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The accompanying data are a subset of data read from a graph in the paper "Ladies First? A Field Study of Discrimination in Coffee Shops" (Applied Economics [April, 2008]). The data are wait times (in seconds) between ordering and receiving coffee for 19 female customers at a Boston coffee shop. \(\begin{array}{rrrrrrr}60 & 80 & 80 & 100 & 100 & 100 & 120 \\ 120 & 120 & 140 & 140 & 150 & 160 & 180 \\ 200 & 200 & 220 & 240 & 380 & & \end{array}\) a. Calculate and interpret the values of the median and interquartile range. b. Explain why the median and interquartile range is an appropriate choice of summary measures to describe center and spread for this data set.

Short Answer

Expert verified
The median offers a measure of central tendency, a single 'typical' wait time. The Interquartile Range displays how spread out the wait times are, i.e. where bulk of the wait times tend to fall. Since the data may have outliers (extremely high or low values), using means as a measure of central tendency could be misleading. Similarly, the standard deviation could also misjudge the spread due to influence of outliers. Hence, in this case, using the median and IQR are the most appropriate measures.

Step by step solution

01

Calculate the Median

Start by sorting out the given data in an order, increasing or decreasing. Once sorted, the middle value in the data set is referred to as the median. If the data set contains an odd number of observations, the median is the middle number. If the data set contains an even number of observations, the median is the average of the two middle numbers.
02

Calculate the Interquartile Range

The Interquartile Range (IQR) expresses the statistical spread in the data set. It is the range within which the central 50% of the values fall. IQR is calculated by subtracting the first quartile from the third quartile. Quartiles are the values that divide a list of numbers into quarters: Lower Quartile (Q1) is at 25 percentile, median (Q2) is at 50 percentile and Upper Quartile (Q3) at 75 percentile.
03

Interpret the Calculated Values

Now interpret these calculated values. The Median provides a measure of central tendency, offering the middle value of wait times. The Interquartile Range gives a measure of how spread out these wait times are. This tells us the range within which the central 50% of the values fall.
04

Explain Why These Measures are Suitable

In this step, explain why using the median and the interquartile range as measures are the appropriate choices for this particular data set.

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Key Concepts

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

Median Calculation
When analyzing a data set, finding the median is a standard method to identify the central tendency of the values. The median is the middle value that divides a data set into two equal halves when the data are arranged in ascending or descending order.

In the context of the exercise, with 19 wait times recorded, the median is particularly useful. Being an odd number of observations, there's a single middle value, so no averaging is required. To identify the median, we first order the data from least to greatest and then locate the central number.

In this case, the median would be the 10th value when the wait times are sorted, which provides a robust indicator of central tendency, unaffected by skewness or extreme values which is characteristic of asymmetrical distributions commonly found in real-world data.
Interquartile Range (IQR)
Moving beyond the center of the dataset, the interquartile range (IQR) is a statistical measure that captures the spread of the middle 50% of the data. It is calculated by subtracting the first quartile (25th percentile) from the third quartile (75th percentile).

To calculate the quartiles, we split the ordered data set into four equal parts. The value at the first quartile, Q1, is greater than 25% of the data, and the value at the third quartile, Q3, is greater than 75% of the data. The IQR is the range of values between Q1 and Q3 and is a measure of variability that is resistant to outliers. This is particularly useful for skewed distributions or data with outliers, as it describes the variability around the central tendency.
Data Set Analysis
An effective data set analysis involves summarizing and understanding the intricacies of the data, which includes the deployment of various statistical measures. These measures include locating the center of the data (central tendency) and assessing the data's spread or variability (dispersion).

When examining the coffee shop wait times, after calculating the median and IQR, we can analyze how the data are distributed. For example, if most customers' wait times are close to the median value, this suggests a higher concentration of data near the center. In contrast, a wide IQR could indicate more variability in the wait times.

Analysis of this data set would also involve observing trends, patterns, and potential outliers, such as the unusually long wait time of 380 seconds, and considering how these may affect the choice of statistical measures used.
Central Tendency
Central tendency is a key concept in statistics, aimed at identifying the single value that best represents all of the values in a data set. It is an estimate of the 'center' around which the data cluster. Common measures of central tendency include the mean (average), median, and mode (most frequent value).

For the coffee shop wait times, the median is favored due to its resistance to outliers. Mean, while commonly used, can be skewed by atypical values. This is pertinent when the spread or range of the data is significant, which, as evidenced by the calculation of the IQR, is indeed the case for the coffee shop wait times. The median, being the middle value, ensures that half of the wait times are above and half are below this central value, regardless of any extreme wait time occurrences, making it a reliable measure of central tendency for this analysis.

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Most popular questions from this chapter

The accompanying data are consistent with summary statistics in the paper "Shape of Glass and Amount of Alcohol Poured: Comparative Study of Effect of Practice and Concentration" (British Medical Journal [2005]: \(1512-1514\) ). The data are the actual amount (in \(\mathrm{ml}\) ) poured into a tall, slender glass for individuals asked to pour a "shot" of alcohol \((44.3 \mathrm{ml}\) or 1.5 ounces). Calculate and interpret the values of the mean and standard deviation. \(\begin{array}{llllllll}44.0 & 49.6 & 62.3 & 28.4 & 39.1 & 39.8 & 60.5 & 73.0\end{array}\) $$ \begin{array}{llllllll} 57.5 & 56.5 & 65.0 & 56.2 & 57.7 & 73.5 & 66.4 & 32.7 \end{array} $$ \(\begin{array}{ll}40.4 & 21.4\end{array}\)

USA Today (May 9,2006 ) published the accompanying weekday circulation numbers for the top 20 newspapers in the country for the 6 -month period ending March 31,2006: \(\begin{array}{rrrrr}2,272,815 & 2,049,786 & 1,142,464 & 851,832 & 724,242 \\\ 708,477 & 673,379 & 579,079 & 513,387 & 438,722 \\ 427,771 & 398,329 & 398,246 & 397,288 & 365,011 \\ 362,964 & 350,457 & 345,861 & 343,163 & 323,031\end{array}\) Explain why the average may not be the best measure of a typical value for this data set.

Research by the U.S. Food and Drug Administration (FDA) shows that acrylamide (a possible cancer-causing substance) forms in high-carbohydrate foods cooked at high temperatures, and that acrylamide levels can vary widely even within the same brand of food (Associated Press, December 6,2002 ). FDA scientists analyzed McDonald's french fries purchased at seven different locations and found the following acrylamide levels: \(\begin{array}{lllll}497 & 193 & 328 & 155 & 326\end{array}\) \(\begin{array}{ll}245 & 270\end{array}\) a. Calculate the mean acrylamide level. For each data value, calculate the deviation from the mean. b. Verify that, except for the effect of rounding, the sum of the seven deviations from the mean is equal to 0 forthis data set. (If you rounded the sample mean or the deviations, your sum may not be exactly zero, but it should still be close to zero.) c. Use the deviations from Part (a) to calculate the variance and standard deviation for this data set.

In August 2009 , Harris Interactive released the results of the Great Schools survey, in which 1,086 parents of children attending a public or private school were asked approximately how much they had spent on school supplies over the last school year. For this sample, the mean amount spent was \(\$ 235.20\) and the median amount spent was \(\$ 150.00 .\) What does the large difference between the mean and median tell you about this data set?

The mean playing time for a large collection of compact discs is 35 minutes, and the standard deviation is 5 minutes. a. What value is 1 standard deviation above the mean? One standard deviation below the mean? What values are 2 standard deviations away from the mean? b. Assuming that the distribution of times is mound shaped and approximately symmetric, approximately what percentage of times are between 25 and 45 minutes? Less than 20 minutes or greater than 50 minutes? Less than 20 minutes? (Hint: See Example 3.19\()\)

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