jany Reply
find the mean mew and the standard devation sigma of the given population 9.8,10.2,10.4,9.8,10.0,10.2,9.6
Vaneza Reply
1. A card is drawn at random from an ordinary deck of 52 playing cards.* *Find the probability that it is* (a). an ace (b). a jack of hearts (c). a three of clubs or a six of diamonds (d). a heart (e). any suit except hearts (f). a ten or a spade (g). neither a four nor a club. *Hint:* 1st determine how many of the following is in a deck of cards. A deck of cards have 52 cards ♠️- Club ♣️ - Spade ♥️ - Heart ♦️- Diamond
Agness Reply
a measure of central tendency is a typical value around which other figures congregate
Anand Reply
hmm mean mode median
Y = alpha0 + alpha1X1 + E what is this equation
Musawenkosi Reply
An estimated linear regression equation
Thank you
simple linear regression .. where Alpho zero is reg constant ( intercept of the reg line) and Alpha1 is the regression coefficient ( slope of the regression line)
null and altarnate hypothesis are the statement about
farri Reply
about any population of interest
don't know
I said we give hypothesis about any population and mean , in null hyp we say sample mean is equal to the population mean where as in alternate hyp we say sample mean is not equal to the pop mean .. to test these things we use students test statistics commonly
ok thnx
probability of getting one black card
kaynat Reply
from a standard deck?
it will be 1/2
coz there are 26 blacks card out of 52 in a deck
so prob of getting a black card out of the deck = 26/52 = 1/5 =0.5
when are not two events mutually exclusive? 1) they overlap in a venn diagram 2) Probability of one affects the other
Junaid Reply
A stenographer claims that she can take dictation at the rate of 120 words per minute can we reject her claim on the basis of 100 trails in which she demonstrates a mean of 116 words with a variance of 225 words
Aromal Reply
the hypothesis to be tested is the claim to be tested
H0= 120,Ha not equal to 120 x bar =116,s=225,n=100

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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