/*! 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 1 A coin is thrown independently 1... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 1

A coin is thrown independently 10 times to test the hypothesis that the probability of heads is \(\frac{1}{2}\) versus the alternative that the probability is not \(\frac{1}{2} .\) The test rejects if either 0 or 10 heads are observed. a. What is the significance level of the test? b. If in fact the probability of heads is \(.1,\) what is the power of the test?

Short Answer

Expert verified
a. \( \frac{1}{512} \) b. \( \approx 0.3487 \)

Step by step solution

01

Understanding the Significance Level

To determine the significance level of the test, we need to calculate the probability of rejecting the null hypothesis when it is true. The null hypothesis here is that the probability of heads is \( \frac{1}{2} \). The test rejects the null hypothesis if we observe either 0 heads or 10 heads in 10 coin tosses.
02

Calculating Significance Level

Under the null hypothesis, the number of heads follows a binomial distribution with \( n = 10 \) and \( p = \frac{1}{2} \). The test's rejection region is \{0, 10\}. Using the binomial probability formula, we compute:\[P(X = 0) = \binom{10}{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \P(X = 10) = \binom{10}{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{0} = \frac{1}{1024} \\text{Thus, the significance level is } \frac{1}{1024} + \frac{1}{1024} = \frac{2}{1024} = \frac{1}{512}.\]
03

Understanding Power of the Test

The power of the test is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true. In this problem, the alternative hypothesis is that the probability of heads is \( 0.1 \). We need to find the probability of observing 0 or 10 heads when \( p = 0.1 \).
04

Calculating Power of the Test

With the alternative hypothesis \( p = 0.1 \), the rejection region is \{0, 10\}. We calculate the probabilities:\[P(X = 0) = \binom{10}{0}(0.1)^0(0.9)^{10} = (0.9)^{10} \P(X = 10) = \binom{10}{10}(0.1)^{10}(0.9)^0 = (0.1)^{10} \\text{Therefore, the power of the test is } (0.9)^{10} + (0.1)^{10} \approx 0.3487 + 0.0000000001 \approx 0.3487.\]

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 91影视!

Key Concepts

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

Understanding Binomial Distribution
In hypothesis testing, the binomial distribution is a powerful tool used to predict the probability of a certain number of successes in a fixed number of independent trials. Each trial must result in one of two possible outcomes. In this exercise, tossing a coin ten times to check if the probability of getting heads is indeed \( \frac{1}{2} \) is modeled using a binomial distribution.

Key properties:
  • The number of trials \( n \) is fixed, which in this case is 10.
  • Each trial is independent, meaning the result of one coin toss does not affect any other toss.
  • The probability \( p \) of a success, which here means getting heads, is assumed to be constant across trials.
The probability mass function (PMF) of the binomial distribution is given by:\[ P(X = k) = \binom{n}{k}p^k(1-p)^{n-k} \]Where \( k \) is the number of successes. In our case, the formula helps to calculate how likely it is to get exactly 0 or 10 heads when the coin's fairness is questioned under these conditions.
Significance Level
The significance level of a test, often denoted by \( \alpha \), is a critical concept in hypothesis testing. It measures the likelihood of rejecting a true null hypothesis. In other words, it tells us how often we might be wrong to dismiss a hypothesis when it is actually correct. The smaller the significance level, the more reliable the test is.

In the context of our exercise, the null hypothesis states that the coin is fair with a probability of heads being \( \frac{1}{2} \). The test is set up to reject this hypothesis only if all coin tosses result in 0 or 10 heads. To find the significance level, we calculate:
  • \( P(X = 0) \) for getting 0 heads with fair coin odds.
  • \( P(X = 10) \) for seeing all 10 as heads.
Adding these probabilities gives us \( \frac{1}{1024} + \frac{1}{1024} = \frac{1}{512} \), indicating a relatively low chance of incorrectly discarding a fair coin.
Power of the Test
The power of a test is a measure of its effectiveness in identifying when a false null hypothesis is actually incorrect. It is defined as the probability that the test will successfully reject a false null hypothesis in favor of the alternative hypothesis when it is true.

To find the power, we consider the alternative hypothesis claiming that the probability of heads is \( 0.1 \) instead of \( \frac{1}{2} \). For our coin toss test, with a rejection criteria of observing 0 or 10 heads, the power is calculated by:
  • Finding \( P(X = 0) \) when the probability of heads is actually \( 0.1 \), resulting in most tosses being tails.
  • Finding \( P(X = 10) \) when the chance of getting heads is \( 0.1 \), reflecting an extremely rare event of all heads.
Adding these probabilities provides an approximate power of \( 0.3487 \), indicating roughly a 34.87% chance of the test correctly identifying when the coin isn't fair.

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

It has been suggested that dying people may be able to postpone their death until after an important occasion, such as a wedding or birthday. Phillips and King (1988) studied the patterns of death surrounding Passover, an important Jewish holiday, in California during the years \(1966-1984\). They compared the number of deaths during the week before Passover to the number of deaths during the week after Passover for 1919 people who had Jewish surnames. Of these, 922 occurred in the week before Passover and \(997,\) in the week after Passover. The significance of this discrepancy can be assessed by statistical calculations. We can think of the counts before and after as constituting a table with two cells. If there is no holiday effect, then a death has probability \(\frac{1}{2}\) of falling in each cell. Thus, in order to show that there is a holiday effect, it is necessary to show that this simple model does not fit the data. Test the goodness of fit of the model by Pearson's \(X^{2}\) test or by a likelihood ratio test. Repeat this analysis for a group of males of Chinese and Japanese ancestry, of whom 418 died in the week before Passover and 434 died in the week after. What is the relevance of this latter analysis to the former?

Let \(X_{1}, \ldots, X_{25}\) be a sample from a normal distribution having a variance of 100. Find the rejection region for a test at level \(\alpha=.10\) of \(H_{0}: \mu=0\) versus \(H_{A}: \mu=1.5 .\) What is the power of the test? Repeat for \(\alpha=.01.\)

True or false: a. The generalized likelihood ratio statistic \(\Lambda\) is always less than or equal to \(1 .\) b. If the \(p\) -value is \(.03,\) the corresponding test will reject at the significance level .02. c. If a test rejects at significance level.06, then the \(p\) -value is less than or equal to .06. d. The \(p\) -value of a test is the probability that the null hypothesis is correct. e. In testing a simple versus simple hypothesis via the likelihood ratio, the \(p\) -value equals the likelihood ratio. f. If a chi-square test statistic with 4 degrees of freedom has a value of \(8.5,\) the \(p\) -value is less than .05.

a. In \(1965,\) a newspaper carried a story about a high school student who reported getting 9207 heads and 8743 tails in 17,950 coin tosses. Is this a significant discrepancy from the null hypothesis \(H_{0}: p=\frac{1}{2} ?\) b. Jack Youden, a statistician at the National Bureau of Standards, contacted the student and asked him exactly how he had performed the experiment (Youden 1974). To save time, the student had tossed groups of five coins at a time, and a younger brother had recorded the results, shown in the following table: $$\begin{array}{cc} \hline \text { Number of Heads } & \text { Frequency } \\\ \hline 0 & 100 \\ 1 & 524 \\ 2 & 1080 \\ 3 & 1126 \\ 4 & 655 \\ 5 & 105 \\\ \hline \end{array}$$ Are the data consistent with the hypothesis that all the coins were fair \(\left(p=\frac{1}{2}\right) ?\) c. Are the data consistent with the hypothesis that all five coins had the same probability of heads but that this probability was not necessarily \(\frac{1}{2} ?\) (Hint: Use the binomial distribution.)

Consider testing goodness of fit for a multinomial distribution with two cells. Denote the number of observations in each cell by \(X_{1}\) and \(X_{2}\) and let the hypothesized probabilities be \(p_{1}\) and \(p_{2} .\) Pearson's chi- square statistic is equal to $$\sum_{i=1}^{2} \frac{\left(X_{i}-n p_{i}\right)^{2}}{n p_{i}}$$ Show that this may be expressed as $$\frac{\left(X_{1}-n p_{1}\right)^{2}}{n p_{1}\left(1-p_{1}\right)}$$ Because \(X_{1}\) is binomially distributed, the following holds approximately under the null hypothesis: $$\frac{X_{1}-n p_{1}}{\sqrt{n p_{1}\left(1-p_{1}\right)}} \sim N(0,1)$$ Thus, the square of the quantity on the left-hand side is approximately distributed as a chi-square random variable with 1 degree of freedom.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation
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.