/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Baseball: Home Run Percentage Th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Baseball: Home Run Percentage The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gave the following data for home run percentages (Reference: The Baseball Encyclopedia, Macmillan). $$ \begin{array}{llllllllll} 1.6 & 2.4 & 1.2 & 6.6 & 2.3 & 0.0 & 1.8 & 2.5 & 6.5 & 1.8 \\ 2.7 & 2.0 & 1.9 & 1.3 & 2.7 & 1.7 & 1.3 & 2.1 & 2.8 & 1.4 \\ 3.8 & 2.1 & 3.4 & 1.3 & 1.5 & 2.9 & 2.6 & 0.0 & 4.1 & 2.9 \\ 1.9 & 2.4 & 0.0 & 1.8 & 3.1 & 3.8 & 3.2 & 1.6 & 4.2 & 0.0 \\ 1.2 & 1.8 & 2.4 & & & & & & & \end{array} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x} \approx 2.29\) and \(s \approx 1.40\). (b) Compute a \(90 \%\) confidence interval for the population mean \(\mu\) of home run percentages for all professional baseball players. Hint: If you use Table 6 of Appendix II, be sure to use the closest \(d . f\). that is smaller. (c) Compute a \(99 \%\) confidence interval for the population mean \(\mu\) of home run percentages for all professional baseball players. (d) Interpretation The home run percentages for three professional players are Tim Huelett, \(2.5 \quad\) Herb Hunter, \(2.0 \quad\) Jackie Jensen, \(3.8\) Examine your confidence intervals and describe how the home run percentages for these players compare to the population average. (e) Check Requirements In previous problems, we assumed the \(x\) distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section \(6.5\).

Short Answer

Expert verified
Home run percentages confirm \( \bar{x} = 2.29 \) and \( s = 1.40 \). The \(90\%\) CI is \((1.89, 2.69)\); the \(99\%\) CI is \((1.70, 2.88)\). Both CIs are applicable as the sample satisfies the CLT requirements.

Step by step solution

01

Verify the Mean and Standard Deviation

Calculate the mean \( \bar{x} \) and standard deviation \( s \) using a statistical calculator or software. Given the data, ensure \( \bar{x} \approx 2.29 \) and \( s \approx 1.40 \) are accurate results.
02

Calculate the 90% Confidence Interval

Use the formula for the confidence interval \( \bar{x} \pm t \, \frac{s}{\sqrt{n}} \), where \( t \) is the t-value from the t-distribution table for \( 90\% \) confidence and \( n-1 \) degrees of freedom. With \( n = 43 \), determine \( t \approx 1.682 \) and compute the interval.
03

Calculate the 99% Confidence Interval

Use the same confidence interval formula: \( \bar{x} \pm t \, \frac{s}{\sqrt{n}} \). Find the \( t \)-value for \( 99\% \) confidence with \( 42 \) degrees of freedom, approximately \( t \approx 2.423 \), and substitute into the formula.
04

Interpret Player Statistics

Compare the home run percentages of Tim Huelett (2.5), Herb Hunter (2.0), and Jackie Jensen (3.8) to the calculated confidence intervals. See if these percentages fall within the intervals to determine if they are typical or unusual.
05

Evaluate Assumptions

Based on the central limit theorem, as the sample size is \( 43 \), which is considered large enough, the sampling distribution of \( \bar{x} \) is approximately normal regardless of the distribution of the data.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Confidence Intervals
Confidence intervals give us a range of values that likely contain the population mean. They are essential in statistics, providing a measure of uncertainty regarding our sample estimates. When constructing a confidence interval, we use the sample mean, which represents our best estimate for the population mean. However, since we're drawing from a sample, we also need to account for variability. This is where the standard deviation and the t-distribution (or sometimes the z-distribution) come into play.

To calculate a confidence interval:
  • Identify your desired confidence level (e.g., 90% or 99%).
  • Find the corresponding t-value from the t-distribution table based on your sample size, minus one (degrees of freedom).
  • Use the formula: \[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]where \( \bar{x} \) is the sample mean, \( s \) is the standard deviation, and \( n \) is the sample size.
This results in a range where the actual population mean is likely to be found. For example, for a 90% confidence interval, if your interval is 2.2 to 2.4, this means you are 90% confident the true mean falls within this range.

Confidence intervals provide a way to quantify uncertainty, helping to make informed decisions based on statistical data.
Central Limit Theorem
The Central Limit Theorem (CLT) is one of the most important ideas in statistics. It states that the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution, as long as the sample size is sufficiently large. This is good news for statisticians because it justifies the use of normal distribution methods in many situations.

Key Points of the Central Limit Theorem:
  • As the sample size increases, the sampling distribution of the sample mean becomes more normally distributed.
  • This theorem holds true no matter what the shape of the population distribution is – it could be skewed or have outliers.
  • For most practical purposes, a sample size of 30 is considered large enough to apply the CLT.
Because of the CLT, we can apply methods that assume a normal distribution, like calculating confidence intervals, even if our data doesn't come from a normal distribution, provided our sample size is large enough. In the context of the baseball player statistics, because we have 43 data points, we can be confident that the sample mean follows a normal distribution, aiding in our confidence interval calculations.
Sample Mean
The sample mean, denoted as \( \bar{x} \), is the average of the data values in a sample and serves as an estimate for the population mean. In our baseball example, the sample mean of home run percentages is calculated by adding together all the percentages and dividing by the number of players in the sample (43, in this case).

Formula to calculate the sample mean:\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \]where \( x_i \) are the data points and \( n \) is the number of observations in the sample.
The sample mean is a crucial component in constructing confidence intervals and conducting hypothesis tests. It's important to remember that the sample mean is an estimation; sometimes, it might differ from the true population mean, which is why confidence intervals are useful. They provide a range based on this sample estimate that we're fairly confident includes the population mean.
Standard Deviation
Standard deviation is a measure of variability or dispersion in a dataset. It quantifies how much the data points deviate from the mean. A low standard deviation means that data points are close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range.

In the formula for the confidence interval, \[ \bar{x} \pm t \frac{s}{\sqrt{n}} \],standard deviation \( s \) plays a key role in determining the "width" of the interval. The larger the standard deviation, the wider the interval, reflecting more uncertainty about the population mean.

Calculating the standard deviation involves:
  • Finding the mean of your dataset.
  • Subtracting the mean from each number to find the deviation for each value.
  • Squaring these deviations, summing them up, and dividing by \( n - 1 \) to find the variance.
  • The standard deviation is the square root of this variance.
This statistic helps in understanding the spread of data and is critical when making statistical inferences, including constructing confidence intervals and testing hypotheses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Answer true or false. Explain your answer. If the sample mean \(\bar{x}\) of a random sample from an \(x\) distribution is relatively small, then the confidence interval for \(\mu\) will be relatively short.

Myers-Briggs: Marriage Counseling Most married couples have two or three personality preferences in common (see reference in Problem 17). Myers used a random sample of 375 married couples and found that 132 had three preferences in common. Another random sample of 571 couples showed that 217 had two personality preferences in common. Let \(p_{1}\) be the population proportion of all married couples who have three personality preferences in common. Let \(p_{2}\) be the population proportion of all married couples who have two personality preferences in common. (a) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(90 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Examine the confidence interval in part (a) and explain what it means in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you about the proportion of married couples with three personality preferences in common compared with the proportion of couples with two preferences in common (at the \(90 \%\) confidence level)?

Suppose \(x\) has a mound-shaped distribution with \(\sigma=3\). (a) Find the minimal sample size required so that for a \(95 \%\) confidence interval, the maximal margin of error is \(E=0.4\). (b) Check Requirements Based on this sample size, can we assume that the \(\bar{x}\) distribution is approximately normal? Explain.

Myers-Briggs: Actors Isabel Myers was a pioneer in the study of personality types. The following information is taken from A Guide to the Development and Use of the Myers-Briggs Type Indicator by Myers and McCaulley (Consulting Psychologists Press). In a random sample of 62 professional actors, it was found that 39 were extroverts. (a) Let \(p\) represent the proportion of all actors who are extroverts. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p .\) Give a brief interpretation of the meaning of the confidence interval you have found. (c) Check Requirements Do you think the conditions \(n p>5\) and \(n q>5\) are satisfied in this problem? Explain why this would be an important consideration.

Campus Life: Coeds What percentage of your campus student body is female? Let \(p\) be the proportion of women students on your campus. (a) If no preliminary study is made to estimate \(p\), how large a sample is needed to be \(99 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of \(0.05\) from \(p ?\) (b) The Statistical Abstract of the United States, 112 th Edition, indicates that approximately \(54 \%\) of college students are female. Answer part (a) using this estimate for \(p\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.