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91Ó°ÊÓ

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) \(480,630,630,740,590,510\)

Short Answer

Expert verified
Option (c) is not a possible bootstrap sample, as 740 is not in the original data set.

Step by step solution

01

Understand Bootstrap Sampling

A bootstrap sample is created by randomly selecting values from the original sample with replacement. The original sample has the values: 480, 510, 590, 670, 520, and 630.
02

Examine Each Option for Validity

Check each proposed sample to see if it can be constructed from the original sample by selecting with replacement. - (a) Only uses 480, which is in the original sample. It is possible through replacement. - (b) Uses only 480 and 670, both are in the sample, and it is possible through replacement. - (c) Includes 740, which is not in the original sample.
03

Identify the Non-Bootstrap Sample

Since a bootstrap sample must exclusively be drawn from the original set of values, check which of the proposed samples contains a value not present in the original: option (c) contains 740, which is not part of the original sample, so it cannot be a bootstrap sample.

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Key Concepts

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

Random Sample
Choosing a random sample means selecting individuals from a larger group, ensuring each individual has an equal chance to be chosen. This technique is crucial in statistics because it helps gather unbiased data to make generalizations about a larger population.
A random sample is a fundamental concept used to minimize bias. Imagine wanting to understand the verbal GRE scores of freshmen at a university: selecting a random sample ensures you get a diverse representation, rather than selecting friends or just the top students.
  • Random sampling preserves the integrity of data collection.
  • It promotes fairness and accuracy in drawing conclusions.
  • Ensures all individuals have an equal opportunity to be selected.
Hence, relying on random samples is crucial for any accurate statistical inference.
Replacement
When performing bootstrap sampling, employing replacement is essential. Replacement means that each individual selected from the sample is placed back into the pool of potential selections before choosing another individual. This contrasts with sampling without replacement, where once an item is picked, it cannot be selected again.
This method allows the same data point to be chosen multiple times, capturing the variability in the population. In the example given, the repeated choice of 480 in option (a) is possible because of replacement. Without replacement, each choice could only be selected once, altering the dynamic of sampling significantly.
  • Replacement ensures each sample has the potential to be similar to the original distribution.
  • It allows for more flexibility and realism in samples.
  • Critical for simulations that require diversity mirroring real distribution variability.
Understanding replacement is key to grasping pivot models like the bootstrap method.
Statistical Analysis
Statistical analysis involves collecting, examining, and interpreting data to uncover patterns and trends. Bootstrap sampling, as a part of statistical analysis, provides insight when carrying out hypothesis testing or determining the accuracy of sample estimates.
The strength of bootstrap lies in its simplicity and power. By repeatedly sampling with replacement, it helps estimate confidence intervals and allows for robust analysis without complex assumptions. For instance, assessing the reliability of verbal GRE scores can guide decision-making for educational strategies.
  • Enhances validity by reinforcing results with multiple sample observations.
  • Reduces over-reliance on assumptions that might not hold true for complex data sets.
  • Aids in making informed decisions by providing a more empirical, data-driven foundation.
With statistical analysis, especially via bootstrap methods, conclusions can be drawn more confidently, si​nce they reflect repeated sampling and real-world variability.

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Most popular questions from this chapter

A 95\% bootstrap percentile confidence interval for the population mean (a) is centered about the original sample mean. (b) is centered about the mean of the bootstrap means. (c) uses the \(2.5\) and the \(97.5\) percentiles of the bootstrap distribution as the endpoints of the confidence interval.

Does taking notes by hand in a statistics course improve performance? Some recent research suggests that this may be the case. 2 To explore this, six volunteers (Doug, Elizabeth, Oksana, Sebastian, Vishal, and Xinyi) agree to take part in an experiment. Four are assigned completely at random to take handwritten notes in class, and the other two are assigned to take notes on their laptops. Total points earned on the two in-class exams and final exam are used to determine course performance. The results are (out of a possible total of 500 points): $$ \begin{array}{ll} \hline \text { Handwritten Notes (Person) } & \text { Notes on Laptop (Person) } \\ \hline 380 \text { (Doug) } & 370 \text { (Elizabeth) } \\ \hline 400 \text { (Oksana) } & 310 \text { (Xinyi) } \\ \hline 420 \text { (Sebastian) } & \\ \hline 360 \text { (Vishal) } & \\ \hline \end{array} $$ (a) There are 15 possible ways the six subjects can be assigned to the two groups, with the handwritten notes group having size 4 and the laptop notes group size 2. List these. (b) For each, determine the difference in mean points (mean number of points for the handwritten notes group minus mean number of points for the laptop notes group). Combine any duplicates and make a table of the possible mean differences and the corresponding probability of each under the null hypothesis of no difference in the effect of the treatments on total points earned. (Each of the 15 possible assignments of subjects to treatments has probability \(1 / 15\) under the null hypothesis.) This is the permutation distribution. (c) Compute the \(P\)-value of the data. Assume the two-sided alternative hypothesis is that the mean number of points is different for the two groups. (d) In this example, is it possible to demonstrate significance at the \(5 \%\) level using the permutation test? Explain. (e) Assume that total number of points is Normally distributed for both groups. Use the two-sample \(t\) procedure to test the hypotheses. Use Option 1 if you have access to software.

The composition of the earth's atmosphere may have changed over time. To try to discover the nature of the atmosphere long ago, we can examine the gas in bubbles inside ancient amber. Amber is tree resin that has hardened and been trapped in rocks. The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on specimens of amber from the late Cretaceous era (75 million to 95 million years ago) give these percentages of nitrogen: \({ }^{15} _{\text { }}\) $$ \begin{array}{lllllllll} 63.4 & 65.0 & 64.4 & 63.3 & 54.8 & 64.5 & 60.8 & 49.1 & 51.0 \end{array} $$ Assume (this is not yet agreed on by experts) that these observations are an SRS from the late Cretaceous atmosphere. (a) Construct a 95\% bootstrap confidence interval for the mean percentage of nitrogen in ancient air (the population). (b) We wonder if ancient air differs significantly from the present atmosphere, which is \(78.1 \%\) nitrogen. Based on your confidence interval in part (a), what do you conclude?

We plan to use the bootstrap method to construct a confidence interval for a population median from a sample of 43 subjects from the population. An important assumption for using the bootstrap method is (a) the sample is a random sample from the population. (b) the sampling distribution for the sample median must not be well approximated by the Normal distribution. (c) there are no outliers in the sample.

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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