/*! 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 80 Speed, size, and strength are th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Speed, size, and strength are thought to be important factors in football performance. The article "Physical and Performance Characteristics of NCAA Division I Football Players" (Research Quarterly for Exercise and Sport \([1990]: 395-401\) ) reported on physical characteristics of Division I starting football players in the 1988 football season. Information for teams ranked in the top 20 was easily obtained, and it was reported that the mean weight of starters on top- 20 teams was \(105 \mathrm{~kg} .\) A random sample of 33 starting players (various positions were represented) from Division I teams that were not ranked in the top 20 resulted in a sample mean weight of \(103.3 \mathrm{~kg}\) and a sample standard deviation of \(16.3 \mathrm{~kg} .\) Is there sufficient evidence to conclude that the mean weight for nontop-20 starters is less than 105 , the known value for top20 teams?

Short Answer

Expert verified
In order to conduct a hypothesis test to see whether the mean weight of non-top-20 football players is less than 105 kg, we first form our hypotheses, then perform a z-test. We calculate and interpret a p-value to make a final decision.

Step by step solution

01

Formulating the Hypotheses

The first step in a hypothesis test is to set up the null and alternative hypotheses. The null hypothesis would be that the mean weight of non-top-20 players is not less than 105 kg. The alternative hypothesis would be that the mean weight of non-top-20 players is less than 105 kg. We can write this as: \(H_{0}: \mu \geq 105 \) and \(H_{a}: \mu < 105\)
02

Conducting the Z-test

After having the hypotheses, we can now conduct the one-sample z-test. The test statistic is calculated as follows: \[z = \frac{(\bar{x}- \mu_0)}{(\sigma/ \sqrt{n})} = \frac{(103.3 - 105)}{(16.3/ \sqrt{33})}\] where \(\bar{x} = 103.3kg\), \(\mu_0 = 105kg\), \(\sigma = 16.3kg\) and \(n = 33\)
03

Calculating the Critical Region or P-value

Now we’re going to use the calculated z-statistic from step 2 to find the critical region or the respective p-value. Looking up this z-score in the z-table (or using a statistical calculator), we determine the probability
04

Decision Making

Finally, we make our decision based on the calculated p-value and the given significance level. If the p-value is lower than our significance level (usually 0.05), we reject the null hypothesis. Otherwise, we do not reject it.

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.

Null and Alternative Hypotheses
Hypothesis testing in statistics starts with formulating two opposing statements about a population parameter: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \) or \( H_1 \)). The null hypothesis represents the status quo or a statement of no effect or no difference. In the context of the given exercise, the null hypothesis is that the average weight of non-top-20 football starters is not less than 105 kg (\( H_0: \mu \geq 105 \) kg).

The alternative hypothesis, on the other hand, is a statement that indicates the presence of an effect or difference. For the given scenario, the alternative hypothesis is that average weight of non-top-20 football starters is less than 105 kg (\( H_a: \mu < 105 \) kg). Essentially, the hypothesis test aims to determine whether there is enough evidence to support the alternative hypothesis over the null hypothesis.
One-sample Z-test
The one-sample z-test is a statistical method used to determine if there is a significant difference between the mean of a single sample and a known population mean. It is appropriate when the population standard deviation is known and the sample size is sufficiently large, which assumes the sampling distribution of the mean is normally distributed due to the Central Limit Theorem.

In the provided example, a one-sample z-test is used to compare the sample mean weight of 103.3 kg from non-top-20 football starters against the known population mean of 105 kg for top-20 teams. The test statistic is calculated using the formula: \[ z = \frac{(\bar{x} - \mu_0)}{(\sigma / \sqrt{n})} \] This results in a numerical value that represents how many standard deviations the sample mean is from the population mean.
P-value Calculation
The p-value is a fundamental concept in hypothesis testing, representing the probability of observing a test statistic as extreme as, or more extreme than, the value obtained from the sample data, assuming that the null hypothesis is true.

After calculating the z-test statistic from the sample data, the p-value is found by looking up this statistic in standard normal distribution tables or using statistical software. This p-value quantifies the evidence against the null hypothesis; a small p-value suggests that such extreme data is unlikely under the null hypothesis and hence provides evidence in favor of the alternative hypothesis.

Example Calculation

In our exercise, once the z-score is obtained, the p-value associated with this z-score gives us the probability that the sample mean weight could be 103.3 kg or less, if the true mean weight were really 105 kg or larger.
Statistical Significance
The concept of statistical significance is closely related to the p-value and is a measure of whether the observed data are likely due to chance. A result is said to be statistically significant if the p-value is less than the predetermined significance level, often set at 0.05.

This threshold of 0.05 is an arbitrary but commonly accepted standard that indicates there is less than a 5% chance the observed results are due to random variation. In our football weight example, if the p-value calculated from the z-test is below 0.05, the finding is considered statistically significant, and we reject the null hypothesis, concluding that there is sufficient evidence to say the mean weight of non-top-20 football starters is less than that of the top 20 teams.

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

According to a Washington Post- \(A B C\) News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bothered if the National Security Agency collected records of personal telephone calls they had made. Is there sufficient evidence to conclude that a majority of U.S. adults feel this way? Test the appropriate hypotheses using a .01 significance level.

Many people have misconceptions about how profitable small, consistent investments can be. In a survey of 1010 randomly selected U.S. adults (Associated Press, October 29,1999 ), only 374 responded that they thought that an investment of \(\$ 25\) per week over 40 years with a \(7 \%\) annual return would result in a sum of over \(\$ 100,000\) (the correct amount is \(\$ 286,640\) ). Is there sufficient evidence to conclude that less than \(40 \%\) of U.S. adults are aware that such an investment would result in a sum of over \(\$ 100,000 ?\) Test the relevant hypotheses using \(\alpha=.05\).

A random sample of \(n=44\) individuals with a B.S. degree in accounting who started with a Big Eight accounting firm and subsequently changed jobs resulted in a sample mean time to change of \(35.02\) months and a sample standard deviation of \(18.94\) months ("The Debate over Post-Baccalaureate Education: One University's Experience," Issues in Accounting Education \([1992]: 18-36)\). Can it be concluded that the true average time to change exceeds 2 years? Test the appropriate hypotheses using a significance level of \(.01\).

The authors of the article "Perceived Risks of Heart Disease and Cancer Among Cigarette Smokers" (Journal of the American Medical Association [1999]: \(1019-1021\) ) expressed the concern that a majority of smokers do not view themselves as being at increased risk of heart disease or cancer. A study of 737 current smokers selected at random from U.S. households with telephones found that of 737 smokers surveyed, 295 indicated that they believed they have a higher than average risk of cancer. Do these data suggest that \(\pi\), the true proportion of smokers who view themselves as being at increased risk of cancer is in fact less than \(.5\), as claimed by the authors of the paper? Test the relevant hypotheses using \(\alpha=.05\).

The desired percentage of silicon dioxide in a certain type of cement is \(5.0 \%\). A random sample of \(n=36\) specimens gave a sample average percentage of \(\bar{x}=5.21\). Let \(\mu\) be the true average percentage of silicon dioxide in this type of cement, and suppose that \(\sigma\) is known to be 0.38. Test \(H_{0}: \mu=5\) versus \(H_{a}: \mu \neq 5\) using a significance level of .01.

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.