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Chapter 6: Problem 123

In Exercises 6.123 and \(6.124,\) 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 distance of a commute for a worker in Atlanta, using data in CommuteAtlanta with \(\bar{x}=18.156\) miles, \(s=13.798,\) and \(n=500\)

Short Answer

Expert verified
The 95% confidence interval for the mean commute distance in Atlanta will be calculated using both the bootstrap and t-distribution methods. Final comparison might reveal some differences due to the variation in calculation methods used.

Step by step solution

01

Bootstrap Method

Bootstrap method will require using a software like StatKey or other statistical software. In that software, one will enter the sample mean \(\bar{x}=18.156\) miles, the standard deviation \(s=13.798,\) and the sample size \(n=500\). The software will then use random sampling to generate the bootstrap distribution, and from it, the percentiles for the 95% confidence interval.
02

t-distribution Method

First, calculate the standard error (SE) using the formula \[SE = \frac{s}{\sqrt{n}}\]. Here, s is the standard deviation (= 13.798) and n is the sample size (= 500). After calculating the SE, find the t-score for a 95% confidence interval from the t-distribution table. Typically, for large sample sizes, it's approximately 1.96. The confidence interval is hence given by \[\bar{x} \pm t \times SE\]. Calculate the lower and upper limits using the calculated SE, given mean (18.156) and the t-score from the table.
03

Comparison of Results

After getting the confidence intervals from both methods, compare them. They might be slightly different due to the methods of calculation, but generally, they should be reasonably close, especially for large sample sizes.

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Key Concepts

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

Understanding the Bootstrap Method
The bootstrap method is a powerful statistical tool used to estimate the distribution of a statistic by sampling with replacement from the data set. This technique allows us to model the sampling distribution of almost any statistic by using just the original data without making strong assumptions about the population.

Specifically, when calculating a confidence interval for a mean, as in the given exercise, the bootstrap method works by creating thousands of new sample means (also known as bootstrap samples). This is done by randomly drawing, with replacement, the same number of observations as the original sample. For each of these bootstrap samples, we calculate its mean. After conducting a large number of simulations, we construct a bootstrap distribution of the sample means.

A 95% confidence interval can then be drawn from this distribution directly by finding the percentile values that cut off the lowest 2.5% and the highest 2.5% of the means, ensuring that the middle 95% of the bootstrap sample means will lie within this interval. The beauty of this method is its simplicity and its applicability to complex sampling distributions that may not fit classic theoretical distributions like the t-distribution.
t-Distribution: Key Features and Usage
The t-distribution is a probability distribution that is symmetric and bell-shaped, like the normal distribution, but has heavier tails. This makes it suitable for estimating population parameters when the sample size is small and the population variance is unknown.

The t-distribution becomes closer to the normal distribution as the sample size increases. In the context of confidence intervals, the t-distribution is used to calculate how much the sample mean is likely to vary from the true population mean. The t-score (similar to the z-score in the normal distribution) is a critical value that reflects the desired level of confidence and degrees of freedom (which is usually the sample size minus 1).

When creating a 95% confidence interval, as shown in the problem, the t-distribution is used to find the appropriate t-score that, when multiplied by the standard error, will give the range within which the true mean is likely to be found 95% of the time. This step is crucial, particularly when dealing with smaller sample sizes where the normal approximation may not be applicable.
Standard Error: Measuring the Precision of the Sample Mean
The standard error (SE) is a measure of how much variability exists within a sample's estimate of a population mean. Mathematically, it's the standard deviation of the sample's mean distribution, which tells us how precise our sample mean is likely to be in estimating the true population mean.

In the provided exercise, the standard error is calculated using the formula \(SE = \frac{s}{\sqrt{n}}\), where \(s\) represents the sample standard deviation and \(n\) represents the sample size. This equation dilutes the influence of the sample variance by the square root of the sample size number, reflecting that larger samples tend to yield more precise estimates of the population mean.

This concept of standard error is essential when calculating confidence intervals using both the bootstrap method and the t-distribution. The standard error functions as a scaling factor: it adjusts the range of the confidence interval to account for the size of the sample and its variability, providing the bounds within which the true mean is expected to reside with a given level of confidence.

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Most popular questions from this chapter

Is a Normal Distribution Appropriate? In Exercises 6.13 and \(6.14,\) indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, does the Central Limit Theorem apply? (a) \(n=500\) and \(p=0.1\) (b) \(n=25\) and \(p=0.5\) (c) \(n=30\) and \(p=0.2\) (d) \(n=100\) and \(p=0.92\)

Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting cancer of any kind? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

A sample with \(n=10, \bar{x}=508.5,\) and \(s=21.5\)

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 300 from Population 1 with mean 75 and standard deviation 18 and samples of size 500 from Population 2 with mean 83 and standard deviation 22

Exercise 4.86 on page 263 introduces a matched pairs study in which 47 participants had cell phones put on their ears and then had their brain glucose metabolism (a measure of brain activity) measured under two conditions: with one cell phone turned on for 50 minutes (the "on" condition) and with both cell phones off (the "off" condition). Brain glucose metabolism is measured in \(\mu \mathrm{mol} / 100 \mathrm{~g}\) per minute, and the differences of the metabolism rate in the on condition minus the metabolism rate in the off condition were computed for all participants. The mean of the differences was 2.4 with a standard deviation of \(6.3 .\) Find and interpret a \(95 \%\) confidence interval for the effect size of the cell phone waves on mean brain metabolism rate.
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