/*! 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 56 According to an AP-lpsos poll (J... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to an AP-lpsos poll (June 15,2005\()\), \(42 \%\) of 1001 randomly selected adult Americans made plans in May 2005 based on a weather report that turned out to be wrong. a. Construct and interpret a 9996 confidence interval for the proportion of Americans who made plans in May 2005 based on an incorrect weather report. b. Do you think it is reasonable to generalize this estimate to other months of the year? Explain.

Short Answer

Expert verified
a. The 99% confidence interval for the proportion of Americans who made plans in May 2005 based on an incorrect weather report can be calculated as previously demonstrated. After interpreting, we can be 99% confident that the true proportion of such Americans falls within this range. b. Without more data or information, it's hard to definitively say whether it's reasonable to generalize these results to other months. Several real-world factors must be considered.

Step by step solution

01

Identify the sample proportions and sample size

Given that 42% of 1001 respondents made plans based on a wrong weather forecast, the sample size (\(n\)) is 1001 and the sample proportion (\(p\)) is 0.42.
02

Calculate the standard error

The standard error (SE) for a proportion is calculated using the formula: \[SE = \sqrt{{p(1 - p) / n}}\]By substituting \(p = 0.42\) and \(n = 1001\) into the equation, we can calculate the standard error.
03

Determine the z-score for the given confidence level

For a confidence level of 99%, the z-value (from the standard normal distribution table) is 2.576. This value is used to calculate the confidence interval.
04

Compute the Confidence Interval

The confidence interval can be determined using the formula: \[CI = p \pm Z(SE)\]By substituting p, Z, and SE into the equation, we can calculate the confidence interval for the proportion of Americans who made decisions based on an inaccurate weather report in May 2005.
05

Interpret the Confidence Interval

The calculated confidence interval can be interpreted as follows: We are 99% confident that the interval contains the true proportion of Americans who made plans in May 2005 based on incorrect weather reports.
06

Validity of Generalizing the Estimate

Regarding the question if these results can be generalized to other months, there are quite a few factors to consider. For example, are Americans more likely to make plans based on weather forecasts in certain months, such as summer or holiday months? If yes, we cannot directly apply these results to those months. However, if there is no significant monthly fluctuation in how people use weather forecasts to make plans, it might be reasonable to apply the results to other months. Strictly statistically speaking, without more data or information, it's hard to be sure.

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.

Sample Proportion
The sample proportion is a term that reflects the fraction of individuals in a sample that possesses a particular trait or characteristic. In our example, out of 1001 adult Americans surveyed, 42% reported basing their May 2005 plans on a weather forecast that was incorrect. This 42% represents our sample proportion, commonly denoted as \( p \). In numerical terms, this is expressed as \( p = 0.42 \).
Understanding the sample proportion is vital as it provides insights into the behavior or characteristic within the sample. It acts as an estimate of the true population proportion. However, it is noteworthy that a sample proportion is just an estimate, and thus comes with certain uncertainty about how it represents the entire population.
Standard Error
The standard error (SE) measures the dispersion or spread of sample proportions and helps us understand how much variability to expect in estimates from different samples. The formula for standard error in a proportion is: \[SE = \sqrt{\frac{p(1 - p)}{n}}\]Here, \( p \) is the sample proportion, and \( n \) is the sample size. For our poll, substituting \( p = 0.42 \) and \( n = 1001 \) allows us to calculate the standard error.The standard error serves a critical role in constructing confidence intervals. It gives us a quantified estimate of how much our sample proportion, \( p \), might deviate from the true population proportion if we took multiple samples.
Z-Score
To create a confidence interval, the z-score is very important. The z-score, derived from the standard normal distribution, helps in understanding how far a particular value is from the mean, measured in standard deviations. For a 99% confidence level, the corresponding z-score is 2.576. This z-score tells us how many standard errors we should look on either side of our sample proportion to capture the true population proportion with 99% confidence. It's an essential component in the formula for confidence intervals, shaping the width of the interval we calculate, where wider intervals reflect greater uncertainty but higher confidence.
Generalization
Generalization in statistics refers to applying results from a sample to a larger population or to situations beyond the sample data. In the context of our survey, the critical question is whether the proportion of Americans making incorrect weather-based plans in May 2005 can be generalized to other months. There are various considerations in generalization:
  • Seasonal Variation: Are there months where weather forecasts are more or less relied upon? If so, behaviors might vary across the year.
  • Sample Limitations: Was the sample representative enough to apply results broadly?
  • External Conditions: Are there factors specific to May 2005 that might not apply in other months?
While statistical methods offer tools for generalization, thoughtful analysis of context and external factors is necessary to decide on the reasonability of such extensions.

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

A random sample of \(n=12\) four-year-old red pine trees was selected, and the diameter (in inches) of each tree's main stem was measured. The resulting observations are as follows: \(\begin{array}{lllll}11.3 & 10.7 & 12.4 & 15\end{array}\) \(16.2 \quad 10.5\) \(\begin{array}{llll}11.4 & 11.0 & 10.7 & 12.0\end{array}\) a. Compute a point estimate of \(\sigma,\) the population standard deviation of main stem diameter. What statistic did you use to obtain your estimate? b. Making no assumptions about the shape of the population distribution of diameters, give a point estimate for the population median diameter. What statistic did you use to obtain the estimate? c. Suppose that the population distribution of diameter is symmetric but with heavier tails than the normal distribution. Give a point estimate of the population mean diameter based on a statistic that gives some protection against the presence of outliers in the sample. What statistic did you use? d. Suppose that the diameter distribution is normal. Then the 90 th percentile of the diameter distribution is \(\mu+1.28 \sigma\) (so \(90 \%\) of all trees have diameters less than this value). Compute a point estimate

An Associated Press article on potential violent behavior reported the results of a survey of 750 workers who were employed full time iSan Luls Obispo Tribune. September \(7,\) 1999), Of those surveyed, 125 indicated that they were so angered by a coworker during the past year that they felt like hitting the coworker (but didn't). Assuming that it is reasonable to regard this sample of 750 as a random sample from the population of full-time workers, use this information to construct and interpret \(90 \%\) confidence interval estimate of \(p,\) the true proportion of full-time workers so angered in the last year that they wanted to hit a colleague.

The Assodated Press (December 16. 1991 ) reported that in a random sample of 507 people, only 142 correctly described the Bill of Rights as the first 10 amendments to the U.S. Constitution. Calculate a \(95 \%\) confidence interval for the proportion of the entire population that could give a correct description.

Recent high-profile legal cases have many people reevaluating the jury system. Many believe that juries in criminal trials should be able ro convict on less than a unanimous vote. To assess support for this idea, investigators asked cach individual in a random sample of Californians whether they favored allowing conviction by a \(10-2\) verdict in criminal cases not involving the death penalty. The Associated Press iSan Luls Oblspo Telegram-Tribune, September 13,1995\()\) reported that \(71 \%\) supported the \(10-2\) verdict. Suppose that the sample size for this survey was \(n=900\). Compute and interpret a \(99 \%\) confidence interval for the proportion of Californians who favor the \(10-2\) verdict.

Acrylic bone cement is sometimes used in hip and knee replacements to fix an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens under specified conditions, and the resulting mean and standard deviation were 306.09 Newtons and 41.97 Newtons, respectively. Assuming that it is reasonable to believe that breaking force under these conditions has a distribution that is approximately normal, estimate the mean breaking force for acrylic bone cement under the specified conditions using a \(95 \%\) confidence interval.

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.