/*! 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 16 In a survey on supernatural expe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a survey on supernatural experiences, 722 of 4013 adult Americans surveyed reported that they had seen or been with a ghost ("What Supernatural Experiences We've Had," USA Today, February 8,2010 ). a. What assumption must be made in order for it to be appropriate to use the formula of this section to construct a confidence interval to estimate the proportion of all adult Americans who have seen or been with a ghost? b. Construct and interpret a \(90 \%\) confidence interval for the proportion of all adult Americans who have seen or been with a ghost. c. Would a \(99 \%\) confidence interval be narrower or wider than the interval computed in Part (b)? Justify your answer.

Short Answer

Expert verified
The assumption that must be made is that the sampled group is representative and large enough to approximate the normal population distribution. The 90% confidence interval for the proportion can be calculated using the confidence interval formula with z-score of 1.645 corresponding to 90% confidence interval. The 99% confidence interval would be wider than the 90% confidence interval as it needs to capture more of the population.

Step by step solution

01

Understanding The Assumption

For it to be appropriate to use the formula to construct a confidence interval to estimate the proportion of all adult Americans who have seen or been with a ghost, the assumption that must be made is that the sample sizes are big enough such that both \(np \geq 5\) and \(n(1 - p) \geq 5\), where \(n = 4013\) (total respondents), \(p = \frac{722}{4013}\) (sample proportion who've had the supernatural experience). It's also assumed that the sample is random and reflects the general population.
02

Constructing a 90% Confidence Interval

Next, construct a 90% confidence interval for the proportion. The formula for a confidence interval is \(p \pm z*SE\), where \(SE\) is the standard error and is calculated as \(SE = \sqrt{ \frac{p*(1-p)}{n}}\), and \(z\) is the z value which corresponds to the level of confidence. For a 90% confidence interval, the z-value is 1.645 (comes from the z-table or z-distribution chart). By substituting the values into the formula, the interval can be calculated.
03

Understanding the Effect of Confidence Interval on Width

The impact of changing the confidence interval from 90% to 99% can now be evaluated. A 99% confidence interval would be wider than the 90% confidence interval as a higher confidence level would require capturing more of the population within the interval, hence it has to be wider in order to do so.

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Sample Proportion
When examining survey results or data from a sample, the concept of a sample proportion is central to making inferences about the larger population. In simple terms, the sample proportion represents the percentage of the surveyed sample who exhibit a certain characteristic. In the supernatural experiences survey, 722 out of 4013 respondents reported having seen or been with a ghost. Therefore, the sample proportion, often denoted by \( \hat{p} \), is calculated as:- \( \hat{p} = \frac{722}{4013} \approx 0.1798 \).This figure means that, in the surveyed group, approximately 17.98% had such an experience. In statistical analysis, using sample proportions helps to estimate the true proportion of a particular characteristic within the entire population. This estimate isn't perfect, as it comes from a sample, but it provides a valuable way to infer what might be true for the larger population.
Calculating the Standard Error
Once you have the sample proportion, the next step is often to calculate the standard error. The standard error measures the accuracy with which the sample proportion approximates the population proportion. Think of it as the variability or potential error in the sample proportion's estimate of the population proportion. The formula for standard error \( SE \) when dealing with proportions is:- \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \).In our example:- \( \hat{p} = 0.1798 \) and - \( n = 4013 \).Plug these numbers in to find the standard error:- \( SE = \sqrt{\frac{0.1798(1 - 0.1798)}{4013}} \approx 0.006 \).This value allows statisticians to understand the variation one might expect in sample proportions from different random samples of the same size.
Interpreting the Z-value
The z-value is an essential component when constructing confidence intervals. It determines how far from the mean a data point is, in terms of standard deviations. When creating confidence intervals, the z-value ensures the interval accurately reflects the confidence level desired. Here's how it works:- For a 90% confidence interval, the z-value is approximately 1.645, coming from the standard normal distribution. This z-value indicates that we want our interval to capture the central 90% of the distribution.For example, when calculating a 90% confidence interval:- You start with the sample proportion and- Add and subtract the product of the z-value and standard error: \( \hat{p} \pm z \times SE \).A higher confidence level, such as 99%, would automatically require a larger z-value, thus resulting in a wider interval. This is because we aim to capture more certainty about the population proportion, necessitating a broader range to ensure accuracy.

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

In the article "Fluoridation Brushed Off by Utah" (Associated Press, August 24, 1998 ), it was reported that a small but vocal minority in Utah has been successful in keeping fluoride out of Utah water supplies despite evidence that fluoridation reduces tooth decay and despite the fact that a clear majority of Utah residents favor fluoridation. To support this statement, the article included the result of a survey of Utah residents that found \(65 \%\) to be in favor of fluoridation. Suppose that this result was based on a random sample of 150 Utah residents. Construct and interpret a \(90 \%\) confidence interval for \(p\), the true proportion of Utah residents who favor fluoridation. Is this interval consistent with the statement that fluoridation is favored by a clear majority of residents?

The formula used to compute a large-sample confidence interval for \(p\) is $$\hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

In a survey of 1000 randomly selected adults in the United States, participants were asked what their most favorite and what their least favorite subject was when they were in school (Associated Press, August 17, 2005 ). In what might seem like a contradiction, math was chosen more often than any other subject in both categories! Math was chosen by 230 of the 1000 as the favorite subject, and it was also chosen by 370 of the 1000 as the least favorite subject. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the favorite subject in school. b. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the least favorite subject.

The article "Career Expert Provides DOs and DON'Ts for Job Seekers on Social Networking" (CareerBuilder.com, August 19, 2009) included data from a survey of 2667 hiring managers and human resource professionals. The article noted that many employers are using social networks to screen job applicants and that this practice is becoming more common. Of the 2667 people who participated in the survey, 1200 indicated that they use social networking sites (such as Facebook, MySpace, and LinkedIn) to research job applicants. For the purposes of this exercise, assume that the sample is representative of hiring managers and human resource professionals. Construct and interpret a \(95 \%\) confidence interval for the proportion of hiring managers and human resource professionals who use social networking sites to research job applicants.

The paper "The Curious Promiscuity of Queen Honey Bees (Apis mellifera): Evolutionary and Behavioral Mechanisms" (Annals of Zoology Fennici [2001]:255- 265) describes a study of the mating behavior of queen honeybees. The following quote is from the paper: Queens flew for an average of \(24.2 \pm 9.21\) minutes on their mating flights, which is consistent with previous findings. On those flights, queens effectively mated with \(4.6 \pm 3.47\) males (mean \(\pm \mathrm{SD})\). a. The intervals reported in the quote from the paper were based on data from the mating flights of \(n=\) 30 queen honeybees. One of the two intervals reported is stated to be a confidence interval for a population mean. Which interval is this? Justify your choice. b. Use the given information to construct a \(95 \%\) confidence interval for the mean number of partners on a mating flight for queen honeybees. For purposes of this exercise, assume that it is reasonable to consider these 30 queen honeybees as representative of the population of queen honeybees.

See all solutions

Recommended explanations on Math Textbooks

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.