/*! 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 68 In 1991, California imposed a "s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In 1991, California imposed a "snack tax" (a sales tax on snack food) in an attempt to help balance the state budget. A proposed alternative tax was a \(12 \phi\) -per-pack increase in the cigarette tax. In a poll of 602 randomly selected California registered voters, 445 responded that they would have preferred the cigarette tax increase to the snack tax (Reno Gazette-Journal, August 26, 1991). Estimate the true proportion of California registered voters who preferred the cigarette tax increase; use a \(95 \%\) confidence interval.

Short Answer

Expert verified
We can be 95% confident that the proportion of California registered voters who would have preferred the cigarette tax increase to the snack tax lies between 0.7038 and 0.7746.

Step by step solution

01

Calculate the Sample Proportion

The first step to estimating the population proportion is to calculate the sample proportion (\(p\)). This is done by dividing the number of people that prefer the cigarette tax increase by the total number of people surveyed. In this case, 445 out of 602 polled voters preferred the cigarette tax increase, so the sample proportion (\(p\)) would be \(\frac{445}{602}\) = 0.7392.
02

Find the Z-score

The z-score associated with a 95% confidence interval is 1.96. This value is obtained from a standard z-score table or statistical text.
03

Calculate the Margin of Error

The next step is to calculate the margin of error using formula \( z \sqrt{\frac{p(1-p)}{n}} \), where \( p \) is the sample proportion, \( n \) is the sample size, and \( z \) is the z-score. This gives us: \( 1.96 \sqrt{\frac{0.7392(1-0.7392)}{602}} \) = 0.0354.
04

Construct the Confidence Interval

The final step is to construct the confidence interval which includes the sample proportion and the margin of error. This gives us: \( (0.7392 - 0.0354, 0.7392 + 0.0354) \) = \( (0.7038, 0.7746) \). This is the 95% confidence interval for the true proportion of California registered voters who would have preferred the cigarette tax increase over the snack tax.

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
Understanding the sample proportion is fundamental in statistics as it helps estimate the characteristics of a population. The sample proportion, denoted by \( p \), is simply the ratio of members in a sample exhibiting a particular trait to the total number of individuals within that sample. In our exercise, the aim was to ascertain the portion of California voters in favor of an alternative tax proposal, and this was achieved by dividing the number of favorable responses (445) by the total surveyed individuals (602), resulting in a sample proportion of \( p = \frac{445}{602} = 0.7392 \). This fraction is central as it lays the groundwork for further estimation of the population's stance using confidence interval methods.

It's important to realize, though, that the sample proportion can vary from the actual population proportion due to the size and variability of the sample. Hence, statisticians use it as an estimate rather than an exact figure—a concept essentially linked with the calculation of confidence intervals.
Z-score
The z-score occupies a pivotal role in statistical analysis when it comes to measuring how far, and in what direction, a data point is from the mean. Regarded as a standard score, it represents the number of standard deviations an element is from the mean of the distribution.

When constructing a confidence interval, such as in our sample exercise, the z-score indicates the required level of confidence. The chosen level, usually expressed as a percentage like 95%, dictates the z-score to be used. For a 95% confidence interval, as in our problem, the z-score is 1.96. This value reflects a point on the standard normal distribution that has 95% of the data within it; meaning that there's a 95% chance that the true population parameter lies within this range. Z-scores are extracted from standard tables or could also be obtained using statistical software or calculators.
Margin of Error
The margin of error is a statistic expressing the amount of random sampling error in a survey's results and is a crucial component in confidence interval estimation. It defines the range within which we can expect the true population parameter to fall, considering a certain level of confidence. In the context of our exercise, the margin of error is calculated using the formula \( z \sqrt{\frac{p(1-p)}{n}} \) where \( p \) is the sample proportion, \( n \) is the total sample size, and \( z \) is the z-score correlating to the desired confidence level.

For our specific example, the margin of error was computed to be 0.0354. This interval conveys the reliability of the sample proportion's representation of the population—essentially, a smaller margin of error suggests a more accurate portrayal whereas a larger one indicates greater uncertainty.
Population Proportion
The population proportion is a parameter that represents the fraction of the entire population that exhibits a certain characteristic. It is denoted by \( \phi \) and is what we aim to estimate through surveying a sample. Unlike the sample proportion, which is a known statistic we calculate from collected data, the population proportion is a fixed but unknown value.

In our exercise, the objective was to estimate the true proportion of California registered voters (/\phi/ in the original equation) who preferred the cigarette tax increase over the snack tax. To infer this from our sample, we construct a confidence interval using the sample proportion (0.7392), along with the margin of error and our predetermined confidence level. In reality, calculating the exact population proportion is often impractical due to the sheer size of many populations, but with a well-designed sample and the application of statistical methods such as confidence interval estimation, we can approximate it with reasonable certainty.

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

An Associated Press article on potential violent behavior reported the results of a survey of 750 workers who were employed full time (San Luis 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 a \(90 \%\) confidence interval estimate of \(\pi\), the true proportion of fulltime workers so angered in the last year that they wanted to hit a colleague.

An article in the Chicago Tribune (August 29,1999 ) reported that in a poll of residents of the Chicago suburbs, \(43 \%\) felt that their financial situation had improved during the past year. The following statement is from the article: "The findings of this Tribune poll are based on interviews with 930 randomly selected suburban residents. The sample included suburban Cook County plus DuPage, Kane, Lake, McHenry, and Will Counties. In a sample of this size, one can say with \(95 \%\) certainty that results will differ by no more than 3 percent from results obtained if all residents had been included in the poll." Comment on this statement. Give a statistical argument to justify the claim that the estimate of \(43 \%\) is within \(3 \%\) of the true proportion of residents who feel that their financial situation has improved.

The article "CSI Effect Has Juries Wanting More Evidence" (USA Today, August 5,2004 ) examines how the popularity of crime-scene investigation television shows is influencing jurors' expectations of what evidence should be produced at a trial. In a survey of 500 potential jurors, one study found that 350 were regular watchers of at least one crime-scene forensics television series. a. Assuming that it is reasonable to regard this sample of 500 potential jurors as representative of potential jurors in the United States, use the given information to construct and interpret a \(95 \%\) confidence interval for the true proportion of potential jurors who regularly watch at least one crime-scene investigation series. b. Would a 99\% confidence interval be wider or narrower than the \(95 \%\) confidence interval from Part (a)?

Example \(9.3\) gave the following airborne times for United Airlines flight 448 from Albuquerque to Denver on 10 randomly selected days: \(\begin{array}{llllllllll}57 & 54 & 55 & 51 & 56 & 48 & 52 & 51 & 59 & 59\end{array}\) a. Compute and interpret a \(90 \%\) confidence interval for the mean airborne time for flight 448 . b. Give an interpretation of the \(90 \%\) confidence level associated with the interval estimate in Part (a). c. Based on your interval in Part (a), if flight 448 is scheduled to depart at \(10 \mathrm{~A} \mathrm{M} .\), what would you recommend for the published arrival time? Explain.

Increases in worker injuries and disability claims have prompted renewed interest in workplace design and regulation. As one particular aspect of this, employees required to do regular lifting should not have to handle unsafe loads. The article "Anthropometric, Muscle Strength, and Spinal Mobility Characteristics as Predictors of the Rating of Acceptable Loads in Parcel Sorting" (Ergonomics [1992]: \(1033-1044\) ) reported on a study involving a random sample of \(n=18\) male postal workers. The sample mean rating of acceptable load attained with a work-simulating test was found to be \(\bar{x}=9.7 \mathrm{~kg}\). and the sample standard deviation was \(s=4.3 \mathrm{~kg}\). Suppose that in the population of all male postal workers, the distribution of rating of acceptable load can be modeled approximately using a normal distribution with mean value \(\mu\). Construct and interpret a \(95 \%\) confidence interval for \(\mu\).

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.