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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.

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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.

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Most popular questions from this chapter

A 2018 Gallup poll of 3635 randomly selected Facebook users found that 2472 get most of their news about world events on Facebook. Research done in 2013 found that only \(47 \%\) of all Facebook users reported getting their news about world events on Facebook. See page 430 for guidance. a. Does this sample give evidence that the proportion of Facebook users who get their world news on Facebook has changed since 2013 ? Carry out a hypothesis test and use a \(0.05\) significance level. b. After conducting the hypothesis test, a further question one might ask is what proportion of all Facebook users got most of their news about world events on Facebook in 2018 . Use the sample data to construct a \(90 \%\) confidence interval for the population proportion. How does your confidence interval support your hypothesis test conclusion?

Samuel Morse determined that the percentage of \(a\) 's in the English language in the 1800 s was \(8 \%\). A random sample of 600 letters from a current newspaper contained 60 a's. Using the \(0.10\) level of significance, test the hypothesis that the proportion of \(a\) 's in this modern newspaper is \(0.09\).

In 2016 the Harris poll estimated that \(3.3 \%\) of American adults are vegetarian. A nutritionist thinks this rate has increased. The nutritionist samples 150 American adults and finds that 11 are vegetarian. a. What is \(\hat{p}\), the sample proportion of vegetarians? b. What is \(p_{0}\), the hypothetical proportion of vegetarians? c. Find the value of the test statistic. Explain the test statistic in context.

p-Values (Example 11) A researcher carried out a hypothesis test using a two- sided alternative hypothesis. Which of the following \(z\) -scores is associated with the smallest p-value? Explain. i. \(z=0.50\) ii. \(z=1.00\) iii. \(z=2.00\) iv. \(z=3.00\)

A proponent of a new proposition on a ballot wants to know the population percentage of people who support the bill. Suppose a poll is taken, and 580 out of 1000 randomly selected people support the proposition. Should the proponent use a hypothesis test or a confidence interval to answer this question? Explain. If it is a hypothesis test, state the hypotheses and find the test statistic, p-value, and conclusion. Use a \(5 \%\) significance level. If a confidence interval is appropriate, find the approximate \(95 \%\) confidence interval. In both cases, assume that the necessary conditions have been met.

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