Question

Better Traffic Flow Exercise 2.155 on page 105 introduces the dataset TrafficFlow, which gives delay time in seconds for 24 simulation runs in Dresden, Germany, comparing the current timed traffic light system on each run to a proposed flexible traffic light system in which lights communicate traffic flow information to neighboring lights. On average, public transportation was delayed 105 seconds under the timed system and 44 seconds under the flexible system. Since this is a matched pairs experiment, we are interested in the difference in times between the two methods for each of the 24 simulations. For the \(n=24\) differences \(D\), we saw in Exercise 2.155 that \(\bar{x}_{D}=61\) seconds with \(s_{D}=15.19\) seconds. We wish to estimate the average time savings for public transportation on this stretch of road if the city of Dresden moves to the new system. (a) What parameter are we estimating? Give correct notation. (b) Suppose that we write the 24 differences on 24 slips of paper. Describe how to physically use the paper slips to create a bootstrap sample. (c) What statistic do we record for this one bootstrap sample? (d) If we create a bootstrap distribution using many of these bootstrap statistics, what shape do we expect it to have and where do we expect it to be centered? (e) How can we use the values in the bootstrap distribution to find the standard error? (f) The standard error is 3.1 for one set of 10,000 bootstrap samples. Find and interpret a \(95 \%\) confidence interval for the average time savings.

Short answer

The parameter we are estimating is \(\mu_{D}\) - the time saving in public transportation delay if the city of Dresden switched to the flexible system. To create a bootstrap sample, we repeatedly randomly select samples (with replacement) from our original 24 differences. We record the mean of each bootstrap sample, and expect the bootstrap distribution to be centered at 61, with a standard deviation (standard error) of 3.1. Using these, a 95% confidence interval is 54.92 to 67.08 seconds.

Step-by-step solution

Identify the Parameter

The parameter estimated in this exercise is the average time difference savings \(\mu_{D}\) in transport delay for the general population of stretches if the flexible system were to replace the timed system. The notation for this parameter is \(\mu_{D}\).

Create a Bootstrap Sample

To create a bootstrap sample, put all your 24 slips of paper (each with a different time difference) into a hat. Then, draw slips out one at a time, writing down the number on the slip each time, but replacing the slip back in the hat after each draw. Repeating this 24 times will give you a bootstrap sample. Remember each draw is independent and each number has the same probability of being drawn.

Determine the Bootstrap Statistic

For each bootstrap sample created, calculate and record the mean. The mean of the bootstrap sample is the bootstrap statistic that we are interested in.

Look at Bootstrap Distribution

The bootstrap distribution of the means should be approximately normally distributed due to the Central Limit Theorem, as long as the number of bootstrap samples taken is sufficiently large. The mean of this distribution will be around the original sample mean, which is 61 seconds in this example.

Calculate the Standard Error

The standard error of the bootstrap distribution is simply the standard deviation of the means of the bootstrap samples. It measures the variability of the sample mean estimate.

Calculate and Interpret the Confidence Interval

With a standard error of 3.1, we can calculate a 95% confidence interval using the formula: \(\bar{x} \pm (1.96 * SE) = 61 \pm (1.96 * 3.1) = [54.92, 67.08]\). This means we can be 95% confident that the true average time savings in delaying transport, if the flexible system is utilized, falls between 54.92 and 67.08 seconds.