91Ó°ÊÓ

Pro Football and Basketball: Heights of Players 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 Online Study Center. $$ \begin{aligned} &\text { Heights (in } \mathrm{ft} \text { ) of pro football players: } x_{1} ; n_{1}=45\\\ &\begin{array}{llllllllll} 6.33 & 6.50 & 6.50 & 6.25 & 6.50 & 6.33 & 6.25 & 6.17 & 6.42 & 6.33 \\ 6.42 & 6.58 & 6.08 & 6.58 & 6.50 & 6.42 & 6.25 & 6.67 & 5.91 & 6.00 \\ 5.83 & 6.00 & 5.83 & 5.08 & 6.75 & 5.83 & 6.17 & 5.75 & 6.00 & 5.75 \\ 6.50 & 5.83 & 5.91 & 5.67 & 6.00 & 6.08 & 6.17 & 6.58 & 6.50 & 6.25 \\ 6.33 & 5.25 & 6.67 & 6.50 & 5.83 & & & & & \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Heights (in } \mathrm{ft} \text { ) of pro basketball players: } x_{2} ; n_{2}=40\\\ &\begin{array}{llllllllll} 6.08 & 6.58 & 6.25 & 6.58 & 6.25 & 5.92 & 7.00 & 6.41 & 6.75 & 6.25 \\ 6.00 & 6.92 & 6.83 & 6.58 & 6.41 & 6.67 & 6.67 & 5.75 & 6.25 & 6.25 \\ 6.50 & 6.00 & 6.92 & 6.25 & 6.42 & 6.58 & 6.58 & 6.08 & 6.75 & 6.50 \\ 6.83 & 6.08 & 6.92 & 6.00 & 6.33 & 6.50 & 6.58 & 6.83 & 6.50 & 6.58 \end{array} \end{aligned} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 6.179, s_{1} \approx 0.366, \bar{x}_{2} \approx 6.453\), and \(s_{2} \approx 0.314 .\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2}\). Find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation 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 \(90 \%\) level of confidence, do professional football players tend to have a higher population mean height than professional basketball players? (d) Check Requirements Which distribution (standard normal or Student's \(t)\) did you use? Why? Do you need information about the height distributions? Explain.

Step by step solution

01

Compute Mean and Standard Deviation for Football Players

Verify the provided mean \(\bar{x}_1\) and standard deviation \(s_1\) using a calculator with relevant functions for the dataset provided. After computation, we obtain \(\bar{x}_1 \approx 6.179\) and \(s_1 \approx 0.366\) as indicated in the problem.

02

Compute Mean and Standard Deviation for Basketball Players

Similarly, use a calculator to verify the mean \(\bar{x}_2\) and standard deviation \(s_2\) for the basketball player's dataset. This confirms \(\bar{x}_2 \approx 6.453\) and \(s_2 \approx 0.314\).

03

Calculate the Standard Error for the Difference in Means

Calculate the standard error (SE) of the difference in means using the formula: \( SE(\bar{x}_1 - \bar{x}_2) = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \), where \(s_1 = 0.366\), \(s_2 = 0.314\), \(n_1 = 45\), and \(n_2 = 40\). This yields \(SE(\bar{x}_1 - \bar{x}_2) \approx 0.078\).

04

Find the Critical Value for a 90% Confidence Interval

For a 90% confidence level and an approximate large sample size, use a standard normal distribution to find the critical value \(z^* \approx 1.645\). In smaller samples where the populations' standard deviations are unknown, a Student's \(t\)-distribution would be more appropriate, but the large sample sizes and similar sample standard deviations suffice for the normal approximation.

05

Calculate the Confidence Interval for Difference in Means

Compute the confidence interval for \(\mu_1 - \mu_2\) using the formula: \[ CI = \left((\bar{x}_1 - \bar{x}_2) - z^* \cdot SE, (\bar{x}_1 - \bar{x}_2) + z^* \cdot SE\right) \]With \(\bar{x}_1 - \bar{x}_2 = 6.179 - 6.453 = -0.274\), this gives the interval \(( -0.403, -0.145 )\).

06

Interpret the Confidence Interval

The confidence interval \((-0.403, -0.145)\) indicates that we are 90% confident the difference in population mean heights (\(\mu_1 - \mu_2\)) is between -0.403 and -0.145 feet. This interval is entirely negative, suggesting basketball players generally are taller than football players.

07

Check Distribution Requirements and Interpretation

The normal distribution is used in this context due to the large sample sizes which adequately approximate a normal distribution according to the Central Limit Theorem. Since sample standard deviations are similar and the population distributions aren't significantly skewed, the approximation suffices without additional distribution details.

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Key Concepts

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

Mean and Standard Deviation

The mean and standard deviation are fundamental statistical concepts used to summarize data sets. The **mean**, often referred to as the average, is calculated by adding up all the numbers in a data set and then dividing by the number of data points. It provides a central value for the dataset. In our scenario:

• For professional football players, the mean height \(\bar{x}_1\) is approximately 6.179 feet.
• For basketball players, the mean height \(\bar{x}_2\) is about 6.453 feet.

These values indicate the average heights of players in each sport.
On the other hand, the **standard deviation** tells us how much the values in the data set deviate from the mean. A low standard deviation means that most numbers in the set are close to the average, whereas a high standard deviation indicates a larger variance from the mean. In the given data:

• The standard deviation for football players \(s_1\) is approximately 0.366 feet.
• For basketball players, the standard deviation \(s_2\) is approximately 0.314 feet.

This suggests that football player heights vary slightly more around their average compared to their basketball counterparts.

Population Mean

Population mean represents the average value for a population, which is a collection of all possible individuals or measurements of a given type. In statistical studies, we often try to estimate this value using sample means due to the impracticality of measuring the entire population.

In the exercise, we refer to the population means of heights for:

• Football players with \(\mu_1\).
• Basketball players with \(\mu_2\).

The sample means \(\bar{x}_1\) for football players and \(\bar{x}_2\) for basketball players serve as estimators for these population means. It's important to remember that the true population mean is often unknown, and our objective is to use samples to make informed estimates about it.
For example, when we compute a 90% confidence interval for the difference in population means \(\mu_1 - \mu_2\), we rely on sample statistics: the means \(\bar{x}_1\) and \(\bar{x}_2\), and their respective standard errors. This confidence interval gives us a range in which we believe the actual difference in population means lies, accounting for sampling variability.

Standard Error

The standard error (SE) is a measure that reflects the variability of a sample statistic, like the mean, when multiple samples are drawn from a population. It essentially provides insight into the accuracy of the sample mean as an estimate of the population mean.
To calculate the standard error of the difference in means for two groups (football and basketball players), use:

• \( SE(\bar{x}_1 - \bar{x}_2) = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \)

Where:

• \(s_1\) and \(s_2\) denote the standard deviations of the two samples.
• \(n_1\) and \(n_2\) are the respective sample sizes.

This formula calculates the combined variability from both groups. In our exercise:

• The calculated SE is approximately 0.078, emphasizing a relatively small sampling variability between the means of the two groups.

Overall, the standard error is pivotal for constructing confidence intervals and conducting hypothesis tests, allowing us to make statistically supported statements about population parameters based on our samples.