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In Exercises 6.99 to \(6.102,\) use StatKey or other technology to generate a bootstrap distribution of sample means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviation as an estimate of the population standard deviation. Mean number of penalty minutes for NHL players using the data in OttawaSenators with \(n=\) \(24, \bar{x}=34.13,\) and \(s=27.26\)

Short Answer

Expert verified
After carrying out the bootstrap distribution and corresponding standard error calculation, compare the value to the standard error calculated from the Central Limit Theorem (\(\frac{s}{\sqrt{n}}\)). The results should closely align, verifying the estimations provided by the Central Limit Theorem for a given population sample.

Step by step solution

01

Bootstrap Distribution

Generate a bootstrap distribution of sample means using StatKey or another technology. This involves repeatedly resampling the original data of the mean number of penalty minutes for NHL players (with a size of \(n = 24, \bar{x}=34.13,\) and \(s=27.26\)) and calculating the sample mean for each subset. You will then get a distribution of these sample means, known as the bootstrap distribution.
02

Standard Error for Bootstrap Distribution

Find the standard error for the bootstrap distribution you generated in Step 1. The standard error is the standard deviation of the sample means. It can be calculated by numerically determining the standard deviation of the sample means obtained from the bootstrap distribution.
03

Central Limit Theorem's Standard Error

Calculate the standard error using the Central Limit Theorem (CLT). According to the CLT, the standard error of a sample mean is \(\frac{s}{\sqrt{n}}\), where `s` is the sample standard deviation (in this case, \(27.26\)), and `n` is the number of samples (in this case, \(24\)).
04

Comparison

Compare the results from Step 2 and Step 3. Our goal here is to see how the result of the bootstrap method corresponds to the theoretical calculation given by the Central Limit Theorem.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics, stating that the distribution of sample means tends to be normal, or bell-shaped, as the sample size becomes large, regardless of the shape of the population distribution. By large, we usually mean a sample size over 30.

In essence, the CLT allows us to make predictions about the behavior of sample means. If you take a large enough sample from a population, the mean of that sample is likely to be close to the mean of the entire population. Furthermore, if you were to take many samples, the means of those samples would form their own distribution — the sampling distribution of the mean.

This concept is crucial for performing hypothesis tests and creating confidence intervals. It allows researchers to infer about populations based on sample data, a standard practice in fields like economics, psychology, and biology. The CLT assures us that even with a sample, we can gain reliable insights into the broader population.
Standard Error
Standard error is a measure of the amount of variability or dispersion in a set of related sample means. It is essentially the standard deviation of the sampling distribution of the sample mean. Put another way, standard error gives us an idea of how far the sample mean is likely to be from the actual population mean.

Why is this important? The smaller the standard error, the more representative the sample mean is likely to be of the population mean. When you're evaluating statistical data, a smaller standard error means you can have more confidence in your sample estimates. It's a crucial concept when dealing with the precision of statistical estimates, and for interpreting results within the context of margin of error in surveys, experiments, and observational studies.
Resampling
Resampling is a statistical method that involves drawing repeated samples from the same sample data. It is used to assess the variability of a statistic (like the mean or median) without needing the data from the entire population.

One common resampling technique is bootstrapping. It involves taking many samples, often thousands, from a single sample dataset, with replacement. Each of these 'bootstrap samples' is the same size as the original dataset, and each one is used to calculate a statistic, such as the mean or standard deviation.

Through resampling, you can create a bootstrap distribution of the statistic and estimate properties like its standard error or build confidence intervals. Resampling is particularly powerful because it does not rely on the normality of the data and can be applied to small sample sizes.
Sample Mean
The sample mean, often denoted as \( \bar{x} \), is the arithmetic average of a set of sample values. It is a central concept in statistics used to estimate the central tendency of a population. Calculating the sample mean is straightforward: sum all the values in the sample and divide by the number of observations.

While the sample mean is a useful estimator, it does have variability when considered as a random variable. Therefore, its precision and reliability can be described using the standard error. Understanding the behavior of the sample mean under repeated sampling is a core aspect of statistical inference, as it allows us to generalize findings from a sample to the larger population from which it was drawn.

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Most popular questions from this chapter

Use the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal. Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the sample results \(\bar{x}_{1}=15.3, s_{1}=11.6\) with \(n_{1}=100\) and \(\bar{x}_{2}=18.4, s_{2}=14.3\) with \(n_{2}=80\).

What Gives a Small P-value? In each case below, two sets of data are given for a two-tail difference in means test. In each case, which version gives a smaller \(\mathrm{p}\) -value relative to the other? (a) Both options have the same standard deviations and same sample sizes but: Option 1 has: \(\quad \bar{x}_{1}=25 \quad \bar{x}_{2}=23\) $$ \text { Option } 2 \text { has: } \quad \bar{x}_{1}=25 \quad \bar{x}_{2}=11 $$ (b) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same sample sizes but: Option 1 has: \(\quad s_{1}=15 \quad s_{2}=14\) $$ \text { Option } 2 \text { has: } \quad s_{1}=3 \quad s_{2}=4 $$ (c) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same standard deviations but: Option 1 has: \(\quad n_{1}=800 \quad n_{2}=1000\) $$ \text { Option } 2 \text { has: } \quad n_{1}=25 \quad n_{2}=30 $$

(a) Find the relevant sample proportions in each group and the pooled proportion. (b) Complete the hypothesis test using the normal distribution and show all details. Test whether males are less likely than females to support a ballot initiative, if \(24 \%\) of a random sample of 50 males plan to vote yes on the initiative and \(32 \%\) of a random sample of 50 females plan to vote yes.

Dark Chocolate for Good Health A study \(^{47}\) examines chocolate's effects on blood vessel function in healthy people. In the randomized, doubleblind, placebo-controlled study, 11 people received 46 grams (1.6 ounces) of dark chocolate (which is naturally flavonoid-rich) every day for two weeks, while a control group of 10 people received a placebo consisting of dark chocolate with low flavonoid content. Participants had their vascular health measured (by means of flow-mediated dilation) before and after the two-week study. The increase over the two-week period was measured, with larger numbers indicating greater vascular health. For the group getting the good dark chocolate, the mean increase was 1.3 with a standard deviation of \(2.32,\) while the control group had a mean change of -0.96 with a standard deviation of 1.58 . (a) Explain what "randomized, double-blind, placebo-controlled study" means. (b) Find and interpret a \(95 \%\) confidence interval for the difference in means between the two groups. Be sure to clearly define the parameters you are estimating. You may assume that neither sample shows significant departures from normality. (c) Is it plausible that there is "no difference" between the two kinds of chocolate? Justify your answer using the confidence interval found in \(\operatorname{part}(\mathrm{b})\)

Use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

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