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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 \(H_{a}: \mu_{1}>\mu_{2}\) using the sample results \(\bar{x}_{1}=56, s_{1}=8.2\) with \(n_{1}=30\) and \(\bar{x}_{2}=51, s_{2}=6.9\) with \(n_{2}=40\).

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 30 successes with \(n=100\) and Sample \(\mathrm{B}\) has a count of 50 successes with \(n=250\).

Find a \(95 \%\) confidence interval for the mean two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the t-distribution and the formula for standard error. Compare the results. Mean price of a used Mustang car online, in \$1000s, using data in MustangPrice with \(\bar{x}=15.98\), \(s=11.11,\) and \(n=25\)

Using the dataset NutritionStudy, we calculate that the average number of grams of fat consumed in a day for the sample of \(n=315\) US adults in the study is \(\bar{x}=77.03\) grams with \(s=33.83\) grams. (a) Find and interpret a \(95 \%\) confidence interval for the average number of fat grams consumed per day by US adults. (b) What is the margin of error? (c) If we want a margin of error of only ±1 , what sample size is needed?

Quiz Timing A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-Sample T-Test and Cl \begin{tabular}{lrrrr} Sample & \(\mathrm{N}\) & Mean & StDev & SE Mean \\ Late & 32 & 22.56 & 5.13 & 0.91 \\ Early & 30 & 19.73 & 6.61 & 1.2 \\ \multicolumn{3}{c} { Difference } & \(=\mathrm{mu}(\) Late \()\) & \(-\mathrm{mu}\) (Early) \end{tabular} Estimate for difference: 2.83 $$ \begin{aligned} &95 \% \mathrm{Cl} \text { for difference: }(-0.20,5.86)\\\ &\text { T-Test of difference }=0(\text { vs } \operatorname{not}=): \text { T-Value }=1.87\\\ &\text { P-Value }=0.066 \quad \mathrm{DF}=54 \end{aligned} $$

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