1. Predict your reaction time.
How long do you think you would wait before you honked? (in seconds)
Section 8.2 Investigation 2.2: Honking Reaction Times
Exercises 8.2.1 The Study
Diekmann, Krassnig, and Lorenz (1996) conducted a field study to explore whether driver characteristics are related to an aggressive response (Thanks to Jeff Sklar for pointing us to this article).
The study was conducted at a busy intersection in Munich, West Germany, on two afternoons (Sunday and Monday) in 1986.The experimenters sat in a Volkswagen Jetta (the "blocking car") and did not accelerate after the traffic light turned green, and timed how long before the driver of the blocked car reacted (either by honking or flashing headlights). The response time (in seconds) is our variable of interest. Some values were "censored" in that the researcher stopped timing before the driver actually honked. This can happen if there is a time limit to the observation period and "success" has not been observed within that time period.
The data for the above study can be found in honking.txt.
2. Explore honking reaction time data.
Aside: Descriptive Statistics Applet.
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Use technology to create a histogram and describe the behavior of the data – shape, center, spread, outliers (suggest an explanation?).
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Then overlay a normal probability model. Do these data behave like a normal distribution? If not, how do they deviate from normality? Does the shape make sense in this context?
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Also examine a normal probability plot and discuss how deviations from the line correspond to the nonnormal shape you are observing.
Hint.
Were the observed response times/quantiles placed on the vertical or the horizontal axis?
Solution.
The distribution of response times are skewed to the right. Most people waited less than 5 seconds, but a couple waited more than 15 seconds. There is one outlier 17.15 seconds.
Example output:
The data do not behave like a normal distribution. The probability plot confirms that the largest response times are larger than we would expect from a normal distribution and the lowest response times are larger than we would expect.
3. Compare mean and median.
Determine and compare the mean and median wait times.
Hint.
Recall that the mean is the numerical average and the median has 50% of the observations on each side.
Mean: seconds
Median: seconds
Solution.
The mean (4.249 s) is larger than the median (3.24 s) as it is pulled towards the larger values. Also note that the skewness statistic equals 2.52.
4. Which is larger.
Which is larger?
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Mean is larger
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Median is larger
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They are equal
Definition: Resistant Statistic.
A numerical summary (statistic) is said to be resistant when it is not strongly influenced by a change in one or two extreme data values.
When data are skewed to the right/left, the mean will be pulled in the direction of the longer tail.
When data are skewed, we might often prefer to report the median as a "typical" value in the data set, rather than the mean because the mean is pulled in the direction of the longer tail. In addition, you might not want to cite the standard deviation as a measure of spread in a skewed distribution.
Definition: Interquartile Range (IQR).
Interquartile range (IQR) = upper quartile – lower quartile
The lower quartile is a value such that roughly one-fourth of all the observations fall below it; the upper quartile is a value such that roughly one-fourth of all the observations fall above it. The IQR then measures the width of the interval containing the middle 50% of the observations.
5. Compute interquartile range.
Use technology to compute the interquartile range.
Hint.
Run the descriptive statistics command and report the lower and upper quartiles and then subtract.
IQR: seconds
Solution.
The interquartile range is 4.835 - 2.590 (Applet) gives 2.245 seconds. The width of the middle 50% of wait times is 2.245 seconds.
6. IQR as resistant measure.
Would you consider the IQR a resistant measure of spread?
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Yes
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No
Solution.
Like the median, the IQR should be resistant.
When the data are skewed, the median and interquartile range are often considered better numerical summaries of the center and variability of the distribution than the mean and standard deviation. When working with the median and interquartile range, we often report the five number summary which consists of the minimum, lower quartile, median, upper quartile, and maximum values.
Definition: Boxplot.
Another graph is based on the five-number summary, called a boxplot (invented by John Tukey in 1970). The box extends from the lower quartile to the upper quartile with a vertical line inside the box at the location of the median. Whiskers then typically extend to the min and max values.
7. Create boxplot by hand.
Create a boxplot for these data by hand. Which display do you prefer, the boxplot or the histogram? Why?
Solution.
Example boxplot:
Preference will vary, but we do lose a fair bit of information in the boxplot.
Although boxplots are a nice visual of the five-number summary, they can sometimes miss interesting features in a data set. In particular, shape can be more difficult to judge in a boxplot, and they usually don’t provide information about the size of a data set.
Another application of the inter-quartile range is as a way to measure how far an observation is from the bulk of the distribution.
Definition: Outlier Criterion and Modified Boxplot.
A value is an outlier according to the 1.5IQR criterion if the value is larger than the upper quartile + 1.5 × box length or smaller than the lower quartile – 1.5 × box length.
Note: The box length = upper quartile – lower quartile, is the interquartile range.
A modified boxplot will display such outliers separately and then extend the whiskers to the most extreme non-outlier observation.
8. Technology Detour — Modified Boxplots.
Create a "modified" boxplot for the honking response time data.
Hint 1. Applet Instructions
Use the Descriptive Statistics applet:
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Paste in the data.
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Check the Add Boxplot box.
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Check the Show outliers box.
Expected output:
Hint 2. R Instructions
Use the
boxplot or iscamboxplot function:
boxplot(responsetime, ylab="time until reaction", # Adds labels
horizontal=TRUE) # Makes horizontal
# OR
iscamboxplot(responsetime, xlab="time until reaction") # Uses quartiles
Expected output:
Try it yourself:
Hint 3. JMP Instructions
In the Distributions window, use the red triangle (hot spot) next to the variable name and select Outlier Boxplot.
Expected output:
Solution.
A modified boxplot displays the five-number summary with whiskers extending to the most extreme non-outlier observations. Points beyond 1.5 × IQR from the quartiles are marked as outliers.
9. Counting outliers.
Are there any outliers according to these criteria?
Solution.
Applet/R/JMP identify four ("high") outliers.
These data are not well modelled by a normal distribution. So can we still make predictions? There are a couple of strategies. One would be to consider whether a rescaling or transformation of the data might create a more normal-looking distribution, allowing us to use the methods from Investigation 2.1: Birth Weights. In this case, we need a transformation that will downsize the large values more than the small values. Log transformations are often very helpful in this regard.
Definition: Data Transformation.
A data transformation applies a mathematical function to each value to re-express the data on an alternative scale. Data transformations can also make the data more closely modeled with a normal distribution, which could then satisfy the conditions the Central Limit Theorem.
Create a new variable which is log(responsetime). (You can use either natural log or log base 10, but so we all do the same thing, let’s use natural log here, which is the default in most software when you say "log.")
10. Technology Detour — Log Transformation.
Create a log transformation of the response time data.
Hint 1. R Instructions
lnresponsetime = log(responsetime)
Hint 2. JMP Instructions
Create a new column and edit the formula. Type or use your mouse to select Transcendental > Log to create
Log(responsetime). Press OK.
11. Log transformation analysis.
Create a histogram of these data and a normal probability plot. Does log(responsetime) approximately follow a normal distribution? What are the mean and standard deviation of this distribution?
Solution.
This histogram of the ln response times is more symmetric and the normal probability plot is more linear. This is definitely more normal looking than before (but we might be able to do better). The mean is 1.292 and the SD = 0.5285. The skewness statistic is closer to zero (0.676).
12. Predict using normal distribution on log scale.
Aside: Normal Probability Calculator.
Use a normal distribution with mean 1.29 ln-sec and standard deviation 0.53 ln-seconds for the logged response times and predict how often someone will honk within the first 2 seconds.
Hint.
What are you going to use for the observation of interest?
Solution.
Using ln(2) ≈ 0.693 as the observed result, we find about 13% of the model distribution falls below.
13. Compare prediction to observed data.
How does this prediction in Question 12 compare to the observed percentage honking within the first 2 seconds in the data set?
Solution.
In the original distribution, 6 of the 57 observations are below 2 (10.5%).
Alternative Probability Models.
Another approach is to fit a different mathematical model to the original data. Use technology to overlay an exponential probability model (often used to model wait times) to these data and/or create a probability plot using the exponential distribution as the reference distribution.
14. Technology Detour — Exponential Distribution.
Overlay an exponential probability model to the honking response time data.
Hint 1. R Instructions
For the qqplot we have to first get the quantiles:
theoquant = qexp(ppoints(12)) # Generates 1/n quantiles for 12 observations
# from exponential distribution
hist(theoquant)
qqplot(responsetime, theoquant) # Your data vs. quantiles. Look for a line.
iscamaddexp(responsetime) # overlay exponential model
Hint 2. JMP Instructions
In the Distribution window, use the hot spot to select Continuous Fit > Exponential.
15. Exponential probability model.
Describe the behavior of the exponential distribution. Does it appear to be a reasonable fit for these data? Describe any deviations.
Solution.
The exponential is skewed to the right with its peak at zero. This does not appear to be as good of a fit. We have more observations around 4-5 seconds than the exponential distribution would predict and fewer observations less than 1 second (we don’t have any, the times could be rounded in the dataset?).
16. Technology Detour — Exponential Probability Calculation.
Use technology to calculate the probability of a wait time under 2 seconds using the exponential distribution with mean 4.25 sec
Hint 1. R Instructions
pexp(2, rate = 1/4.25) calculates P(X < 2) when X ~ Exp(mean = 4.25)
Try it yourself:
Hint 2. JMP Instructions
Using the Distribution Calculator and select the Exponential distribution. Specify a scale parameter of 1/4.25. Choose X <= Qa and enter 2 for Qa (we don’t need to worry about strict vs. non-strict inequalities here)
17. Compare exponential probability to previous estimates.
How does it compare to your estimate in Question 12 and the result in Question 13?
Solution.
This way overpredicts how often we will get a response time less than 2 seconds (0.3754).
18. Lognormal distribution.
Repeat Question 12 and Question 13 with a "lognormal" distribution.
Note: This is equivalent to fitting the normal distribution on the log-transformed data!
Hint 1. In R
qqplot(responsetime, qlnorm(ppoints(12)))
iscamaddlnorm(responsetime)
plnorm(2, meanlog = 1.292, sdlog = 0.5238)
Hint 2. In JMP
Scroll down the Distribution list and choose the Lognormal distribution with location = 1.292 and scale = 0.5238.
Solution.
The log normal has the skewed right behavior but not as severely as the exponential and provides a better fit to our data. Now the prediction is 0.1326.
Discussion.
There are of course, many other probability models we could look into. One limitation of the exponential distribution is that it assumes the same value for the mean and the standard deviation, clearly not the case for these data. There are other more flexible distributions (e.g., Gamma and Weibull) that use two parameters to characterize the distribution rather than only one.
19. Generalization of results.
To what population are you willing to generalize these results? Explain.
Hint.
Think about how the data were collected and what you learned about sampling methods in Chapter 1.
Solution.
We don’t have a lot of information about how these participants were selected other than two afternoons in West Germany, so we should be cautious in generalizing these results too far beyond that time period and country. "Waiting time" can be very distinct across cultures.
20. Future research directions.
If you were to continue to explore this research area/data set, what would you be interested in investigating next?
Solution.
Answers will vary.
Study Conclusions.
In this study, we found that the amount of time a blocked driver waits before responding follows a skewed right distribution with a mean of 4.25 seconds and a median of 3.24 seconds, with a few drivers waiting more than 10 seconds. Although the dataset is small, we might consider using these data to build a mathematical model for these skewed data, either using a transformation or a probability model other than the normal distribution, to help predict future results (e.g., a wait time of less than 2 seconds). These researchers were actually interested in whether the "social status" of the blocked car was related to how long it took before people honked. They found that "the mean and median response times decreased monotonically with the status of car, except when the blocked car was very small." Similar results had been found in an earlier study by Doob and Gross in the United States ( 1968) which varied the status of the blocking car. However, neither study was replicated by a Swiss study (Jann, Suhner, and Marioni, 1995), perhaps due to cultural differences.
Subsection 8.2.2 Practice Problem 2.2A
A group of Cal Poly students (Sasscer, Mease, Tanenbaum, and Hansen, 2009) conducted the following study: volunteers were to say "go," and then to say "stop" when they believed 30 seconds had passed. The researchers asked the participants to not count in their heads and recorded how much time had actually passed. The data are in 30seconds.txt.
Checkpoint 8.2.1. Examine distribution of 30-second estimates.
Examine graphical and numerical summaries of these data. Describe the shape of the distribution. How do the mean and median compare? Is the skewness statistic positive or negative? What does this tell you about whether people tend to over or underestimate the length of 30 seconds?
Checkpoint 8.2.2. Assess normal model appropriateness.
Does a normal model seem appropriate here?
Checkpoint 8.2.3. Evaluate log transformation.
Does a log transformation succeed in creating a normal distribution? Explain how you might have predicted this answer based on the first graph you looked at in Checkpoint 8.2.1.
Checkpoint 8.2.4. Generalization of results.
To what population would you be willing to generalize these data?
Subsection 8.2.3 Practice Problem 2.2B
Checkpoint 8.2.5. Compare Weibull and Gamma fits.
Below are distribution fits for the honking data with a Weibull distribution and a Gamma distribution.
Weibull
Gamma
Which do you think indicates a better fit for these data? Explain how you are deciding.
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