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Chapter 4: Problem 1

Recall For the baseball data (bb.rda), 15 out of 59 ballplayers are lefthanded. Let \(p\) be the probability that a professional baseball player is left-handed. Determine an exact \(90 \%\) confidence interval for \(p .\) Show first that the equations to be solved are: $$ \sum_{j=0}^{14}\left(\begin{array}{c} n \\ j \end{array}\right) \underline{\theta}^{j}(1-\underline{\theta})^{n-j}=0.95 \text { and } \sum_{j=0}^{15}\left(\begin{array}{c} n \\ j \end{array}\right) \bar{\theta}^{j}(1-\bar{\theta})^{n-j}=0.05 $$ Then do the following steps to obtain the confidence interval. (a) Show that \(0.10\) and \(0.17\) bracket the solution to the first equation. (b) Show that \(0.34\) and \(0.38\) bracket the solution to the second equation. (c) Then use the \(\mathrm{R}\) function binomci. \(\mathrm{r}\) to solve the equations.

Short Answer

Expert verified
The 90% confidence interval for the probability \(p\) that a professional baseball player is left-handed can be calculated using the equations given and with the help of the binomci function in R programming language.

Step by step solution

01

- Setup the two equations

First, setup the two equations based on the problem. The equations are given in the problem as follows: \[ \sum_{j=0}^{14}\left(\begin{array}{c} n \ j \end{array}\right) \underline{\theta}^{j}(1-\underline{\theta})^{n-j}=0.05\] \[ \sum_{j=0}^{15}\left(\begin{array}{c} n \ j \end{array}\right) \bar{\theta}^{j}(1-\bar{\theta})^{n-j}=0.05 \]
02

- Find the brackets for the solutions

Next, show that 0.10 and 0.17 bracket the solution to the first equation, and that 0.34 and 0.38 bracket the solution to the second equation. This can be done by checking that the equations hold true at the end points, and that there is a sign change between the end points which would imply there is at least one solution in between.
03

- Use binomci function in R

Finally, use the binomci function in R to solve the equations. With the R function 'binomci', a 90% confidence interval for \(p\) can be computed. 'binomci' needs as argument the number of left-handed players (which is 15), the total number of players (which is 59), and the confidence level (which is 0.9). The output will be the exact confidence interval for \(p\).

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Key Concepts

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

Binomial Distribution
To understand the binomial distribution, imagine a series of yes-or-no questions, such as flipping a coin and asking, 'Is it heads?' If you flip it 5 times, how many times can you expect heads? Every flip, there are two outcomes: heads or tails. If each outcome is as likely as the other, the probability (\( p \)) of getting heads is 0.5.

The binomial distribution describes the number of successes (\( x \) heads) in a fixed number of independent Bernoulli trials (\( n \) flips), each with the same probability of success (\( p \) for heads). The probability of observing exactly \( x \) successes in \( n \) trials is determined by a specific formula, where \( p \) is the probability of success on an individual trial, and \( 1 - p \) is the probability of failure.

When we say 'binomial distribution,' think of it as a more formal way of understanding sequences of independent events with two possible outcomes on each event.
Probability
Probability, widely used in statistics, is a measure of how likely an event is to occur. It ranges from 0 (impossible) to 1 (certain). Picture it as the chance of rain on a cloudy day. The probability plays a key role in determining binomial distributions because it represents the likelihood of each individual outcome.

For instance, the probability of a baseball player being left-handed (denoted as \( p \) in the problem) is crucial as it influences the expected number of left-handed players out of a given sample of baseball players. Understanding probability enables us to predict outcomes and make more educated guesses in uncertain situations.
Hypothesis Testing
Hypothesis testing is a formal methodology for determining whether a certain premise is plausible by using sample data. Think of it like detective work, hunting for clues (evidence) to support or refute an initial theory (the hypothesis).

In the context of our baseball example, we are essentially testing the hypothesis that the observed proportion of left-handed players represents the true proportion within the entire population of professional players. By calculating the confidence interval, we use hypothesis testing to determine if our sample data likely reflects the larger population or if the observed results are just due to random variation.
R Statistical Software
R is a powerful tool for statistical computing and graphics, like a high-tech calculator that statisticians and data scientists use to analyze data and make sense of numbers. It's designed to handle complex data analysis with ease and can perform operations ranging from simple calculations to advanced modeling.

In our baseball case, R's 'binomci' function is like a specialized assistant, one that precisely calculates confidence intervals for binomial distributions given a few inputs like number of successes, total trials, and the desired confidence level (the \( 90\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text/lgpltext{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text}\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\textğpercent}}} \) confidence level). It takes the crunching of numbers out of human hands and ensures precise, quick results which are essential for informed decision making in statistics.

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Most popular questions from this chapter

To illustrate Exercise 4.2.24, let \(X_{1}, X_{2}, \ldots, X_{9}\) and \(Y_{1}, Y_{2}, \ldots, Y_{12}\) represent two independent random samples from the respective normal distributions \(N\left(\mu_{1}, \sigma_{1}^{2}\right)\) and \(N\left(\mu_{2}, \sigma_{2}^{2}\right) .\) It is given that \(\sigma_{1}^{2}=3 \sigma_{2}^{2}\), but \(\sigma_{2}^{2}\) is unknown. Define a random variable that has a \(t\) -distribution that can be used to find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

In Exercise \(4.2 .27\), in finding a confidence interval for the ratio of the variances of two normal distributions, we used a statistic \(S_{1}^{2} / S_{2}^{2}\), which has an \(F\) distribution when those two variances are equal. If we denote that statistic by \(F\), we can test \(H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}\) against \(H_{1}: \sigma_{1}^{2}>\sigma_{2}^{2}\) using the critical region \(F \geq c\). If \(n=13, m=11\), and \(\alpha=0.05\), find \(c .\)

Let \(X_{1}, X_{2}, \ldots, X_{n}, X_{n+1}\) be a random sample of size \(n+1, n>1\), from a distribution that is \(N\left(\mu, \sigma^{2}\right) .\) Let \(\bar{X}=\sum_{1}^{n} X_{i} / n\) and \(S^{2}=\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2} /(n-1)\). Find the constant \(c\) so that the statistic \(c\left(\bar{X}-X_{n+1}\right) / S\) has a \(t\) -distribution. If \(n=8\), determine \(k\) such that \(P\left(\bar{X}-k S

Two different teaching procedures were used on two different groups of students. Each group contained 100 students of about the same ability. At the end of the term, an evaluating team assigned a letter grade to each student. The results were tabulated as follows. \begin{tabular}{ccccccc} \hline & \multicolumn{5}{c} { Grade } & \\ \cline { 2 - 6 } Group & A & B & C & D & F & Total \\ \hline I & 15 & 25 & 32 & 17 & 11 & 100 \\ II & 9 & 18 & 29 & 28 & 16 & 100 \\ \hline \end{tabular} If we consider these data to be independent observations from two respective multinomial distributions with \(k=5\), test at the \(5 \%\) significance level the hypothesis

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be two independent random samples from the respective normal distributions \(N\left(\mu_{1}, \sigma_{1}^{2}\right)\) and \(N\left(\mu_{2}, \sigma_{2}^{2}\right)\), where the four parameters are unknown. To construct a confidence interval for the ratio, \(\sigma_{1}^{2} / \sigma_{2}^{2}\), of the variances, form the quotient of the two independent \(\chi^{2}\) variables, each divided by its degrees of freedom, namely, $$ F=\frac{\frac{(m-1) S_{2}^{2}}{\sigma_{2}^{2}} /(m-1)}{\frac{(n-1) S_{1}^{2}}{\sigma_{1}^{2}} /(n-1)}=\frac{S_{2}^{2} / \sigma_{2}^{2}}{S_{1}^{2} / \sigma_{1}^{2}} $$ where \(S_{1}^{2}\) and \(S_{2}^{2}\) are the respective sample variances. (a) What kind of distribution does \(F\) have? (b) Critical values \(a\) and \(b\) can be found so that \(P(F
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