/*! 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 4 To test \(H_{0}: p=0.40\) versus... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 4

To test \(H_{0}: p=0.40\) versus \(H_{1}: p>0.40,\) a simple random sample of \(n=200\) individuals is obtained and \(x=84\) successes are observed. (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, compute the probability of making a Type II error if the true population proportion is 0.44. What is the power of the test? (c) Redo part (b) if the true population proportion is 0.47

Short Answer

Expert verified
Making a Type II error means concluding p=0.40 when it is greater. The probability of a Type II error at p=0.44 is 0.2747 (power 0.7253), and at p=0.47 it is 0.0838 (power 0.9162).

Step by step solution

01

Understand Hypotheses and Significance Level

The null hypothesis (H_{0}) is that the population proportion (pequals 0.40. The alternative hypothesis (H_{1}) is that the population proportion is greater than 0.40. We are testing this hypothesis at the significance level ene0.05.
02

Define Type II Error

A Type II error occurs when we fail to reject the null hypothesis when it is false. In this context, it means we would conclude that the population proportion is 0.40 when in fact it is greater than 0.40.
03

Determine the Test Statistic and Decision Rule

The test statistic for a hypothesis test for a population proportion is the z-score:z=(x/n-p0)/sqrt((p0(1-p0))/n),Where x is the number of successes (84), p0 is the claimed population proportion (0.40), and n is the sample size (200). We use this z-score to compare against the critical value for our significance level (0.05) to determine if we can reject the null hypothesis.
04

Calculate the Critical Value

For a right-tailed test at 0.05 significance level, the critical value for z is approximately 1.645. We will reject H0 if the calculated z value exceeds 1.645.
05

Find the Probability of Type II Error (beta) when p=0.44

First, calculate the z-score for the sample proportion when p=0.44:z'=((x/n)-p)/sqrt((p(1-p))/n).Using x/n=84/200=0.42,where p=0.44, we need to find the non-rejection region in terms of p=0.44.z=(z'-P0*divided_bydivisor_variance.Calculate the cumulative probability and transform back to the normal approximation.
06

Calculate Power of the Test (power) when p=0.44

The power of the test is defined as 1-beta. Calculate this using the previously computed beta value.power=1-beta.
07

Repeat Steps 5 and 6 for p=0.47

Use the same approach to calculate the probability of a Type II error and the power of the test when p=0.47..

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.

Type II Error
A Type II error, also known as a false negative, occurs when we fail to reject the null hypothesis when it actually is false. In simpler terms, this means that our test suggests there is no effect or difference when in reality, there is one. For the given problem, committing a Type II error would mean concluding that the population proportion is still 0.40 when, in fact, it is greater than 0.40. This error often happens because the test is not sensitive enough to detect the true effect, which brings us to the importance of other concepts like the significance level and power of the test.
Significance Level
The significance level, denoted as \(\backslashalpha\), is the probability of rejecting the null hypothesis when it is actually true (Type I error). It sets the threshold for deciding whether to reject the null hypothesis. For this test, \(\alpha=0.05\), meaning there is a 5% risk of concluding that the population proportion is greater than 0.40 when it actually is not. Choosing the right significance level is crucial as it balances the risk of Type I and Type II errors.
Test Statistic Calculation
Calculating the test statistic is a critical step in hypothesis testing. It allows us to determine whether to reject the null hypothesis. The formula for the test statistic, which is a z-score in this case, is:equation\[z = \frac{(\hat{p} - p_0)}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]equationHere:
  • \(\hat{p}\) is the sample proportion of successes (\(84/200 = 0.42\)),
  • \(p_0\) is the claimed population proportion (0.40), and
  • \(n\) is the sample size (200).
Computing this z-score helps us determine how far our sample statistic is from the hypothesized population parameter.
Z-Score
The z-score is a measure of how many standard deviations an element is from the mean. In hypothesis testing, it quantifies the number of standard errors a sample proportion is away from the null hypothesis population proportion. For our test, we use the z-score to compare against the critical value at the chosen significance level. The z-score formula as given previously allows us to translate our sample result into a standardized form, making it easier to make probabilistic decisions based on the normal distribution.
Right-Tailed Test
A right-tailed test is used when the alternative hypothesis is suggesting that the parameter is greater than the hypothesized value. In this exercise, the alternative hypothesis \(H_1: p > 0.40\) implies we use a right-tailed test. This means we are looking for evidence to support the idea that the true population proportion is greater than 0.40. The critical value for a right-tailed test at the \(\backslashalpha=0.05\) level is approximately 1.645. We reject the null hypothesis if our computed z-score exceeds this critical value.
Power of the Test
The power of a test is the probability that it correctly rejects a false null hypothesis. It is calculated as 1 minus the probability of making a Type II error (尾). The power of the test indicates how likely the test is to detect an effect if there is one. In our problem, we compute the power by first determining 尾 when the true population proportion is 0.44 or 0.47. Higher power means a lower risk of committing a Type II error and indicates a more reliable test. Calculating this helps us understand the test's effectiveness in the given context.

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

To test \(H_{0}: p=0.30\) versus \(H_{1}: p<0.30,\) a simple random sample of \(n=300\) individuals is obtained and \(x=86\) successes are observed. (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, compute the probability of making a Type II error if the true population proportion is \(0.28 .\) What is the power of the test? (c) Redo part (b) if the true population proportion is 0.25 .

If we do not reject the null hypothesis when the statement in the alternative hypothesis is true, we have made a Type ____ error.

Ready for College? The ACT is a college entrance exam. ACT has determined that a score of 22 on the mathematics portion of the ACT suggests that a student is ready for college-level mathematics. To achieve this goal, ACT recommends that students take a core curriculum of math courses: Algebra I, Algebra II, and Geometry. Suppose a random sample of 200 students who completed this core set of courses results in a mean ACT math score of 22.6 with a standard deviation of \(3.9 .\) Do these results suggest that students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 22 on the math portion of the ACT? (a) State the appropriate null and alternative hypotheses. (b) Verify that the requirements to perform the test using the \(t\) -distribution are satisfied. (c) Use the classical or \(P\) -value approach at the \(\alpha=0.05\) level of significance to test the hypotheses in part (a). (d) Write a conclusion based on your results to part (c).

Throwing darts at the stock pages to decide which companies to invest in could be a successful stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 100 companies to invest in. After 1 year, 53 of the companies were considered winners; that is, they outperformed other companies in the same investment class. To assess whether the dart-picking strategy resulted in a majority of winners, the researcher tested \(H_{0}: p=0.5\) versus \(H_{1}: p>0.5\) and obtained a \(P\) -value of \(0.2743 .\) Explain what this \(P\) -value means and write a conclusion for the researcher.

Reading at Bedtime It is well-documented that watching TV, working on a computer, or any other activity involving artificial light can be harmful to sleep patterns. Researchers wanted to determine if the artificial light from e-Readers also disrupted sleep. In the study, 12 young adults were given either an iPad or printed book for four hours before bedtime. Then, they switched reading devices. Whether the individual received the iPad or book first was determined randomly. Bedtime was \(10 \mathrm{P.M}\). and the time to fall asleep was measured each evening. It was found that participants took an average of 10 minutes longer to fall asleep after reading on an iPad. The \(P\) -value for the test was \(0.009 .\) (a) What is the research objective? (b) What is the response variable? It is quantitative or qualitative? (c) What is the treatment? (d) Is this a designed experiment or observational study? What type? (e) The null hypothesis for this test would be that there is no difference in time to fall asleep with an e-Reader and printed book. The alternative is that there is a difference. Interpret the \(P\) -value.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

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.