Question

Data on tipping percent for 20 restaurant tables, consistent with summary statistics given in the paper "Beauty and the Labor Market: Evidence from Restaurant Servers"(unpublished manuscript by Matt Parrett, 2007), are: \(\begin{array}{rrrrrrr}0.0 & 5.0 & 45.0 & 32.8 & 13.9 & 10.4 & 55.2 \\ 50.0 & 10.0 & 14.6 & 38.4 & 23.0 & 27.9 & 27.9 \\ 105.0 & 19.0 & 10.0 & 32.1 & 11.1 & 15.0 & \end{array}\) a. Calculate the mean and standard deviation for this data set. b. Delete the observation of 105.0 and recalculate the mean and standard deviation. How do these values compare to the values from Part (a)? What does this suggest about using the mean and standard deviation as measures of center and spread for a data set with outliers?

Short answer

The mean will be significantly lower and the standard deviation will be noticeably smaller once the outlier (105.0) is removed. This is because the outlier being exceptionally high skews the mean and increases the standard deviation. This serves as a notable lesson about how outliers can offer a distorted view of data.

Step-by-step solution

Calculation of Mean and Standard Deviation (Part a)

First, calculate the sum of all the values in the data set. Then divide this sum by the total number of values (N) to get the mean. Mean = \(\frac{{\text{{Sum of all observations}}}}{{\text{{Total number of observations}}}}\)After the mean is calculated, use it to calculate the standard deviation which measures how spread out the numbers are from the mean. Standard Deviation = \(\sqrt{\frac{{\text{{Sum}} ((\text{{observation}} - \text{{mean}})^2)}}{{\text{{Total number of observations}} - 1}}}\)

Remove the Outlier and Recalculate (Part b)

Remove the outlier (105.0) from the data set. Repeat the same steps as in Step 1 to calculate the new mean and standard deviation for this reduced data set.

Compare the Values

Compare the new mean and standard deviation calculated after the outlier's removal with the ones calculated in Step 1 before its removal. This can help us understand how an outlier affects the measures of central tendency (mean) and spread (standard deviation).

Key concepts

Mean Calculation

The mean represents the average value of a data set and is a fundamental aspect of descriptive statistics. It's calculated by adding all the numerical values together and then dividing by the count of values. For example, to calculate the mean tipping percent for the given data set:

\text{Mean} = \( \frac{{\text{{Sum of all observations}}}}{{\text{{Total number of observations}}}} \)

To start, sum the given tipping percentages: 0.0, 5.0, 45.0, 32.8, and so on. If we let the sum of these numbers be 'S' and the total number of observations be 'N', the mean or average tipping percentage is just 'S/N'. It's a simple yet powerful way to determine the central value of the data set.

However, the mean alone doesn't tell us everything. It can be greatly affected by extreme values, or 'outliers', which can distort the representation of the typical value in the data set. That's why it's also important to look at other measures like the median and the mode, alongside the mean, to get a full picture of the data's central tendency. Always remember that the mean is only appropriate when the data is symmetrically distributed without outliers.

Standard Deviation

The standard deviation is a measure of the amount of variation or dispersion in a set of values. By calculating the standard deviation, we can understand how much the values in a data set deviate from the mean on average.

Standard Deviation = \( \sqrt{\frac{{\text{{Sum}} ((\text{{observation}} - \text{{mean}})^2)}}{{\text{{Total number of observations}} - 1}}} \)

This equation requires us to subtract the mean from each observation, square the result, sum all the squared differences, and finally take the square root of the sum divided by one less than the total number of observations. The reason we use N-1, known as Bessel's correction, is to get a better estimate of the standard deviation for the entire population from our sample. A high standard deviation indicates that values are generally far from the mean, while a low standard deviation means they are close to the mean.

Understanding the standard deviation is essential for interpreting data correctly. For instance, in contexts such as test scores or heights of individuals, it allows us to see the variability and predict the range where most values lie.

Outliers Impact on Data

Outliers are data points that differ significantly from other observations. They can have a profound impact on descriptive statistics, particularly on the mean and standard deviation. For instance, an extremely high value, like the 105.0 tipping percent in our exercise, can inflate the mean, suggesting that the average tipping percent is higher than what is typical for the majority of the data.

When we remove the outlier and recalculate the mean, the result is often a more representative value for the central tendency of the data set. The standard deviation also gets affected; without the outlier, it typically decreases, indicating that the remaining data points are closer to the mean than originally calculated.

Outliers can arise due to measurement errors, data entry errors, or they can be just a result of the natural variability in data. It is important for analysts to examine outliers and decide whether to include them in the analysis. Sometimes, if justified, removing outliers can provide a clearer picture of the data's distribution and tendency. However, blindly removing outliers without proper justification can lead to biased results, so careful consideration is always necessary.