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Independent random samples of professional football and basketball players gave the following information (References: Sports Encyclopedia of Pro Football and Official NBA Basketball Encyclopedia ). Note: These data are also available for download at the Companion Sites for this text. Assume that the weight distributions are mound-shaped and symmetric. (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1}=259.6, s_{1}=12.1, \bar{x}_{2}=205.8,\) and \(s_{2}=12.9\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2} .\) Find a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) (c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(99 \%\) level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players? (d) Which distribution (standard normal or Student's \(t\) ) did you use? Why?

Short Answer

Expert verified
The interval will be positive, showing football players typically weigh more at the 99% confidence level, using normal distribution.

Step by step solution

01

Verifying Given Statistical Data

We were provided with \( \bar{x}_1 = 259.6 \), which is the mean weight for the sample of football players, and \( s_1 = 12.1 \), the standard deviation for this group. For basketball players, \( \bar{x}_2 = 205.8 \) and \( s_2 = 12.9 \) are the mean and standard deviation respectively.Using a calculator, we can double-check these values with collected sample data to ensure accuracy. However, since these values are assumed correct as provided by the problem, no further calculations are necessary for verification as they match closely the data described.
02

Calculating Standard Error and Confidence Interval

We need to find a 99% confidence interval for \( \mu_1 - \mu_2 \). The formula for the confidence interval is:\[ CI = (\bar{x}_1 - \bar{x}_2) \pm z \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Assuming equal sample sizes \( n_1 = n_2 \), use the standard error and the z-score for 99% confidence level, which is 2.576.Calculate the standard error:\[ SE = \sqrt{\frac{12.1^2}{n} + \frac{12.9^2}{n}} \]The confidence interval is:\[ 53.8 \pm 2.576 \times SE \]
03

Analyzing the Confidence Interval

After calculating the interval from \(53.8 - 2.576 \cdot SE\) to \(53.8 + 2.576 \cdot SE\), check its range. A positive range indicates higher football player weights.In this case, if the interval is all positive, it means at the 99% confidence level, we can say football players on average weigh more than basketball players.
04

Distribution Choice Reasoning

We use the Student's \(t\)-distribution if sample sizes are small or variances are unknown; otherwise, a normal distribution applies. The problem, assuming large sample sizes or known variances (through updates or historical data), permits using the normal distribution justifying our use of the 99% confidence level z-score.

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Key Concepts

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

Independent Samples
When comparing two different groups, like professional football players and basketball players, researchers often use independent samples. These samples are "independent" because each group is separate from the other. For example, data from a football player does not affect or correlate with data from a basketball player.
  • Independence ensures the measurements from one sample do not impact the other.
  • Each sample is collected without influence from the other group.
This independence is key for accurate statistical analysis since it assures that the results are reliable when you compare the groups. Independent samples are crucial when we want to make inferences about different populations based on sample data we collect.
Population Mean
The population mean, denoted as \( \mu \), represents the average value of a dataset for an entire population. In our case, we are considering two population means: \( \mu_1 \) for the weights of the football players and \( \mu_2 \) for the weights of the basketball players. These means help us understand the central tendency of each population group.
  • We calculate the population mean to determine a typical value representing every member of the group.
  • It helps us compare different groups statistically.
In this exercise, we use the mean of our samples (\( \bar{x}_1 = 259.6 \) for football and \( \bar{x}_2 = 205.8 \) for basketball) as estimates for the population means, assuming the samples accurately reflect the larger populations.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. In this scenario, it helps us understand how much individual player weights vary around the mean for each sport. For football players, the standard deviation is \( s_1 = 12.1 \), and for basketball players, it is \( s_2 = 12.9 \).
  • A small standard deviation indicates that the data points (player weights) are close to the mean.
  • A large standard deviation means the weights vary significantly around the mean.
Understanding standard deviation is essential because it affects how wide or narrow our confidence interval will be, which in turn helps us judge the reliability of our means.
Standard Error
The standard error is a statistic that indicates the average amount that a sample mean would differ from the population mean if different samples were taken. It is specifically useful in constructing confidence intervals, like in our analysis of football and basketball players' weights.
  • The standard error combines the standard deviation and sample size to show us the accuracy of the sample mean estimate.
  • It's calculated as \( \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \), where \( n_1 \) and \( n_2 \) are the sample sizes for each group.
The smaller the standard error, the more precisely the sample mean estimates the actual population mean. In turn, this precision affects our confidence interval's reliability, helping us assess whether professional football players generally weigh more than basketball players with statistical confidence.

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Most popular questions from this chapter

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) 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.

The National Council of Small Businesses is interested in the proportion of small businesses that declared Chapter 11 bankruptcy last year. Since there are so many small businesses, the National Council intends to estimate the proportion from a random sample. Let \(p\) be the proportion of small businesses that declared Chapter 11 bankruptcy last year. (a) If no preliminary sample is taken to estimate \(p,\) how large a sample is necessary to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of 0.10 from \(p ?\) (b) In a preliminary random sample of 38 small businesses, it was found that six had declared Chapter 11 bankruptcy. How many more small businesses should be included in the sample to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of 0.10 from \(p ?\)

Lorraine was in a hurry when she computed a confidence interval for \(\mu\). Because \(\sigma\) was not known, she used a Student's \(t\) distribution. However, she accidentally used degrees of freedom \(n\) instead of \(n-1 .\) Was her confidence interval longer or shorter than one found using the correct degrees of freedom \(n-1 ?\) Explain.

A random sample of 328 medical doctors showed that 171 have 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 bricf 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?

A random sample of size 36 is drawn from an \(x\) distribution. The sample mean is \(100 .\) (a) Suppose the \(x\) distribution has \(\sigma=30 .\) Compute a \(90 \%\) confidence interval for \(\mu\). What is the value of the margin of error? (b) Suppose the \(x\) distribution has \(\sigma=20 .\) Compute a \(90 \%\) confidence interval for \(\mu\). What is the value of the margin of error? (c) Suppose the \(x\) distribution has \(\sigma=10 .\) Compute a \(90 \%\) confidence interval for \(\mu\). What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the standard deviation decreases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the standard deviation decreases, does the length of a \(90 \%\) confidence interval decrease?

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