Q 3.54. Outliers and trimmed means. Some... [FREE SOLUTION]

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Chapter 3: Q 3.54. (page 105)

Outliers and trimmed means. Some data sets contain outliers, observation that fall well outside the overall pattern of the data(We discuss outliers in more detail in section 3.4) Suppose, for instance that you are interested in the ability o high school algebra student to compute square roots. You decide to give a square root exam to 10 of these students Unfortunately, one of the student had a fight with his girlfriend and cannot concentrate he gets a 0. The 10 score are displayed in increasing order in the following table. The score of 0 is an outlier.
Statisticians have a systematic method for avoiding extreme observation and outliers when they calculated means. They compute trimmed means, in which high and low observation are deleted or "trimmed off" before the mean is calculated . For instance, to compute the 10% trimmed mean of the test score data. we first delete both the bottom 10% and the top 10% of the ordered data. that is,0 and 80. Then we calculated the mean of the remaining data. Thus the10% trimmed mean of the test score data is
The following table displays a set of score for a 40 question algebra final
Part (a) Do any the score look like outliers?
Part (b) Compute the usual mean of the data.
Part (c) Compute the 5% trimmed mean of the data.
Part (d) Compute the 10% trimmed mean of the data.
Part (e) Compare the means you obtained in parts (b) (d) which of the three provides the best measure of center for the data?

Short Answer

Expert verified

Part (a) 2 and 4 are outliers.

Part (b) 19.25

Part (c) 19.72

Part (d) 20.25

Part (e) 10% trimmed value is better measure of center

Step by step solution

Part (a) Step 1. Given information.

The provided data set is:

$(5, 15, 16, 16, 19, 21, 21, 25, 26, 27, 4, 15, 16, 17, 20, 21, 24, 25, 27, 28)$

Part (a) Step2. If any of the scores appear to be outliers.

Outliers are observations that deviate significantly from the overall pattern of the data.

The sorted data is written as,

(2,,4,15,15,16,16,16,17,19,20,21,21,21,24,25,25,26,27,27,28)

Because the other scores are all two digits, and scores 2 and 4 are single digits. As a result, they act as an outlier.

As a result, the scores that appear to be outliers are 2 and 4.

Part (b) Step 1. Given information.

The provided data set is:

$(5, 15, 16, 16, 19, 21, 21, 25, 26, 27, 4, 15, 16, 17, 20, 21, 24, 25, 27, 28)$

Part (b) Step2. The given data's mean.

Mean is expressed as follows:

$\overset{炉}{x} = \frac{\underset{}{鈭�} x}{n}$

The sorted data is written as,

(2,,4,15,15,16,16,16,17,19,20,21,21,21,24,25,25,26,27,27,28)

The mean of the given data set is computed as follows:

$\overset{炉}{x} = \frac{((2 + 4 + 15 + 15 + 16 + 16 + 16 + 17 + 19 + 20 + 21 + 21 + 21 + 2 + + 25 + 25 + 26 + 27 + 27 + 28)}{20} \overset{炉}{x} = \frac{385}{20} \overset{炉}{x} = 19.25$

Part (c) Step 1. Given information.

The provided data set is:

$(5, 15, 16, 16, 19, 21, 21, 25, 26, 27, 4, 15, 16, 17, 20, 21, 24, 25, 27, 28)$

Part (c) Step 2. The trimmed mean of the given data by 5%.

For the 5% trimmed mean, the 5% beginning and 5% ending data are removed.

5% trimmed data = 4,15,15,,16,16,16,17,,19,20,21,21,,21,24,25,26,27,27

Mean is expressed as follows:

$\overset{炉}{x} = \frac{\underset{}{鈭�} x}{n}$

The mean of the given data set is computed as follows:

$\overset{炉}{x} = \frac{(4 + 15 + 15 + 16 + 16 + 16 + 17 + 19 + 20 + 21 + 21 + 21 + 2 + + 25 + 25 + 26 + 27 + 27)}{18} \overset{炉}{x} = \frac{355}{18} \overset{炉}{x} = 19.7222 \overset{炉}{x} 鈮� 19.72$

Part (d) Step 1. Given information.

The provided data set is:

$(5, 15, 16, 16, 19, 21, 21, 25, 26, 27, 4, 15, 16, 17, 20, 21, 24, 25, 27, 28)$

Part (d) Step 2. The trimmed mean of the given data by 10%.

For the 10% trimmed mean, the 10% beginning and 10% ending data are removed.

10% trimmed data = 15,15,,16,16,16,17,,19,20,21,21,,21,24,25,26,27

Mean is expressed as follows:

$\overset{炉}{x} = \frac{\underset{}{鈭�} x}{n}$

The mean of the given data set is computed as follows:

$\overset{炉}{x} = \frac{(15 + 15 + 16 + 16 + 16 + 17 + 19 + 20 + 21 + 21 + 21 + 2 + + 25 + 25 + 26 + 27)}{16} \overset{炉}{x} = \frac{324}{16} \overset{炉}{x} = 20.25$

Part (e) Step 1. Given information.

The provided data set is:

$(5, 15, 16, 16, 19, 21, 21, 25, 26, 27, 4, 15, 16, 17, 20, 21, 24, 25, 27, 28)$

Part (d) Step 2. The mean of parts (b) to (d) that provides the best measure of center.

The true mean is 19.25.

19.72 is the trimmed mean.

20.25 is the trimmed mean.

As a result, 10% trimmed mean is greater than 5% trimmed mean, which is greater than the actual mean.

The mean with a 10% trimmed value is a better measure of center because both outlier values have been removed.

As a result, the 10% trimmed mean is greater than the 5% trimmed mean, which is greater than the actual mean, and the 10% trimmed value is a better measure of centre because both outlier values have been removed.