/*! 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 84 A poll done for Newsweek found t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 84

A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent鈥檚 survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

Short Answer

Expert verified
No, the data suggests that the true proportion is likely less than 13%.

Step by step solution

01

Define the Null and Alternative Hypotheses

Let the population proportion who have seen or sensed the presence of an angel be denoted by \( p \). The null hypothesis (\( H_0 \)) is that \( p = 0.13 \). The alternative hypothesis (\( H_a \)) is that \( p eq 0.13 \), as we test if the new survey provides enough evidence to doubt the Newsweek poll.
02

Calculate the Sample Proportion

From the contingent鈥檚 survey, out of 76 Americans surveyed, 2 claimed to have seen or sensed the presence of an angel. Thus, the sample proportion \( \hat{p} \) is \( \hat{p} = \frac{2}{76} \approx 0.0263 \).
03

Calculate the Standard Error

The standard error (SE) of \( \hat{p} \) is given by the formula \( SE = \sqrt{\frac{p(1-p)}{n}} \), where \( n \) is the sample size. Substituting the values, \( SE = \sqrt{\frac{0.13 \times (1-0.13)}{76}} \approx 0.0395 \).
04

Compute the Test Statistic

The test statistic (z-score) is computed as \( z = \frac{\hat{p} - p}{SE} \). Substituting the known values, \( z = \frac{0.0263 - 0.13}{0.0395} \approx -2.63 \).
05

Determine the P-Value and Make a Decision

The calculated z-score is -2.63. Consulting a standard normal distribution table, the corresponding p-value for \( z = -2.63 \) is approximately 0.0043. Since the p-value is less than the common significance level of 0.05, we reject the null hypothesis.
06

Conclusion

We have sufficient evidence to doubt the Newsweek poll's claim that 13% of Americans have sensed the presence of an angel, based on the contingent survey.
07

Discuss Possible Reasons for Different Results

1. **Sampling Variability:** Different sample sizes and potentially different demographics could lead to varying results. 2. **Survey Methodology:** The method of data collection (e.g., in-person interviews, surveys, phone calls) may influence people鈥檚 willingness to disclose personal experiences. 3. **Reporting Clarity:** Differences in how questions were phrased or understood might result in variations in responses.

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.

Sample Proportion
In statistical terms, when we collect data from a sample, we are interested in estimating parameters about the whole population. A sample proportion is one such estimate.
In our exercise, we surveyed 76 individuals. From these, 2 people reported having seen or sensed an angel. To find the sample proportion, we use the formula:
  • Sample proportion, denoted as \( \hat{p} \)
  • \( \hat{p} = \frac{\text{number of successes}}{\text{sample size}} = \frac{2}{76} \approx 0.0263 \)
This means that 2.63% of the surveyed participants reported the presence of an angel. This sample proportion is a small piece of the bigger population puzzle.
Null Hypothesis
The null hypothesis is a critical starting point in hypothesis testing. It represents a statement of no effect or no difference.
In our case study, the null hypothesis, denoted as \( H_0 \), assumes the population proportion of people who have seen or sensed an angel is exactly as initially suggested by Newsweek: \( p = 0.13 \).
Essentially, our hypothesis test begins by assuming that the Newsweek claim is correct, unless we find enough evidence to prove otherwise. Our goal is to test whether our survey's results provide such evidence.
P-Value
The p-value helps us decide whether to reject the null hypothesis. It represents the probability of observing a sample statistic as extreme as the test statistic, under the assumption the null hypothesis is true.
In our analysis, we calculated a p-value of approximately 0.0043. This value reflects the likelihood of seeing such a drastic deviation from the expected proportion under \( H_0 \), just by random chance.
  • If the p-value is less than the chosen significance level (usually 0.05), we reject the null hypothesis.
  • Given the low p-value, it indicates the observed sample proportion is unlikely to occur if the Newsweek claim were true.
Hence, we have reason to doubt the initial claim.
Standard Error
The standard error (SE) measures the variability or spread of the sampling distribution of a statistic.
For proportions, this is calculated using the formula:
  • SE = \( \sqrt{\frac{p(1-p)}{n}} \)
  • Where \( p \) is the null hypothesis population proportion, and \( n \) is the sample size.
In our situation, \( SE \approx 0.0395 \).
This small SE suggests that even slight differences between our sample proportion and the null hypothesis proportion can result in a statistically significant finding.
Z-Score
The z-score helps us understand where our sample statistic lies in relation to the population parameter assumed by the null hypothesis.
In the outlined steps, we computed the z-score as:
  • \( z = \frac{\hat{p} - p}{SE} \)
  • For our data: \( z \approx -2.63 \)
A z-score tells us how many standard deviations away our sample proportion is from the null hypothesis proportion. A value of -2.63 implies a substantial deviation to the left, indicating a lower observed proportion than expected.
In practical terms, such a z-score supports rejecting the null hypothesis, implying that the actual proportion is likely different from the Newsweek poll's claim.

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

Instructions: For the following ten exercises, Hypothesis testing: For the following ten exercises, answer each question. a. State the null and alternate hypothesis. b. State the p-value. c. State alpha. d. What is your decision? e. Write a conclusion. f. Answer any other questions asked in the problem. The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the ? = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if ? = 0.01? Assume the amount of summer rainfall follows a normal distribution.

It is believed that Lake Tahoe Community College (ITC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students students a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5\(\%\) , do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The Type II error is not to reject that the mean number of sleep LTCC students get per night is at lest seven when, in fact, the mean number of hours a. is more than seven hours. b. is at most seven hours. c. is at least seven hours. d. is less than seven hours.

Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal. What symbol represents the random variable for this test?

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

The mean entry level salary of an employee at a company is $58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

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.