/*! 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 9 A person who claims to be psychi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A person who claims to be psychic says that the probability \(p\) that he can correctly predict the outcome of the roll of a die in another room is greater than \(1 / 6,\) the value that applies with random guessing. If we want to test this claim, we could use the data from an experiment in which he predicts the outcomes for \(n\) rolls of the die. State hypotheses for a significance test, letting the alternative hypothesis reflect the psychic's claim.

Short Answer

Expert verified
Test \(H_0: p = \frac{1}{6}\) against \(H_1: p > \frac{1}{6}\).

Step by step solution

01

Understand the Problem

We need to determine whether a person, who claims to be psychic, is more likely to predict a die roll than random guessing, which has a probability \( p = \frac{1}{6} \). Random guessing is our null hypothesis, and the psychic's claim is our alternative hypothesis.
02

State the Null Hypothesis (H0)

The null hypothesis is the assumption that there is no effect or difference. In this context, it states that the probability that the person can correctly predict the outcome of a die roll is equal to the probability of a random guess: \( H_0: p = \frac{1}{6} \).
03

State the Alternative Hypothesis (H1)

The alternative hypothesis reflects the claim that the person is psychic and can predict die rolls better than random guessing. This hypothesis states that the probability is greater than that of a random guess: \( H_1: p > \frac{1}{6} \).
04

Conclusion

In a significance test, we test the null hypothesis against the alternative hypothesis. If we have enough evidence to reject the null hypothesis, we might support the psychic's claim of having a higher probability \( p > \frac{1}{6} \) to predict the outcomes.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis plays a critical role. It represents a default position or a statement that there is no effect or no difference. In the context of the exercise, the null hypothesis (\( H_0 \)) is concerned with the probability of predicting a die roll outcome being equal to random guessing. A standard die has six sides, making the probability of guessing correctly as \( \frac{1}{6} \).The idea of a null hypothesis is to provide something testable. We assume it to be true until proven otherwise. Essentially, it acts as the status quo, and while it is often a statement of 'no difference' or 'no effect', it can be more complex. However, let's stick to the simplicity of our exercise. To proceed with the test, the aim is to collect data, assess the evidence, and determine whether this presumption can be overturned. The calculation and comparison within a significance test will eventually either retain the null hypothesis or reject it based on statistical evidence collected from the experiment.
Alternative Hypothesis
The alternative hypothesis (\( H_1 \)) is the claim that challenges the null hypothesis. In this context, it represents the psychic's claim of predicting outcomes better than random guessing. Unlike the null hypothesis, which states that the probability \( p \)is equal to \( \frac{1}{6} \), the alternative hypothesis proposes that this probability is greater.The alternative hypothesis is what researchers generally hope to prove true. It suggests that there is a real effect or difference, which, in our example, would mean the psychic does have an ability that exceeds random chance.It is crucial to note that hypothesis testing does not inherently 'prove' an alternative hypothesis. Instead, it assesses whether there is sufficient statistical evidence to reject the null hypothesis in favor of the alternative. This potential shift from the status quo to accepting the alternative is what hypothesis tests aim to investigate through data and analysis.
Significance Test
A significance test is a statistical tool used to decide whether the evidence in your data is strong enough to support a specific claim. This process addresses the core question: Is there enough evidence to reject the null hypothesis? The significance test involves calculating a p-value. This value tells us the probability of observing data as extreme as, or even more extreme than, those observed under the assumption that the null hypothesis is true. If this p-value is small (typically less than a predefined threshold, such as 0.05), it implies that such an extreme outcome is unlikely under the null hypothesis and suggests that we reject the null hypothesis in favor of the alternative. Performing a significance test:
  • Start by assuming the null hypothesis is true.
  • Collect your data and analyze it.
  • Determine the p-value based on this analysis.
  • Compare this p-value to your significance level (α). If p-value < α, reject the null hypothesis.
Conducting a significance test carefully helps ensure that whatever conclusion we make about the psychic’s abilities is based on sound statistical reasoning rather than random chance or error. This is fundamental in not just hypothesis testing, but in making reliable decisions in scientific research.

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Most popular questions from this chapter

Get P-value from \(z\) For a test of \(\mathrm{H}_{0}: p=0.50,\) the \(z\) test statistic equals 1.04 a. Find the P-value for \(\mathrm{H}_{a} \cdot p>0.50\). b. Find the P-value for \(\mathrm{H}_{a}: p \neq 0.50\). c. Find the P-value for \(\mathrm{H}_{c}: p<0.50 .\) (Hint: The P-values for the two possible one-sided tests must sum to \(1 .)\) d. Do any of the P-values in part a, part b, or part c give strong evidence against \(\mathrm{H}_{0}\) ? Explain.

Two sampling distributions A study is designed to test \(\mathrm{H}_{0}: p=0.50\) against \(\mathrm{H}_{a}: p>0.50,\) taking a random sample of size \(n=100,\) using significance level \(0.05 .\) a. Show that the rejection region consists of values of \(\hat{p}>0.582\) b. Sketch a single picture that shows (i) the sampling distribution of \(\hat{p}\) when \(\mathrm{H}_{0}\) is true and (ii) the sampling distribution of \(\hat{p}\) when \(p=0.60 .\) Label each sampling distribution with its mean and standard error, and highlight the rejection region. c. Find \(\mathrm{P}\) (Type II error) when \(p=0.60\).

Practical significance A study considers if the mean score on a college entrance exam for students in 2010 is any different from the mean score of 500 for students who took the same exam in \(1985 .\) Let \(\mu\) represent the mean score for all students who took the exam in \(2010 .\) For a random sample of 25,000 students who took the exam in $$ 2010, \bar{x}=498 \text { and } s=100 . $$ a. Show that the test statistic is \(t=-3.16\). b. Find the P-value for testing \(\mathrm{H}_{0}: \mu=500\) against \(\mathrm{H}_{a^{2}}: \mu \neq 500\) c. Explain why the test result is statistically significant but not practically significant.

Testing a headache remedy Studies that compare treatments for chronic medical conditions such as headaches can use the same subjects for each treatment. This type of study is commonly referred to as a crossover design. With a crossover design, each person crosses over from using one treatment to another during the study. One such study considered a drug (a pill called Sumatriptan) for treating migraine headaches in a convenience sample of children. The study observed each of 30 children at two times when he or she had a migraine headache. The child received the drug at one time and a placebo at the other time. The order of treatment was randomized and the study was double-blind. For each child, the response was whether the drug or the placebo provided better pain relief. Let \(p\) denote the proportion of children having better pain relief with the drug, in the population of children who suffer periodically from migraine headaches. Can you conclude that \(p>0.50,\) with more than half of the population getting better pain relief with the drug, or that \(p<0.50\), with less than half getting better pain relief with the drug (i.e., the placebo being better)? Of the 30 children, 22 had more pain relief with the drug and 8 had more pain relief with the placebo. a. For testing \(\mathrm{H}_{0}: p=0.50\) against \(\mathrm{H}_{a}: p \neq 0.50\), show that the test statistic \(z=2.56\). b. Show that the P-value is 0.01 . Interpret. c. Check the assumptions needed for this test, and discuss the limitations due to using a convenience sample rather than a random sample.

Examples of hypotheses Give an example of a null hypothesis and an alternative hypothesis about a (a) population proportion and (b) population mean.

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