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Numerical identification of outliers: calculating s And finding outliers manually

If you do not have the function LinRegTTest, then you can calculate the outlier in the first example by doing the following.

First, square each | y ŷ |

The squares are

  • 35 2
  • 17 2
  • 16 2
  • 6 2
  • 19 2
  • 9 2
  • 3 2
  • 1 2
  • 10 2
  • 9 2
  • 1 2

Then, add (sum) all the | y ŷ | squared terms using the formula

Σ i   =   1 11 ( | y i y ^ i | ) 2 = Σ i   =   1 11 ε i 2 (Recall that y i ŷ i = ε i .)

= 35 2 + 17 2 + 16 2 + 6 2 + 19 2 + 9 2 + 3 2 + 1 2 + 10 2 + 9 2 + 1 2

= 2440 = SSE . The result, SSE is the Sum of Squared Errors.

Next, calculate s , the standard deviation of all the y ŷ = ε values where n = the total number of data points.

The calculation is s = SSE n 2 .

For the third exam/final exam problem, s = 2440 11 2 = 16.47 .

Next, multiply s by 2:
(2)(16.47) = 32.94
32.94 is 2 standard deviations away from the mean of the y ŷ values.

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 is at least 2 s , then we would consider the data point to be "too far" from the line of best fit. We call that point a potential outlier .

For the example, if any of the | y ŷ | values are at least 32.94, the corresponding ( x , y ) data point is a potential outlier.

For the third exam/final exam problem, all the | y ŷ |'s are less than 31.29 except for the first one which is 35.

35>31.29 That is, | y ŷ | ≥ (2)(s)

The point which corresponds to | y ŷ | = 35 is (65, 175). Therefore, the data point (65,175) is a potential outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.)

Note

When outliers are deleted, the researcher should either record that data was deleted, and why, or the researcher should provide results both with and without the deleted data. If data is erroneous and the correct values are known (e.g., student one actually scored a 70 instead of a 65), then this correction can be made to the data.



The next step is to compute a new best-fit line using the ten remaining points. The new line of best fit and the correlation coefficient are:

ŷ = –355.19 + 7.39 x and r = 0.9121

Using this new line of best fit (based on the remaining ten data points in the third exam/final exam example ), what would a student who receives a 73 on the third exam expect to receive on the final exam? Is this the same as the prediction made using the original line?

Using the new line of best fit, ŷ = –355.19 + 7.39(73) = 184.28. A student who scored 73 points on the third exam would expect to earn 184 points on the final exam.

The original line predicted ŷ = –173.51 + 4.83(73) = 179.08 so the prediction using the new line with the outlier eliminated differs from the original prediction.

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The data points for the graph from the third exam/final exam example are as follows: (1, 5), (2, 7), (2, 6), (3, 9), (4, 12), (4, 13), (5, 18), (6, 19), (7, 12), and (7, 21). Remove the outlier and recalculate the line of best fit. Find the value of ŷ when x = 10.

ŷ = 1.04 + 2.96 x ; 30.64

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The Consumer Price Index (CPI) measures the average change over time in the prices paid by urban consumers for consumer goods and services. The CPI affects nearly all Americans because of the many ways it is used. One of its biggest uses is as a measure of inflation. By providing information about price changes in the Nation's economy to government, business, and labor, the CPI helps them to make economic decisions. The President, Congress, and the Federal Reserve Board use the CPI's trends to formulate monetary and fiscal policies. In the following table, x is the year and y is the CPI.

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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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