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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 \(n=25\)

Short Answer

Expert verified
The 95% confidence interval using the bootstrapping method will depend on the results from the bootstrapping software, while the 95% confidence interval using the t-distribution and standard error method is approximately (11.43, 20.53). If the bootstrap interval is similar to this, both methods have provided comparable results. Otherwise, this indicates some asymmetry or skewness in the bootstrap distribution.

Step by step solution

01

Bootstrapping Method

Using the bootstrap method means taking repeated samples from the observed dataset and computing the mean for each sample. By doing this, a distribution of means is created which can be used to find percentiles and thereby compute the 95% confidence interval. Bootstrapping uses a software, so the specific steps will depend on the software or technology being used. In general, the steps involve selecting the data, choosing a bootstrap sampling (with replacement), selecting the mean as the statistic, running the bootstrap for thousands of iterations, and finding the 2.5th and 97.5th percentiles of the bootstrap distribution.
02

T-distribution and Standard Error Method

First, the standard error (SE) is calculated using the formula \( SE = s / \sqrt{n} \) where s is the standard deviation and n is the sample size. In this case, \( SE = 11.11 / \sqrt{25} = 2.222 \). The t-distribution can then be used to find the critical t-value (t*) corresponding to a 95% confidence level with 24 degrees of freedom (n-1). Checking the t-table, t* is approximately 2.064. The confidence interval is then given by \( \bar{x} \pm t* * SE = 15.98 \pm 2.064 * 2.222 \). Therefore, the 95% confidence interval for the mean using the t-distribution and standard error is approximately (11.43, 20.53).
03

Compare the Results

Compare the 95% confidence intervals obtained from the bootstrapping and t-distribution methods. Consider whether they are approximately equal, which would suggest that the data follow a normal distribution and both methods yield similar results.

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Key Concepts

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

Bootstrapping
Bootstrapping is a modern statistical method that uses resampling to approximate the sampling distribution of a statistic. It is a flexible and powerful technique, particularly useful when traditional assumptions about the population distribution are not met. The essence of bootstrapping lies in using the available data to make inferences about the population without relying heavily on parametric assumptions.

Here's how bootstrapping works:
  • Resampling: Random samples are drawn from the original dataset with replacement, meaning the same data points can be chosen multiple times in each iteration.
  • Sample Statistic: After each sampling, the desired statistic (such as the mean) is calculated for the sample.
  • Distribution: This process is repeated many times, often thousands, to create an empirical distribution of the statistic.
  • Percentiles: From this empirical distribution, you can find percentiles. For a 95% confidence interval, you typically look at the 2.5th and 97.5th percentiles.
By leveraging this resampling technique, bootstrapping can provide insights into the variability and distribution of sample statistics. It is particularly advantageous when the sample size is small or the data do not meet normality assumptions.
T-distribution
The t-distribution is a probability distribution used in statistics to estimate population parameters when the sample size is small or the population standard deviation is unknown. It is similar to the normal distribution but has heavier tails, making it more suitable for data samples where there's more variability or uncertainty.

Key features of the t-distribution include:
  • Degrees of Freedom: Its shape depends on the degrees of freedom, typically calculated as the sample size minus one (-1). With more degrees of freedom, the t-distribution resembles the normal distribution.
  • Critical Values: For constructing confidence intervals, critical t-values are used. These values depend on the desired confidence level and degrees of freedom.
When calculating a confidence interval using the t-distribution, you often follow these steps:Use the calculated mean and standard error of the sample.Determine the critical t-value from a t-table, appropriate for your degrees of freedom and confidence level.Construct the interval using the formula: \( \bar{x} \pm t^* \times SE \), where \(t^*\) is the critical t-value.The t-distribution provides reliable estimates for population means, particularly when sample sizes are limited.
Standard Error
The standard error (SE) is a key concept in statistics, providing a measure of how much sample means can be expected to vary. It quantifies the precision of the sample mean as an estimate of the population mean. The smaller the standard error, the more reliable the sample mean is likely to be.

Here's what you need to know about standard error:
  • Calculation: The standard error is calculated as the sample standard deviation divided by the square root of the sample size, \( SE = \frac{s}{\sqrt{n}} \).
  • Relation to Sample Size: As the sample size increases, the standard error generally decreases, indicating more reliable estimates of the population mean.
  • Role in Confidence Intervals: When calculating confidence intervals, the standard error helps determine the margin of error, providing insights into the uncertainty of the sample mean.
Standard error is a critical component when using the t-distribution to create confidence intervals. It provides insight into how representative the sample mean is concerning the true population mean.

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Most popular questions from this chapter

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 invasive breast cancer? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

Rrefer 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 cardiovascular disease? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the area in a t-distribution above 2.1 if the samples have sizes \(n_{1}=12\) and \(n_{2}=12\).

Percent of Free Throws Made Usually, in sports, we expect top athletes to get better over time. We expect future athletes to run faster, jump higher, throw farther. One thing has remained remarkably constant, however. The percent of free throws made by basketball players has stayed almost exactly the same for 50 years. \(^{5}\) For college basketball players, the percent is about \(69 \%,\) while for players in the NBA (National Basketball Association) it is about \(75 \%\). (The percent in each group is also very similar between male and female basketball players.) In each case below, find the mean and standard deviation of the distribution of sample proportions of free throws made if we take random samples of the given size. (a) Samples of 100 free throw shots in college basketball (b) Samples of 1000 free throw shots in college basketball (c) Samples of 100 free throw shots in the \(\mathrm{NBA}\) (d) Samples of 1000 free throw shots in the \(\mathrm{NBA}\)

In the mid-1990s a Nabisco marketing campaign claimed that there were at least 1000 chips in every bag of Chips Ahoy! cookies. A group of Air Force cadets collected a sample of 42 bags of Chips Ahoy! cookies, bought from locations all across the country, to verify this claim. \({ }^{41}\) The cookies were dissolved in water and the number of chips (any piece of chocolate) in each bag were hand counted by the cadets. The average number of chips per bag was \(1261.6,\) with standard deviation 117.6 chips. (a) Why were the cookies bought from locations all over the country? (b) Test whether the average number of chips per bag is greater than 1000 . Show all details. (c) Does part (b) confirm Nabisco's claim that every bag has at least 1000 chips? Why or why not?

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