/*! 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 118 Exercises 4.117 to 4.122 give nu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Exercises 4.117 to 4.122 give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p<0.5\) Sample data: \(\hat{p}=38 / 100=0.38\) with \(n=100\)

Short Answer

Expert verified
The p-value could either be less-than or greater-than 0.05. If less-than 0.05, we would reject the null hypothesis and accept the alternative hypothesis which would suggest \( p < 0.5 \). If it is greater-than 0.05, we do not reject the null hypothesis hence cannot conclude that \( p < 0.5 \).

Step by step solution

01

State the Hypotheses

The null hypothesis ( \(H_{0}\) ) is that the population proportion \( p = 0.5 \). On the contrary, the alternative hypothesis ( \(H_{a}\) ) states that \( p < 0.5 \). Hypotheses are essential in hypothesis testing because they are what are being tested.
02

Calculate Sample Proportion

Given the sample data, the sample proportion \( \hat{p} = 38 / 100 = 0.38 \). Note that the calculation of the sample proportion is necessary for the next steps.
03

Perform the Test Using StatKey

Typically, to perform the test using a tool like StatKey, select 'Test for Single Proportion' and 'Edit Data' to input the sample data. You then set \( H_{0} \) to 0.5 and lower-tailed because the alternative hypothesis is \( p < 0.5 \). After inputting all parameters, execute the test.
04

Calculate the p-value

The result will provide the randomization distribution and the p-value. The p-value is the probability of obtaining the sample data or more extreme, given that the null hypothesis is true. Comparison of the p-value with a set significance level (generally 0.05) will determine if the null hypothesis should be rejected or not.

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 begins with establishing two contrasting statements: the null hypothesis (\(H_{0}\)) and the alternative hypothesis (\(H_{a}\)). The null hypothesis is typically a statement of no effect or no difference, serving as a starting point for statistical analysis. It posits that any observed effect in the data is due to random chance alone. In our exercise, the null hypothesis is given by \(H_{0}: p = 0.5\), positing that the population proportion, or the true rate of a certain outcome within the whole population, is 50%.
The alternative hypothesis (\(H_{a}\)), on the other hand, is a statement that suggests a new effect or difference exists. In contrast to the null, it posits that the observed data is not solely by chance. In the given example, the alternative hypothesis is \(H_{a}: p < 0.5\), which suggests that the actual population proportion is less than 50%. These hypotheses are mutually exclusive and exhaust all possible scenarios.
Population Proportion
The population proportion is a measure that represents the fraction of the population that has a particular characteristic. In statistical testing, it is referred to by the symbol \(p\). When researchers want to make inferences about the population proportion based on sampled data, they calculate the sample proportion, denoted as \(\hat{p}\). The sample proportion is found by dividing the number of individuals in the sample with the desired trait by the total sample size, as shown with the sample data \(\hat{p} = 38 / 100 = 0.38\) from the exercise.
Knowing the sample proportion is crucial because it's the basis for estimating the true population proportion and for determining statistical significance through hypothesis testing.
Randomization Distribution
When conducting hypothesis testing, we need to compare the observed sample statistic to a distribution of what we would expect if the null hypothesis were true. This is known as the randomization distribution, or sometimes the sampling distribution, under the null hypothesis. The randomization distribution is created by simulating many samples under the assumption that the null hypothesis is true. Using software like StatKey, we can generate thousands of randomized samples to create this distribution. The shape and spread of this distribution offer insight into the range of outcomes we might observe purely by random chance, and thus it forms a benchmark against which our actual sample statistic is compared.
p-value Calculation
The p-value is a pivotal concept in hypothesis testing. It quantifies how extreme the observed sample data is, assuming that the null hypothesis is true. Essentially, it is the probability of observing a sample statistic as extreme or more extreme than the sample result, given that there is no real effect (the null is true). Calculating the p-value involves comparing the sample proportion to the randomization distribution. In the given exercise, a p-value would be calculated by determining the proportion of simulated samples that have a proportion less or equal to \(0.38\), assuming the population proportion is \(0.5\). A small p-value indicates that such an extreme result is unlikely due to chance alone, which can lead to rejecting the null hypothesis.
StatKey Statistical Software
StatKey is a web-based software tool designed specifically for teaching and learning statistics. It is intuitive and user-friendly, making it an ideal choice for students performing tasks such as hypothesis testing. Users can easily input their sample data, as in our example, by selecting 'Test for Single Proportion' and then 'Edit Data'. The tool then generates the randomization distribution, calculates the p-value, and provides visual representations of this. Using StatKey allows students to perform complex statistical tests without the need for advanced mathematics or a deep understanding of the underlying algorithms, thus making statistical concepts more accessible and understandable.

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

Exercises 4.117 to 4.122 give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) Sample data: \(\hat{p}=42 / 100=0.42\) with \(n=100\)

A situation is described for a statistical test and some hypothetical sample results are given. In each case: (a) State which of the possible sample results provides the most significant evidence for the claim. (b) State which (if any) of the possible results provide no evidence for the claim. Testing to see if there is evidence that the proportion of US citizens who can name the capital city of Canada is greater than \(0.75 .\) Use the following possible sample results: Sample A: \(\quad 31\) successes out of 40 Sample B: \(\quad 34\) successes out of 40 Sample C: \(\quad 27\) successes out of 40 Sample \(\mathrm{D}: \quad 38\) successes out of 40

Penalty Shots in Soccer A recent article noted that it may be possible to accurately predict which way a penalty-shot kicker in soccer will direct his shot. \({ }^{23}\) The study finds that certain types of body language by a soccer player-called "tells"-can be accurately read to predict whether the ball will go left or right. For a given body movement leading up to the kick, the question is whether there is strong evidence that the proportion of kicks that go right is significantly different from one-half. (a) What are the null and alternative hypotheses in this situation? (b) If sample results for one type of body movement give a p-value of \(0.3184,\) what is the conclusion of the test? Should a goalie learn to distinguish this movement? (c) If sample results for a different type of body movement give a p-value of \(0.0006,\) what is the conclusion of the test? Should a goalie learn to distinguish this movement?

Paul the Octopus In the 2010 World Cup, Paul the Octopus (in a German aquarium) became famous for being correct in all eight of the predictions it made, including predicting Spain over Germany in a semifinal match. Before each game, two containers of food (mussels) were lowered into the octopus's tank. The containers were identical, except for country flags of the opposing teams, one on each container. Whichever container Paul opened was deemed his predicted winner. \(^{32}\) Does Paul have psychic powers? In other words, is an 8 -for-8 record significantly better than just guessing? (a) State the null and alternative hypotheses. (b) Simulate one point in the randomization distribution by flipping a coin eight times and counting the number of heads. Do this five times. Did you get any results as extreme as Paul the Octopus? (c) Why is flipping a coin consistent with assuming the null hypothesis is true?

Flaxseed and Omega-3 Exercise 4.29 on page 234 describes a company that advertises that its milled flaxseed contains, on average, at least \(3800 \mathrm{mg}\) of ALNA, the primary omega-3 fatty acid in flaxseed, per tablespoon. In each case below, which of the standard significance levels, \(1 \%\) or \(5 \%\) or \(10 \%,\) makes the most sense for that situation? (a) The company plans to conduct a test just to double-check that its claim is correct. The company is eager to find evidence that the average amount per tablespoon is greater than 3800 (their alternative hypothesis) and is not really worried about making a mistake. The test is internal to the company and there are unlikely to be any real consequences either way. (b) Suppose, instead, that a consumer organization plans to conduct a test to see if there is evidence against the claim that the product contains at least \(3800 \mathrm{mg}\) per tablespoon. If the organization finds evidence that the advertising claim is false, it will file a lawsuit against the flaxseed company. The organization wants to be very sure that the evidence is strong, since there could be very serious consequences if the company is sued incorrectly.

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.