/*! 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 50 The null hypothesis on true/fals... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The null hypothesis on true/false tests is that the student is guessing, and the proportion of right answers is \(0.50 .\) A student taking a five-question true/false quiz gets 4 right out of 5 . She says that this shows that she knows the material, because the one-tailed p-value from the one-proportion \(z\) -test is \(0.090\), and she is using a significance level of \(0.10 .\) What is wrong with her approach?

Short Answer

Expert verified
The issue is with the interpretation of the p-value and the setup of the alternative hypothesis. A p-value less than the chosen significance level just evidences against the specified null hypothesis, not proving that she knows the material. To make such a claim, she needs to set up a specific alternative hypothesis and then perform a suitable statistical test to provide evidence for it.

Step by step solution

01

Understanding the Hypothesis

The null hypothesis in this context would be that the student is merely guessing the answers, hence the probability of a correct answer being 0.50. This null hypothesis is established with the assumption that the student does not have knowledge of the material.
02

Assessing the Test Results

The student gets 4 out of 5 questions correct, indicating a success rate of 0.80, which is much higher than the hypothesized 0.50. This forms the basis of the student's argument.
03

Evaluating the p-value

The p-value of 0.090 means that there's a 9% chance of obtaining a success rate as extreme as 0.80 or higher, given that the null hypothesis is true (i.e., the student is merely guessing).
04

Interpreting the Significance Level

The student has chosen a significance level of 0.10 (10%), meaning she is willing to accept a 10% chance of rejecting the null hypothesis even if it is true (Type I error). A lower p-value would provide stronger evidence against the null hypothesis.
05

Identifying the Error

The issue here is with the interpretation of the p-value and the setup of the alternative hypothesis. Even though the p-value 0.090 is less than the significance level of 0.10, rejecting the null hypothesis doesn't necessarily prove that she knows the material. It only provides evidence against the hypothesis that she's just guessing the answers. To claim she knows the material, the student needs to set up a specific alternative hypothesis, i.e., She knows the material so her success rate would be significantly higher than 0.50. Then perform a suitable statistical test to provide evidence for this claim.

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.

Hypothesis Testing
When we engage in hypothesis testing, we're essentially weighing evidence to assess the validity of a claim. The null hypothesis, often denoted as H0, is the default statement we test against; it usually asserts 'no effect' or 'no difference.' In the context of true/false tests, the null hypothesis assumes a student is guessing, with a 50% chance of getting the right answer - represented mathematically as p = 0.50. The alternative hypothesis, or Ha, represents what we believe could be true if the null hypothesis is indeed false. This might, for example, assert a student's proportion of correct answers is significantly different from 0.50.

The process of hypothesis testing involves selecting a suitable statistical test based on our data and hypothesis, calculating a test statistic, and then comparing this to critical values or calculating a p-value to gauge the evidence against the null hypothesis. If the evidence is strong enough, we may reject the null hypothesis in favor of the alternative one. However, it's crucial to note that rejecting the null hypothesis doesn't confirm the alternative hypothesis; it simply suggests there's enough evidence to consider it more seriously.
One-Proportion Z-Test
The one-proportion z-test is a statistical tool used when we want to see if the proportion from our sample, , is significantly different from a hypothesized population proportion, p0. In our scenario, a student's success rate of 0.80 (4 out of 5 right answers) is being compared to the hypothesized rate of 0.50 using this z-test.

The test's formula is given by z = (p̂ - p0) / sqrt[(p0*(1-p0))/n], where is the sample proportion, p0 is the hypothesized proportion, and n is the sample size. This calculation provides a z-score, reflecting how many standard deviations the sample proportion lies from the hypothesized proportion. We then consult the standard normal distribution to determine how likely this result would be if the null hypothesis were true.
P-Value Interpretation
The p-value is a crucial concept in hypothesis testing. It quantifies the probability of observing a result as extreme as, or more extreme than, the one obtained from the experiment if the null hypothesis were true. Essentially, it's a measure of the evidence against the null hypothesis: a smaller p-value suggests stronger evidence.

Let's unpack the student's situation. With a p-value of 0.090, there's a 9% chance that a student could achieve a proportion of correct responses as high as 0.80 by guessing alone. Often, if the p-value is smaller than the chosen significance level, we might reject the null hypothesis. However, the interpretation should always be cautious, as a p-value does not measure the probability that the null hypothesis is true or false, nor does it indicate the magnitude of an effect.
Significance Level
The significance level, denoted by α, is a threshold set before data collection that determines how much evidence we require to reject the null hypothesis. Common significance levels are 0.05 (5%), 0.01 (1%), and 0.10 (10%). Choosing a significance level is a balance—lower levels like 0.01 reduce the chance of incorrectly rejecting the null hypothesis (Type I error), but they can increase the chance of failing to reject a false null hypothesis (Type II error).

In our exercise, the student used a significance level of 0.10. This means she is willing to accept a 10% probability of committing a Type I error. While her p-value of 0.090 is below 0.10, suggesting that her result is statistically significant, caution is needed in interpretation. A proper alternative hypothesis and a clear understanding of the limitations of statistical outputs are essential to avoid misconceptions and to draw accurate conclusions from tests.

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

In the Pew Research social media survey, television viewers were asked if it would be very hard to give up watching television. In \(2002,38 \%\) responded yes. In \(2018,31 \%\) said it would be very hard to give up watching television. a. Assume that both polls used samples of 200 people. Do a test to see whether the proportion of people who reported it would be very hard to give up watching television was significantly different in 2002 and 2018 using a \(0.05\) significance level. b. Repeat the problem, now assuming the sample sizes were both 2000 . (The actual sample size in 2018 was \(2002 .\) ) c. Comment on the effect of different sample sizes on the p-value and on the conclusion.

Choosing a Test and Giving the Hypotheses Give the null and alternative hypotheses for each test, and state whether a one-proportion z-test or a two- proportion z-test would be appropriate. a. You test a person to see whether he can tell tap water from bottled water. You give him 20 sips selected randomly (half from tap water and half from bottled water) and record the proportion he gets correct to test the hypothesis. b. You test a random sample of students at your college who stand on one foot with their eyes closed and determine who can stand for at least 10 seconds, comparing athletes and nonathletes.

Suppose you wanted to test the claim that the majority of U.S. voters are satisfied with the government response to the opioid crisis. State the null and alternative hypotheses you would use in both words and symbols.

According to a 2018 survey by Timex reported in Shape magazine, \(73 \%\) of Americans report working out one or more times each week. A nutritionist is interested in whether this percentage has increased. A random sample of 200 Americans found 160 reported working out one or more times each week. Carry out the first two steps of a hypothesis test to determine whether the proportion has increased. Explain how you would fill in the required TI calculator entries for \(p_{0}, x\), and \(n .\)

Choosing a Test and Naming the Population(s) For each of the following, state whether a one-proportion \(z\) -test or a two-proportion \(z\) -test would be appropriate, and name the population(s). a. A polling agency takes a random sample of voters in California to determine if a ballot proposition will pass. b. A researcher asks a random sample of residents from coastal states and a random sample of residents of non-coastal states whether they favor increased offshore oil drilling. The researcher wants to determine if there is a difference in the proportion of residents who support off-shore drilling in the two regions.

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.