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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consistent with summary statistics given in the paper "Beauty and the Labor Market: Evidence from Restaurant Servers" (unpublished manuscript by Matt Parrett, 2007 ), are: $$ \begin{array}{rrrrrrr} 0.0 & 5.0 & 45.0 & 32.8 & 13.9 & 10.4 & 55.2 \\ 50.0 & 10.0 & 14.6 & 38.4 & 23.0 & 27.9 &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 variability for a data set with outliers?

Short Answer

Expert verified
The mean and standard deviation for the original data set are approximately 26.815 and 22.738, respectively. After removing the observation of 105.0, the new mean and standard deviation are approximately 22.705 and 11.784, respectively. This shows that the presence of an outlier greatly affects the mean and standard deviation, making them less representative of the data set. Alternative measures, such as median and interquartile range, may be more appropriate in such cases.

Step by step solution

01

Calculate the mean

To find the mean of this data set, add up all the values and divide by the total number of observations (20). In this case, the sum of the values is: \(0.0 + 5.0 + 45.0 + 32.8 + 13.9 + 10.4 + 55.2 + 50.0 + 10.0 + 14.6 + 38.4 + 23.0 + 27.9 + 27.9 + 105.0 + 19.0 + 10.0 + 32.1 + 11.1 + 15.0 = 536.3\) So the mean is: \( \frac{536.3}{20} = 26.815 \)
02

Calculate the standard deviation

To find the standard deviation, first calculate the variance, which is the average of the squared differences from the mean. The steps are: 1. Subtract the mean from each data point and square the result. 2. Add up these squared differences. 3. Divide the sum by the total number of observations. 4. Take the square root of the result to obtain the standard deviation. For this data set, after calculating the squared differences and adding them up, we get: \(\sum_{i=1}^{20}(x_i - \bar{x})^2 = 10440.0475\) Now, divide the sum by the total number of observations and take the square root: \( \sqrt{ \frac{10440.0475}{20}}= 22.738 \) Thus, the standard deviation for this data set is approximately 22.738.
03

Remove the observation of 105.0 and recalculate the mean and standard deviation

After removing the observation of 105.0, the sum of the remaining values is: \(536.3 - 105.0 = 431.3\) Now, divide the new sum by the total number of observations (now 19): \( \frac{431.3}{19} = 22.705 \) So, the new mean is approximately 22.705. Next, we need to recalculate the standard deviation. The new sum of squared differences, with 105.0 removed, is 2637.1395. Now, divide the new sum by the total number of observations (19) and take the square root: \(\sqrt{ \frac{2637.1395}{19}}= 11.784 \) Thus, the new standard deviation is approximately 11.784.
04

Compare the values and discuss the implications

Comparing the values calculated in parts a and b, we see that removing the outlier of 105.0 significantly reduced both the mean (from 26.815 to 22.705) and the standard deviation (from 22.738 to 11.784). This suggests that the presence of an outlier can greatly affect the mean and standard deviation, making them less representative of the overall data set. In such cases, alternative measures of center (e.g., median) and variability (e.g., interquartile range) might be more appropriate in describing the data set.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Calculation
Calculating the mean is one of the most common ways to summarize a data set. The mean gives you the average value, offering a central point around which all data points revolve. To find the mean of a data set, simply add up all the data values and divide the total by the number of data points. It's as simple as doing basic division. For instance, if you have 20 restaurant tips recorded and their total sum is 536.3, the mean or average tip amount would be obtained by dividing 536.3 by 20, resulting in a mean of 26.815.

Remember, while the mean is a useful summary measure, it can be easily swayed by extremely high or low values, known as outliers. So, always be cautious of outliers when interpreting the mean as your only measure of central tendency.
Standard Deviation
The standard deviation is a measure of how spread out your data is. It tells you how much the individual data points differ from the mean on average. To calculate the standard deviation, you need to compute the variance first, which is the average of the squared differences between each data point and the mean.

Here's how you calculate it:
  • Subtract the mean from each data point to get the deviation.
  • Square each deviation to make them positive values.
  • Find the average of these squared values to get the variance.
  • Take the square root of the variance to get the standard deviation.
In our exercise, we calculated a standard deviation of about 22.738 initially. A higher standard deviation implies more variability in the data. Like mean, though, the standard deviation is sensitive to outliers, which can inflate the perceived variability.
Outliers Effect
Outliers are data points that are significantly higher or lower than the rest of the dataset. They can have a dramatic effect on your results, particularly when using mean and standard deviation. For example, the initial dataset includes a tipping value of 105.0, which is much higher than the rest.

Outliers like this can skew the mean upwards and increase the standard deviation, making your data seem more spread out than it actually is. In the exercise, removing the outlier of 105.0 brought the mean down from 26.815 to 22.705 and the standard deviation from 22.738 to 11.784. This demonstrates how sensitive these measures are to outliers.

Consider using other measures like the median or interquartile range if outliers drastically affect your data's interpretation.
Variance
Variance provides a measure of how data points differ from the mean. It's the precursor to standard deviation and reflects on how spread out the data points are. Calculating it involves:
  • Finding the difference between each data point and the mean.
  • Squaring each difference.
  • Calculating the average of these squared differences.
Though it sounds complex, variance helps quantify the overall spread in a dataset. In our calculations, the variance came from dividing the sum of squared deviations by the number of observations. A variance, like 10440.0475 in the original dataset, might seem large, but remember, it's the square root of this number that becomes more relevant (standard deviation). Variance is a helpful metric, especially when comparing different datasets analytically.
Data Analysis
Data analysis is the process of inspecting, cleaning, and modeling data to extract useful information, which is often used to support decision-making. In our case, analyzing tipping data involves calculating descriptive statistics like the mean, variance, and standard deviation, and understanding how they can be influenced by outliers.

This analysis allows us to see the patterns in tipping behavior, helping us draw conclusions. It can show trends, such as typical tipping percentages, and reveal any anomalies.

Always keep in mind that real-world data isn't perfect; it's often messy and contains outliers. Although outliers can skew your results, they're also interesting points that might warrant further investigation into why they occurred.

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Most popular questions from this chapter

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