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The one-sample \(t\) CI for \(\mu\) is also a confidence interval for the population median \(\tilde{\mu}\) when the population distribution is normal. We now develop a CI for \(\tilde{\mu}\) that is valid whatever the shape of the population distribution as long as it is continuous. Let \(X_{1}, \ldots, X_{n}\) be a random sample from the distribution and \(Y_{1}, \ldots, Y_{n}\) denote the corresponding order statistics (smallest observation, second smallest, and so on). a. What is \(P\left(X_{1}<\tilde{\mu}\right) ?\) What is \(P\left(\left\\{X_{1}<\tilde{\mu}\right\\} \cap\right.\) \(\left.\left\\{X_{2}<\tilde{\mu}\right\\}\right) ?\) b. What is \(P\left(Y_{n}<\tilde{\mu}\right)\) ? What is \(P\left(Y_{1}>\tilde{\mu}\right)\) ? [Hint: What condition involving all of the \(X_{i}\) 's is equivalent to the largest being smaller than the population median?] c. What is \(P\left(Y_{1}<\tilde{\mu}

Short Answer

Expert verified
The confidence interval for the median using order statistics is \((28.7, 42.0)\) with about 99.804% confidence level.

Step by step solution

01

Compute Probability for Single Sample Less Than Median

The probability that a single observation from a continuous distribution is less than the median \( \tilde{\mu} \) is given by \( P(X_1 < \tilde{\mu}) = 0.5 \). This is because the median splits the distribution such that half the observations lie below it.
02

Compute Probability for Two Independent Samples Less Than Median

The probability that two independent observations are simultaneously less than the median is found by multiplying their individual probabilities: \( P(X_1 < \tilde{\mu} \cap X_2 < \tilde{\mu}) = 0.5 \times 0.5 = 0.25 \).
03

Probability that Maximum Order Statistic is Less Than Median

If the largest order statistic \( Y_n \) is less than \( \tilde{\mu} \), then all observations are less than the median. Hence, \( P(Y_n < \tilde{\mu}) = 0.5^n \) where \( n \) is the sample size. In our case \( n = 10 \), so \( P(Y_{10} < \tilde{\mu}) = 0.5^{10} \approx 0.00098 \).
04

Probability that Minimum Order Statistic is Greater Than Median

If the smallest order statistic \( Y_1 \) is greater than \( \tilde{\mu} \), then all observations are greater than the median. Therefore, \( P(Y_1 > \tilde{\mu}) = 0.5^n \). For \( n = 10 \), this also equals \( 0.5^{10} \approx 0.00098 \).
05

Probability that Median Falls Between Minimum and Maximum Order Statistics

The event \( Y_1 < \tilde{\mu} < Y_n \) expresses that the true median falls between the smallest and largest observations. For a continuous distribution, the confidence interval \( (Y_1, Y_n) \) for \( \tilde{\mu} \) has the probability given by \( 1 - P(Y_n < \tilde{\mu}) - P(Y_1 > \tilde{\mu}) = 1 - 2\times0.5^{n} \). This results in a confidence level of approximately 99.804\% for \( n=10 \).
06

Calculation of Order Statistics from Provided Data

The given data points are: 28.7, 29.9, 31.2, 31.5, 33.3, 35.4, 36.0, 37.2, 39.3, 42.0. Upon ordering, the smallest value \( Y_1 = 28.7 \) and the largest value \( Y_{10} = 42.0 \).
07

Determining Confidence Interval from Order Statistics

According to the probabilities calculated, the confidence interval for the population median \( \tilde{\mu} \) is given by the interval \((28.7, 42.0)\) with a confidence level of approximately 99.804\%.
08

Calculation of One-sample t Confidence Interval

To calculate the one-sample t confidence interval, find the sample mean \( \bar{x} \) and standard deviation \( s \). Then, use the critical value from the t-distribution for \( n-1 = 9 \) degrees of freedom. Calculate \( \bar{x} \pm t_{(0.01/2, 9)} \times \frac{s}{\sqrt{n}} \).
09

Compare CI from t-distribution with Non-parametric CI

Calculate the one-sample t confidence interval using confidence level corresponding to 99.804%. Compare the range and overlap with the interval \( (28.7, 42.0) \) derived non-parametrically from order statistics to see which one is wider or narrower.

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Key Concepts

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

Order Statistics
Order statistics are a powerful concept in statistics that allow us to understand the distribution of sample observations. When we arrange observations from a sample in ascending order, each individual position represents an order statistic. For instance, the smallest observation is known as the first order statistic, while the largest is the nth order statistic, where n is the total number of observations.

Order statistics are useful in estimating population parameters, like the median, because they directly reflect the distribution of the sample data. For example, in a sample containing observations like 28.7, 29.9, 31.2, etc., the order statistics help to systematically identify the minimum and maximum values in the sorted list. This setup is crucial when constructing non-parametric confidence intervals, which we'll explore in subsequent sections.
Population Median
The population median, often denoted as \( \tilde{\mu} \), is a central measure of the distribution that divides a probability distribution or dataset into two equal halves. Understanding the median is essential because it represents the middle value of a distribution, which is particularly valuable when the distribution is skewed or contains outliers.

In statistical analyses, the median provides a more robust central location than the mean, especially for non-normal distributions. To compute the probability that a sample value, like an order statistic, is less or greater than the population median involves evaluating cumulative probabilities, typically resulting in a value of 0.5 for a continuous uniform distribution. Knowing this probability assists in developing accurate confidence intervals as it assumes an equal likelihood of any single observation being below or above the median.
Non-parametric Statistics
Non-parametric statistics are statistical techniques that do not assume a specific distribution form for the underlying population. This is particularly useful when dealing with real-world data that doesn't meet the assumptions of normality required by parametric tests. Non-parametric methods use ranks or order statistics to analyze data, making them more flexible and applicable to a variety of datasets.

In the context of constructing confidence intervals for a population median, non-parametric methods rely on order statistics to have wider applicability. For instance, when developing a confidence interval for the median that doesn't depend on the data being normally distributed, we base it on the probabilities involving the smallest and largest order statistics (i.e., \( Y_1 \) and \( Y_n \)). This approach provides a robust confidence interval regardless of the population's distribution shape.
t-distribution
The t-distribution is a type of probability distribution that resembles the normal distribution but has heavier tails. This distribution is especially useful when you have a small sample size, typically with less than 30 observations, or when the population standard deviation is unknown.

One common use of the t-distribution is in constructing confidence intervals for the sample mean. In our exercise, to build a confidence interval using the t-distribution, we first calculated the sample mean and standard deviation from the data. We then applied the t-statistic corresponding to the desired confidence level, adjusted for the sample size, which is degrees of freedom \( n-1 \).

This calculation provides an interval that estimates the true population mean and is often narrower compared to non-parametric intervals. Understanding the differences in intervals obtained from the t-distribution as compared to non-parametric methods helps students recognize when each method is optimal.

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Most popular questions from this chapter

According to the article "Fatigue Testing of Condoms" (Polymer Testing, 2009: 567-571), "tests currently used for condoms are surrogates for the challenges they face in use", including a test for holes, an inflation test, a package seal test, and tests of dimensions and lubricant quality (all fertile territory for the use of statistical methodology!). The investigators developed a new test that adds cyclic strain to a level well below breakage and determines the number of cycles to break. A sample of 20 condoms of one particular type resulted in a sample mean number of 1584 and a sample standard deviation of 607 . Calculate and interpret a confidence interval at the \(99 \%\) confidence level for the true average number of cycles to break. [Note: The article presented the results of hypothesis tests based on the \(t\) distribution; the validity of these depends on assuming normal population distributions.]

Chronic exposure to asbestos fiber is a wellknown health hazard. The article "The Acute Effects of Chrysotile Asbestos Exposure on Lung Function" (Envir. Res., 1978: 360-372) reports results of a study based on a sample of construction workers who had been exposed to asbestos over a prolonged period. Among the data given in the article were the following (ordered) values of pulmonary compliance \(\left(\mathrm{cm}^{3} / \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\right)\) for each of 16 subjects 8 months after the exposure period (pulmonary compliance is a measure of lung elasticity, or how effectively the lungs are able to inhale and exhale): \(\begin{array}{llllll}167.9 & 180.8 & 184.8 & 189.8 & 194.8 & 200.2 \\ 201.9 & 206.9 & 207.2 & 208.4 & 226.3 & 227.7 \\ 228.5 & 232.4 & 239.8 & 258.6 & & \end{array}\) a. Is it plausible that the population distribution is normal? b. Compute a \(95 \%\) CI for the true average pulmonary compliance after such exposure.

Nine Australian soldiers were subjected to extreme conditions, which involved a 100 -min walk with a 25 -lb pack when the temperature was \(40^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right)\). One of them overheated (above \(39^{\circ} \mathrm{C}\) ) and was removed from the study. Here are the rectal Celsius temperatures of the other eight at the end of the walk ("Neural Network Training on Human Body Core Temperature Data," Combatant Protection and Nutrition Branch, Aeronautical and Maritime Research Laboratory of Australia, DSTO TN-0241, 1999): \(\begin{array}{llllllll}38.4 & 38.7 & 39.0 & 38.5 & 38.5 & 39.0 & 38.5 & 38.6\end{array}\) We would like to get a \(95 \%\) confidence interval for the population mean. a. Compute the \(t\)-based confidence interval of Section 8.3. b. Check for the validity of part (a). c. Generate a bootstrap sample of 999 means. d. Use the standard deviation for part (c) to get a \(95 \%\) confidence interval for the population mean. e. Investigate the distribution of the bootstrap means to see if part (d) is valid. f. Use part (c) to form the \(95 \%\) confidence interval using the percentile method. g. Compare the intervals and explain your preference. h. Based on your knowledge of normal body temperature, would you say that body temperature can be influenced by environment?

The article "Evaluating Tunnel Kiln Performance" (Amer. Ceramic Soc. Bull., Aug. 1997: 59-63) gave the following summary information for fracture strengths (MPa) of \(n=169\) ceramic bars fired in a particular kiln: \(x=89.10\), \(s=3.73\). a. Calculate a (two-sided) confidence interval for true average fracture strength using a confidence level of \(95 \%\). Does it appear that true average fracture strength has been precisely estimated? b. Suppose the investigators had believed a priori that the population standard deviation was about \(4 \mathrm{MPa}\). Based on this supposition, how large a sample would have been required to estimate \(\mu\) to within \(.5 \mathrm{MPa}\) with \(95 \%\) confidence?

If you go to a major league baseball game, how long do you expect the game to be? From the 2,429 games played in 2001 , here is a random sample of 25 times in minutes: \(\begin{array}{llllllllll}352 & 150 & 164 & 167 & 225 & 159 & 142 & 182 & 229 & 163 \\ 188 & 197 & 189 & 235 & 161 & 195 & 177 & 166 & 195 & 160 \\ 154 & 130 & 189 & 188 & 225 & & & & & \end{array}\) This is one of those rare instances in which we can do a confidence interval and compare with the true population mean. The mean of all 2,429 lengths is \(178.29\) (almost \(3 \mathrm{~h}\) ). a. Compute the \(t\)-based confidence interval of Section \(8.3 .\) b. Use a normal plot to see if part (a) is valid. c. Generate a bootstrap sample of 999 means. d. Use the standard deviation for part (c) to get a \(95 \%\) confidence interval for the population mean. e. Investigate the distribution of the bootstrap means to see if the CI of part (d) is valid. f. Use part (c) to form the \(95 \%\) confidence interval using the percentile method. g. Say which interval should be used and explain why. Does your interval include the true value, \(178.29\) ?

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