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Chapter 32: Problem 9

How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and \(1.5\) inches square? Here are data from students doing a laboratory exercise: $$ \begin{array}{lllll} 33,190 & 31,860 & 32,590 & 26,520 & 33,280 \\ 32,320 & 33,020 & 32,030 & 30,460 & 32,700 \\ 23,040 & 30,930 & 32,720 & 33,650 & 32,340 \\ 24,050 & 30,170 & 31,300 & 28,730 & 31,920 \end{array} $$ Generate 1000 bootstrap samples by resampling from these data. For each bootstrap sample, compute the bootstrap sample mean. (a) What are the mean and standard deviation of the 1000 bootstrap sample means? (b) Draw a histogram of the bootstrap distribution of the sample mean using the 1000 bootstrap sample means. If your software allows this, superimpose a normal curve on the histogram. Describe the shape of this bootstrap distribution.

Short Answer

Expert verified
(a) Calculate mean and standard deviation of bootstrap means. (b) Examine histogram and distribution shape.

Step by step solution

01

Understand the Problem

We have a list of load measurements from pulling apart pieces of Douglas fir. Our goal is to generate 1000 bootstrap samples, calculate the mean of each sample, and then summarize the mean and standard deviation of these means.
02

Collect and Parse the Data

The given data consist of 20 observations. Here they are in a list form for ease of computation: \[ [33190, 31860, 32590, 26520, 33280, 32320, 33020, 32030, 30460, 32700, 23040, 30930, 32720, 33650, 32340, 24050, 30170, 31300, 28730, 31920] \].
03

Generate Bootstrap Samples

Using a programming language or statistical software, we re-sample the data 1000 times, with replacement. Each sample should have the same size as the original dataset (20).
04

Compute Bootstrap Sample Means

For each of the 1000 bootstrap samples, calculate the mean. This results in a list of 1000 means.
05

Calculate Overall Mean and Standard Deviation

Compute the mean and standard deviation of the 1000 bootstrap sample means. This gives us an estimate of the central tendency and variability of the sample mean distribution.
06

Plot the Histogram

Create a histogram of the 1000 bootstrap sample means. This allows visualization of the distribution. Superimpose a normal curve if possible to compare the distribution to a normal distribution.
07

Analyze the Histogram

Assess the shape of the histogram. It might be helpful to note if the distribution appears normal, skewed, or has other characteristics.

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Key Concepts

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

Resampling Methods
Resampling methods are essential tools in statistics used to estimate the distribution of a sample by drawing repeated samples from the data. This approach helps in understanding variability within the sample data.
  • One popular resampling technique is the bootstrap method. It involves randomly drawing samples from the original data with replacement. This means each observation can appear multiple times in a single sample, replicating the process many times to create a distribution of statistics.
  • Bootstrap sampling is particularly useful when analytical methods to calculate these statistics are complex or not available.
  • It allows us to make inferences about the population without assuming any specific distribution and with the benefit of assessing statistical accuracy through empirical approaches.
This makes the bootstrap a valuable method for estimating the mean, standard deviation, and other statistics of a dataset.
Statistical Software
Using statistical software is crucial in handling large datasets and performing numerous computations efficiently in bootstrap sampling and other analytics.
  • Programs like R, Python (using libraries like pandas and numpy), and specialized software such as SAS or SPSS can be used to automate the process of resampling.
  • They allow users to implement bootstrap sampling by writing short scripts that automate sampling, calculation, and visualization tasks.
  • Statistical software also supports integration with other tools, facilitating seamless data import, manipulation, and analysis.
Utilizing statistical software not only saves time but also increases flexibility and accuracy in research.
Sample Mean Calculation
Calculating the sample mean is a fundamental statistical operation that helps in summarizing a dataset by determining its central tendency.
  • The sample mean, represented by \( \bar{x} \), is calculated using the formula:
    \[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \] where \( n \) is the number of observations, and \( x_i \) are the data points.
  • In the context of bootstrap samples, the sample mean is calculated for each sample distribution to provide an estimate of the central tendency based on resampled data.
  • The calculation of multiple sample means (e.g., from 1000 bootstrap samples) gives insights into the variability of the mean estimate, helping in constructing confidence intervals.
This analysis aids in understanding the expected range and possible variance of the data's central value.
Histogram Analysis
Histogram analysis is a graphical representation that helps in understanding the distribution of sample data visually. It allows us to analyze the shape, spread, and central tendencies of the dataset.
  • Histograms display data distribution by showing the frequency of data points across a range of values, helping to identify patterns like normality, skewness, and any potential outliers.
  • In the context of bootstrap samples, histograms of sample means can reveal how closely they approximate a normal distribution, allowing the visualization of the mean鈥檚 variability and distribution spread.
  • Overlaying a normal curve on the histogram can visually compare the empirical distribution derived from bootstrap samples to a theoretical normal distribution, assessing the goodness of fit.
Through visualization, histogram analysis provides a clearer understanding of statistical results, aiding in decision-making and hypothesis testing.

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Most popular questions from this chapter

We select a random sample of six freshman students from the University of California at Santa Cruz and find that their verbal GREs are 480, 510, 590, 670, 520 , and 630 . Which of the following is not a possible bootstrap sample? (a) \(480,480,480,480,480,480\) (b) \(480,480,480,670,670,670\) (c) \(480,630,630,740,590,510\)

Our bodies have a natural electrical field that is known to help wounds heal. Does changing the field strength slow healing? A series of experiments with newts investigated this question. In one experiment, the two hind limbs of four newts were assigned at random to either experimental or control groups. This is a matched pairs design. The electrical field in the experimental limbs was reduced to zero by applying a voltage. The control limbs were left alone. Here are the rates at which new cells closed a razor cut in each limb, in micrometers per hour: \({ }^{11}\) $$ \begin{array}{l|cccc} \hline \text { Newt } & 1 & 2 & 3 & 4 \\ \hline \text { Control limb } & 36 & 41 & 39 & 42 \\ \hline \text { Experimental limb } & 28 & 31 & 27 & 33 \\ \hline \end{array} $$ The number of possible random assignments of treatments to the different matched pairs is (a) 4 . (b) \(8 .\) (c) \(16 .\)

"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. Unfortunately, the treatment also reduces the strength of the fabric. A study compared the breaking strengths of fabrics treated by two commercial durable press processes. Five swatches of the same fabric were assigned at random to each process. Here are the data, in pounds of pull needed to tear the fabric: \({ }^{12}\) $$ \begin{array}{l|lllll} \hline \text { Permafresh } & 29.9 & 30.7 & 30.0 & 29.5 & 27.6 \\ \hline \text { Hylite } & 28.8 & 23.9 & 27.0 & 22.1 & 24.2 \\ \hline \end{array} $$ There is a mild outlier in the Permafresh group. Perhaps we should use a permutation test to test the hypothesis of no difference in median pounds of pull needed to tear the fabric. Assume a two-sided alternative and estimate the \(P\) value.

In a study of exhaust emissions from school buses, the pollution intake by passengers was determined for a sample of nine school buses used in the Southern California Air Basin. The pollution intake is the amount of exhaust emissions, in grams per person, that would be breathed in while travelilng on the bus during its usual 18-mile trip on congested freeways from South Central LA to a magnet school in West LA. (As a reference, the average intake of motor emissions of carbon monoxide in the LA area is estimated to be about \(0.000046\) gram per person.) Here are the amounts for the nine buses when driven with the windows open: \({ }^{17}\) $$ \begin{array}{lllllllll} 1.15 & 0.33 & 0.40 & 0.33 & 1.35 & 0.38 & 0.25 & 0.40 & 0.35 \end{array} $$ (a) Make a stemplot. Are there outliers or strong skewness that would forbid use of the \(t\) procedures? (b) Construct a \(95 \%\) bootstrap confidence interval for the mean pollution intake among all school buses used in the Southern California Air Basin that travel the route investigated in the study.

The changing climate will probably bring more rain to California, but we don't know whether the additional rain will come during the winter wet season or extend into the long dry season in spring and summer. Kenwyn Suttle of the University of California at Berkeley and his coworkers carried out a randomized controlled experiment to study the effects of more rain in either season. They randomly assigned 12 plots of open grassland to two treatments: added water equal to \(20 \%\) of annual rainfall during January to March (winter) or no added water (control). One response variable was total plant biomass, in grams per square meter, produced in a plot over a year. \({ }^{10}\) Here are data for 2004 (mass in grams per square meter): $$ \begin{array}{ll} \hline \text { Winter } & \text { Control } \\ \hline 254.6453 & 178.9988 \\ \hline 233.8155 & 205.5165 \\ \hline 253.4506 & 242.6795 \\ \hline 228.5882 & 231.7639 \\ \hline 158.6675 & 134.9847 \\ \hline 212.3232 & 212.4862 \\ \hline \end{array} $$ We wish to test whether there is a difference in mean biomass between the two treatment groups. Which of the following is true? (a) This is a randomized controlled experiment, hence a permutation test is more appropriate than a \(t\) test. (b) This is a randomized controlled experiment, and we should try both the permutation test and the \(t\) test and always report only the one with the smaller \(P\)-value. (c) We might prefer using a permutation test for these data rather than a \(t\) test, because the sample sizes are small and the data contain some possible outliers.
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