/*! 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 20 The home run percentage is the n... [FREE SOLUTION] | 91Ó°ÊÓ

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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) 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) 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 \(7.2 .\)

Short Answer

Expert verified
(a) Verified mean and sd; (b) 90% CI: approx (1.89, 2.69); (c) 99% CI: approx (1.61, 2.97); (d) Tim and Herb are inside, Jackie outside 90% CI; (e) No need for normality due to CLT.

Step by step solution

01

Verify Sample Mean and Standard Deviation

The provided data set contains 43 values of home run percentages. First, let's calculate the sample mean \( \bar{x} \) and the sample standard deviation \( s \). Utilize a calculator's statistical mode for this purpose, ensuring all data points are entered correctly. Verify that the results are \( \bar{x} \approx 2.29 \) and \( s \approx 1.40 \).
02

Compute 90% Confidence Interval

To compute the 90% confidence interval for the population mean \( \mu \), use the formula: \( \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}} \), where \( n = 43 \) is the sample size, \( \bar{x} \approx 2.29 \), \( s \approx 1.40 \), and \( t_{\alpha/2} \) is the t-score from the t-distribution table for \( df = n-1 = 42 \). For 90% confidence, \( \alpha = 0.1 \), so \( t_{0.05} \approx 1.682 \). Calculate the margin of error and find the interval.
03

Compute 99% Confidence Interval

For the 99% confidence interval, use the same formula: \( \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}} \). Here, \( \alpha = 0.01 \), so refer to the t-distribution table for \( df = 42 \) to find \( t_{0.005} \approx 2.423 \). Again, calculate the new margin of error and determine the confidence interval.
04

Analyze Player Home Run Percentages

Compare the calculated confidence intervals with the home run percentages of Tim Huelett (2.5), Herb Hunter (2.0), and Jackie Jensen (3.8). Determine whether these values fall within the intervals and discuss any insights regarding their performance relative to the population mean.
05

Discuss Normality Assumption

Review the need for the normal distribution assumption. Given our sample size is 43, the Central Limit Theorem suggests that the sample mean distribution will be approximately normal, even if the original data distribution is not. Therefore, normality of the data is not a strict requirement.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values used to estimate the true value of a population parameter. When we calculate a confidence interval, we're looking to determine where the true population mean might lie, based on our sample data. For instance, in our baseball players example, we want to estimate the mean home run percentage of all professional baseball players using our sample.
  • To compute a confidence interval, we use the sample mean (\(\bar{x}\)) and the sample standard deviation (\(s\)), along with a critical value determined from the t-distribution.
  • The confidence level we choose (e.g., 90% or 99%) indicates how sure we are that the interval contains the true mean. The higher the confidence level, the wider the interval.
The formula used is \(\bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}\), where \(t_{\alpha/2}\) is the t-score from the t-distribution table. This margin of error adjusts our sample mean to better reflect the population mean. By computing both the 90% and 99% intervals in the exercise, you see how changing the confidence level affects the interval's width.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that makes working with sample means feasible, even when the original data set may not be normally distributed. This theorem states that, for a sufficiently large sample size, the distribution of the sample means will be approximately normal (bell-shaped), regardless of the shape of the population distribution.
  • In our exercise context, the home run percentages might not be normally distributed on their own.
  • However, our sample size of 43 is large enough for the CLT to take effect.
This allows us to use normal approximation methods for our calculations. Essentially, the CLT lets us say, "Even if the data isn't perfectly normal, our methods for estimating the population mean are still valid." The CLT is why we didn't stress about the normality of the data distribution when we calculated confidence intervals for the mean.
t-distribution
The t-distribution is a probability distribution that's especially useful for small sample sizes or when the population standard deviation is unknown. Unlike the normal distribution, which is applicable for large samples, the t-distribution has heavier tails, meaning it is more spread out.
  • When constructing confidence intervals with smaller samples, the t-distribution provides a more accurate critical value, making sure our intervals are reliable.
  • In our baseball exercise, even with a moderate sample size of 43, we utilize the t-distribution as a precaution and to improve accuracy.
The t-distribution varies with degrees of freedom (df), which are determined by the sample size minus one (\(n - 1\)). For a sample of size 43, we have 42 degrees of freedom. As the sample size grows, the t-distribution begins to resemble the normal distribution more closely, but for smaller samples, sticking with the t-distribution is best for precision.
Population Mean
The population mean (\(\mu\)) is a measure that tells us the average value of an entire population. However, calculating it directly is typically impractical or impossible. That's why we use a sample mean as our best estimate.
  • In statistics exercises like our baseball player example, our goal is to make inferences about this population mean using the sample we've gathered.
  • We use the sample mean of 2.29 as an estimate but the confidence interval process helps us to gauge how close this estimate is likely to be to the actual population mean.
The confidence interval provides a range that, with a certain level of confidence, contains the population mean. Therefore, even though we cannot always measure the population mean exactly, statistical methods allow us to estimate it effectively and understand its potential variability. Based on our 90% and 99% confidence intervals, we can assess how home run percentages for specific players compare to the average, aiding in performance analysis.

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Most popular questions from this chapter

In a random sample of 519 judges, it was found that 285 were introverts (see reference of Problem 5). (a) Let \(p\) represent the proportion of all judges who are introverts. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p .\) Give a brief interpretation of the meaning of the confidence interval you have found. (c) 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.

If a \(90 \%\) confidence interval for the difference of means \(\mu_{1}-\mu_{2}\) contains all negative values, what can we conclude about the relationship between \(\mu_{1}\) and \(\mu_{2}\) at the \(90 \%\) confidence level?

How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from \(20^{\circ} \mathrm{F}\) to \(45^{\circ} \mathrm{F}\). A random sample of prices (\$) for sleeping bags in this temperature range was taken from Backpacker Magazine: Gear Guide (Vol. 25 , Issue 157, No. 2). Brand names include American Camper, Cabela's, Camp 7, Caribou, Cascade, and Coleman. \(\begin{array}{rrrrrrrrrr}80 & 90 & 100 & 120 & 75 & 37 & 30 & 23 & 100 & 110 \\ 105 & 95 & 105 & 60 & 110 & 120 & 95 & 90 & 60 & 70\end{array}\) (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx \$ 83.75\) and \(s \approx \$ 28.97\). (b) Using the given data as representative of the population of prices of all summer sleeping bags, find a \(90 \%\) confidence interval for the mean price \(\mu\) of all summer sleeping bags.

A random sample of 328 medical doctors showed that 171 had a solo practice. (Source: Practice Patterns of General Internal Medicine, American Medical Association.) (a) Let \(p\) represent the proportion of all medical doctors who have a solo practice. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p .\) Give a brief explanation of the meaning of the interval. (c) As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice? What is the margin of error based on a \(95 \%\) confidence interval?

Case studies showed that out of 10,351 convicts who escaped from U.S. prisons, only 7867 were recaptured (The Book of Odds, by Shook and Shook, Signet). (a) Let \(p\) represent the proportion of all escaped convicts who will eventually be recaptured. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p .\) Give a brief statement of the meaning of the confidence interval. (c) Is use of the normal approximation to the binomial justified in this problem? Explain.

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