/*! 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 69 Exercises 4.67 to 4.70 give a p-... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Exercises 4.67 to 4.70 give a p-value. State the conclusion of the test based on this p-value in terms of " Reject \(H_{0} "\) or "Do not reject \(H_{0} "\) if we use a \(5 \%\) significance level. $$ \text { p-value }=0.2531 $$

Short Answer

Expert verified
Do not reject \(H_{0}\)

Step by step solution

01

Compare the p-value to the significance level

Firstly, compare the given p-value to the significance level. The p-value for this test is 0.2531 and the given significance level (\(\alpha\)) is 0.05.
02

Making the decision to reject or not reject the null hypothesis

If the p-value is less than or equal to the significance level (\(0.2531 \leq 0.05\)), reject \(H_{0}\). If the p-value is more than the significance level (\(0.2531 > 0.05\)), do not reject \(H_{0}\). In this case, as 0.2531 is greater than 0.05, the conclusion is to not reject the null hypothesis.

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.

Understanding the Significance Level
The significance level, often denoted by \( \alpha \), plays a pivotal role in hypothesis testing. It acts like a threshold that determines the degree of confidence you want in your results.
The most commonly used significance level is 5% or 0.05.
This means that there is a 5% risk of rejecting the null hypothesis when it is actually true. When you set a significance level:
  • You're deciding how much evidence you require to reject the null hypothesis.
  • You're balancing the cost of a Type I error (false positive) with finding an effect if there is one.
In practice, researchers choose a significance level based on the context and consequences of their decision. With a 5% level, you're usually aiming to confidently claim a result as significant, without too much risk of making a wrong decision.
Decoding the P-Value
The p-value is a fundamental concept in statistical testing. It tells you how probable it is to observe your data, or something more extreme, assuming the null hypothesis is true.
In simple terms, it provides a measure of the evidence against the null hypothesis. Here's how p-values work:
  • A small p-value (typically ≤ 0.05) suggests rejecting the null hypothesis, indicating strong evidence against it.
  • A large p-value (> 0.05) implies weak evidence against the null, supporting not rejecting it.
In our example, the p-value was 0.2531. This value is significantly larger than the commonly used significance level of 0.05. Therefore, it suggests insufficient evidence to reject the null hypothesis.
Explaining the Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a key part of hypothesis testing. It is a statement that implies no effect or no difference, and serves as a starting point for testing.Why is it so important?
  • It provides a baseline that can be tested statistically.
  • Allows you to determine if your observed data can be due to random chance.
When you're conducting a hypothesis test, your primary goal is to decide whether you should reject the null hypothesis. In many cases, rejecting \( H_0 \) suggests that there is an effect or difference that is statistically significant. In our case, the p-value was not small enough to reach this conclusion, so you would "Do not reject \( H_0 \)". This means the data did not provide strong enough evidence to support an alternative claim.

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

Desipramine vs Placebo in Cocaine Addiction In this exercise, we see that it is possible to use counts instead of proportions in testing a categorical variable. Data 4.7 describes an experiment to investigate the effectiveness of the two drugs desipramine and lithium in the treatment of cocaine addiction. The results of the study are summarized in Table 4.9 on page \(267 .\) The comparison of lithium to the placebo is the subject of Example \(4.29 .\) In this exercise, we test the success of desipramine against a placebo using a different statistic than that used in Example \(4.29 .\) Let \(p_{d}\) and \(p_{c}\) be the proportion of patients who relapse in the desipramine group and the control group, respectively. We are testing whether desipramine has a lower relapse rate than a placebo. (a) What are the null and alternative hypotheses? (b) From Table 4.9 we see that 20 of the 24 placebo patients relapsed, while 10 of the 24 desipramine patients relapsed. The observed difference in relapses for our sample is $$ \begin{aligned} D &=\text { desipramine relapses }-\text { placebo relapses } \\ &=10-20=-10 \end{aligned} $$ If we use this difference in number of relapses as our sample statistic, where will the randomization distribution be centered? Why? (c) If the null hypothesis is true (and desipramine has no effect beyond a placebo), we imagine that the 48 patients have the same relapse behavior regardless of which group they are in. We create the randomization distribution by simulating lots of random assignments of patients to the two groups and computing the difference in number of desipramine minus placebo relapses for each assignment. Describe how you could use index cards to create one simulated sample. How many cards do you need? What will you put on them? What will you do with them?

Do iPads Help Kindergartners Learn: A Series of Tests Exercise 4.163 introduces a study in which half of the kindergarten classes in a school district are randomly assigned to receive iPads. We learn that the results are significant at the \(5 \%\) level (the mean for the iPad group is significantly higher than for the control group) for the results on the HRSIW subtest. In fact, the HRSIW subtest was one of 10 subtests and the results were not significant for the other 9 tests. Explain, using the problem of multiple tests, why we might want to hesitate before we run out to buy iPads for all kindergartners based on the results of this study.

Classroom Games Two professors \(^{18}\) at the University of Arizona were interested in whether having students actually play a game would help them analyze theoretical properties of the game. The professors performed an experiment in which students played one of two games before coming to a class where both games were discussed. Students were randomly assigned to which of the two games they played, which we'll call Game 1 and Game \(2 .\) On a later exam, students were asked to solve problems involving both games, with Question 1 referring to Game 1 and Question 2 referring to Game 2 . When comparing the performance of the two groups on the exam question related to Game 1 , they suspected that the mean for students who had played Game 1 ( \(\mu_{1}\) ) would be higher than the mean for the other students \(\mu_{2},\) so they considered the hypotheses \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) (a) The paper states: "test of difference in means results in a p-value of \(0.7619 . "\) Do you think this provides sufficient evidence to conclude that playing Game 1 helped student performance on that exam question? Explain. (b) If they were to repeat this experiment 1000 times, and there really is no effect from playing the game, roughly how many times would you expect the results to be as extreme as those observed in the actual study? (c) When testing a difference in mean performance between the two groups on exam Question 2 related to Game 2 (so now the alternative is reversed to be \(H_{a}: \mu_{1}<\mu_{2}\) where \(\mu_{1}\) and \(\mu_{2}\) represent the mean on Question 2 for the respective groups), they computed a p-value of \(0.5490 .\) Explain what it means (in the context of this problem) for both p-values to be greater than \(0.5 .\)

In Exercises 4.150 to \(4.152,\) a confidence interval for a sample is given, followed by several hypotheses to test using that sample. In each case, use the confidence interval to give a conclusion of the test (if possible) and also state the significance level you are using. A \(90 \%\) confidence interval for \(p_{1}-p_{2}: 0.07\) to 0.18 (a) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1} \neq p_{2}\) (b) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) (c) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}

In Exercises 4.112 to \(4.116,\) the null and alternative hypotheses for a test are given as well as some information about the actual sample(s) and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. In addition, indicate whether the test is a left-tail test, a right-tail test, or a twotailed test. Hypotheses: \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) Sample: \(\hat{p}_{1}=0.3, n_{1}=20\) and \(\hat{p}_{2}=0.167, n_{2}=12\) Randomization statistic \(=\hat{p}_{1}-\hat{p}_{2}\)

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.