/*! 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 13 Throwing darts at the stock page... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 13

Throwing darts at the stock pages to decide which companies to invest in could be a successful stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 100 companies to invest in. After 1 year, 53 of the companies were considered winners; that is, they outperformed other companies in the same investment class. To assess whether the dart-picking strategy resulted in a majority of winners, the researcher tested \(H_{0}: p=0.5\) versus \(H_{1}: p>0.5\) and obtained a \(P\) -value of \(0.2743 .\) Explain what this \(P\) -value means and write a conclusion for the researcher.

Short Answer

Expert verified
The P-value of 0.2743 indicates there is not enough evidence to reject the null hypothesis. The dart-picking strategy does not result in a majority of winners according to this test.

Step by step solution

01

- State the Null and Alternative Hypotheses

The null hypothesis ( H_0) is that the proportion of companies considered winners is 0.5 ( p = 0.5). The alternative hypothesis ( H_1) is that the proportion of companies considered winners is greater than 0.5 ( p > 0.5).
02

- Interpret the P-Value

The P -value represents the probability of observing the test results, or more extreme results, under the null hypothesis. In this case, a P -value of 0.2743 indicates that there is a 27.43% chance of obtaining a sample proportion of winners at least as extreme as 0.53, assuming the null hypothesis is true.
03

- Evaluate the P-Value Against the Significance Level

Typically, a significance level ( 伪) of 0.05 is used to decide whether to reject the null hypothesis. If the P -value is greater than 伪, we do not reject the null hypothesis. Here, the P -value (0.2743) is greater than 0.05.
04

- Draw a Conclusion

Since the P -value is greater than the significance level of 0.05, we do not reject the null hypothesis. This means there is not enough evidence to support the claim that the dart-picking strategy results in a majority of winners.

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 91影视!

Key Concepts

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

P-Value Interpretation
The P-value is a key aspect of hypothesis testing. It tells us the probability of observing our data, or something more extreme, if the null hypothesis is true.

In our example, the P-value is 0.2743. This means that if the dart-picking strategy had no effect (i.e., the true proportion of winners is 0.5), there is a 27.43% chance of randomly selecting a sample with 53 or more winners out of 100.

When we evaluate a P-value, we compare it to our chosen significance level, denoted as \( \alpha \). If the P-value is lower than this threshold, we have enough evidence to reject the null hypothesis. Otherwise, we do not reject it.
Null and Alternative Hypotheses
Hypothesis testing begins with formulating two contrasting hypotheses: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_1 \)).

In this example,
the null hypothesis is \( H_0: p = 0.5 \), indicating that the proportion of winning companies is 0.5;
while the alternative hypothesis is \( H_1: p > 0.5 \), suggesting that the strategy results in more than 50% winners.

These hypotheses set the stage for the test. The null hypothesis represents a default position that there is no effect or no difference. The alternative hypothesis, on the other hand, reflects the claim we are testing for.

By conducting a hypothesis test and using the observed P-value, we can determine whether the data provides sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
Significance Level (Alpha)
The significance level is a critical value that helps us decide if a result is statistically significant. It is denoted as \( \alpha \) and typically set at 0.05.

In our dart-picking example, the significance level \( \alpha \) is 0.05. This means we are willing to accept a 5% chance of concluding that the strategy results in a majority of winners when it actually does not.

We compare our P-value to \( \alpha \) to make a decision. If the P-value is less than \( \alpha \), we reject the null hypothesis, suggesting that the observed effect is statistically significant.
In this case, the P-value of 0.2743 is greater than 0.05, so we do not reject the null hypothesis.

This indicates that the observed proportion of 53 winners is not significantly higher than 50%, meaning there's not enough evidence to prove that the dart-picking strategy is effective.

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

Explain the difference between statistical significance and practical significance.

Based on historical birthing records, the proportion of males born worldwide is \(0.51 .\) In other words, the commonly held belief that boys are just as likely as girls is false. Systematic lupus erythematosus (SLE), or lupus for short, is a disease in which one's immune system attacks healthy cells and tissue by mistake. It is well known that lupus tends to exist more in females than in males Researchers wondered, however, if families with a child who had lupus had a lower ratio of males to females than the general population. If this were true, it would suggest that something happens during conception that causes males to be conceived at a lower rate when the SLE gene is present. To determine if this hypothesis is true, the researchers obtained records of families with a child who had SLE A total of 23 males and 79 females were found to have SLE. The 23 males with SL \(E\) had \(a\) total of 23 male siblings and 22 female siblings The 79 females with SLE had a total of 69 male siblings and 80 female siblings Source L.N. Moorthy, M.G.E. Peterson, K.B. Onel, and T.J.A. Lehman. "Do Children with Lupus Have Fewer Male Siblings" Laprus 2008 \(17: 128-131,2008\) (a) Explain why this is an observational study. (b) Is the study retrospective or prospective? Why? (c) There are a total of \(23+69=92\) male siblings in the study How many female siblings are in the study? (d) Draw a relative frequency bar graph of gender of the siblings. (e) Find a point estimate for the proportion of male siblings in families where one of the children has SLE (f) Does the sample evidence suggest that the proportion of male siblings in families where one of the children has SLE is less than 0.51 , the accepted proportion of males born in the general population? Use the \(\alpha=0.05\) level of significance. (g) Construct a \(95 \%\) confidence interval for the proportion of male siblings in a family where one of the children has SLE.

Simulation Simulate drawing 100 simple random samples of size \(n=15\) from a population that is normally distributed with mean 100 and standard deviation 15 . (a) Test the null hypothesis \(H_{0}: \mu=100\) versus \(H_{1}: \mu \neq 100\) for each of the 100 simple random samples. (b) If we test this hypothesis at the \(\alpha=0.05\) level of significance, how many of the 100 samples would you expect to result in a Type I error? (c) Count the number of samples that lead to a rejection of the null hypothesis. Is it close to the expected value determined in part (b)? (d) Describe how we know that a rejection of the null hypothesis results in making a Type I error in this situation.

Ready for College? The ACT is a college entrance exam. ACT has determined that a score of 22 on the mathematics portion of the ACT suggests that a student is ready for college-level mathematics. To achieve this goal, ACT recommends that students take a core curriculum of math courses: Algebra I, Algebra II, and Geometry. Suppose a random sample of 200 students who completed this core set of courses results in a mean ACT math score of 22.6 with a standard deviation of \(3.9 .\) Do these results suggest that students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 22 on the math portion of the ACT? (a) State the appropriate null and alternative hypotheses. (b) Verify that the requirements to perform the test using the \(t\) -distribution are satisfied. (c) Use the classical or \(P\) -value approach at the \(\alpha=0.05\) level of significance to test the hypotheses in part (a). (d) Write a conclusion based on your results to part (c).

A manufacturer of high-strength, lowalloy steel beams requires that the standard deviation of yield strength not exceed 7000 pounds per square inch (psi). The quality-control manager selected a sample of 20 steel beams and measured their yield strength. The standard deviation of the sample was 7500 psi. Assume that yield strengths are normally distributed. Does the evidence suggest that the standard deviation of yield strength exceeds 7000 psi at the \(\alpha=0.01\) level of significance?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation
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.