/*! 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 45 Gloria Chavez and Ronald Flynn a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Gloria Chavez and Ronald Flynn are the candidates for mayor in a large city. We want to estimate the proportion \(p\) of all registered voters in the city who plan to vote for Chavez with \(95 \%\) confidence and a margin of error no greater than 0.03 . How large a random sample do we need? Show your work.

Short Answer

Expert verified
A sample size of 1068 is needed.

Step by step solution

01

Identify the Formula for Sample Size

To determine the sample size needed for a proportion with a given confidence level and margin of error, we use the formula: \[ n = \left( \frac{Z^2 \times p \times (1-p)}{E^2} \right) \] Where \( n \) is the sample size, \( Z \) is the Z-score, \( p \) is the estimated proportion, and \( E \) is the margin of error.
02

Determine the Z-Score

For a 95% confidence interval, the Z-score (critical value) is approximately 1.96. This value corresponds to the number of standard deviations from the mean for a 95% confidence level.
03

Assume Proportion

Since no preliminary estimate of \( p \) is given, we use 0.5 as a conservative estimate. This is because it maximizes the product \( p \times (1-p) \) and thus ensures the sample size calculation is sufficient regardless of actual proportion.
04

Calculate the Necessary Sample Size

Substitute into the formula: \( Z = 1.96 \), \( p = 0.5 \), \( E = 0.03 \).Calculate the sample size:\[ n = \left( \frac{(1.96)^2 \times 0.5 \times (1-0.5)}{(0.03)^2} \right) \]\[ n = \left( \frac{3.8416 \times 0.25}{0.0009} \right) \]\[ n = \left( \frac{0.9604}{0.0009} \right) \]\[ n \approx 1067.111 \]Since the sample size must be a whole number, round up to 1068.

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 Confidence Intervals
When conducting surveys or experiments, it's crucial to estimate how a sample relates to its population.
A confidence interval is a range of values used to estimate the true value of a population parameter.
In simpler terms, it gives us an interval in which we believe the true proportion or mean lies. For example, in our case, we want to know the proportion of voters supporting Chavez. A confidence interval includes a specific level of confidence, commonly 95%.
This indicates we are 95% confident the true proportion falls within the calculated interval. To construct a confidence interval, we:
  • Determine the sample statistic (mean or proportion).
  • Calculate the margin of error (more on this next).
  • Apply the calculated margin of error to get the range around the sample statistic.
Confidence intervals help us make educated guesses about the larger population without surveying every member.
Decoding the Margin of Error
The margin of error represents the amount by which our sample results might differ from the true population value.
It's the buffer or "wiggle room" we allow when making inferences from our sample data. In our example, we have set a margin of error of 0.03 or 3%.
This means we want our estimated proportion of voters to be within ±3% of the actual proportion in the entire population. A smaller margin of error requires a larger sample size.
This is essential because increasing the sample size provides more accurate results, reducing variability. The margin of error depends on:
  • The desired confidence level (higher confidence levels increase the margin).
  • The standard deviation or variability in the sample data.
  • The sample size (larger samples lead to smaller margins).
Understanding the margin of error is key to balancing precision and practical constraints like time and cost.
What is the Z-Score and Why is it Important?
A Z-score is a measure used in statistics to determine how far (in standard deviations) a data point is from the mean of the distribution.
In the context of confidence intervals, the Z-score represents the "critical value" that corresponds to our desired confidence level. For a 95% confidence interval, the Z-score is approximately 1.96.
This value suggests that our sample mean or proportion is 1.96 standard deviations away from the true population mean or proportion. The Z-score is determined based on the standard normal distribution:
  • A higher Z-score is needed for higher confidence levels (e.g., 2.58 for 99%).
  • It ensures we cover the desired probability in the tails of the distribution.
  • Accurate Z-scores are crucial for ensuring our confidence intervals truly reflect the desired level of certainty.
In our exercise, using the correct Z-score ensures we achieve our 95% confidence in the estimated voter proportion while accounting for sampling variability.
Estimating Proportion Accurately
Proportion estimation is crucial when you need to understand how a certain trait (like voter preference) is distributed in a population.
In our exercise, we want to estimate the proportion of registered voters favoring Chavez.When no specific preliminary data is available, it's conservative to assume a proportion of 0.5.
This is necessary because it maximizes the product of the proportion (\( p \times (1-p) \)) and ensures that the calculated sample size is adequate even if the actual proportion is different.Calculating proportions involves:
  • Collecting sample data that represents the population.
  • Determining the fraction of the sample exhibiting the desired trait (e.g., voting preference).
  • Using this fraction to estimate the true population proportion.
Proper proportion estimation provides the necessary insight into population characteristics, allowing for informed decision-making and strategic planning.

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

How strongly do physical characteristics of sisters and brothers correlate? Here are data on the heights (in inches) of 11 adult pairs: \({ }^{8}\) $$\begin{array}{llllllllllll}\hline \text { Brother: } & 71 & 68 & 66 & 67 & 70 & 71 & 70 & 73 & 72 & 65 & 66 \\\\\text { Sister: } & 69 & 64 & 65 & 63 & 65 & 62 & 65 & 64 & 66 & 59 & 62 \\\\\hline \end{array}$$ (a) Construct a scatterplot using brother's height as the explanatory variable. Describe what you see. (b) Use your calculator to compute the least-squares regression line for predicting sister's height from brother's height. Interpret the slope in context. (c) Damien is 70 inches tall. Predict the height of his sister Tonya. (d) Do you expect your prediction in (c) to be very accurate? Give appropriate evidence to support your answer.

Breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers measured the percent change in bone mineral content (BMC) of the spines of 47 randomly selected mothers during three months of breast-feeding. \({ }^{23}\) The mean change in BMC was \(-3.587 \%\) and the standard deviation was \(2.506 \%\). (a) Construct and interpret a \(99 \%\) confidence interval to estimate the mean percent change in BMC in the population. (b) Based on your interval from part (a), do these data give good evidence that on the average nursing mothers lose bone mineral? Explain.

Gambling is an issue of great concern to those involved in college athletics. Because of this concern, the National Collegiate Athletic Association (NCAA) surveyed randomly selected student athletes concerning their gamblingrelated behaviors. \({ }^{17}\) Of the 5594 Division I male athletes in the survey, 3547 reported participation in some gambling behavior. This includes playing cards, betting on games of skill, buying lottery tickets, betting on sports, and similar activities. A report of this study cited a \(1 \%\) margin of error. (a) The confidence level was not stated in the report. Use what you have learned to find the confidence level, assuming that the NCAA took an SRS. (b) The study was designed to protect the anonymity of the student athletes who responded. As a result, it was not possible to calculate the number of students who were asked to respond but did not. How does this fact affect the way that you interpret the results?

Determine the point estimator you would use and calculate the value of the point estimate. How many pairs of shoes, on average, do female teens have? To find out, an AP Statistics class conducted a survey. They selected an SRS of 20 female students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are the data: $$\begin{array}{rrrrrrrrrr}50 & 26 & 26 & 31 & 57 & 19 & 24 & 22 & 23 & 38 \\\13 & 50 & 13 & 34 & 23 & 30 & 49 & 13 & 15 & 51\end{array}$$

A researcher plans to use a random sample of families to estimate the mean monthly family income for a large population. The researcher is deciding between a \(95 \%\) confidence level and a \(99 \%\) confidence level. Compared to a \(95 \%\) confidence interval, a \(99 \%\) confidence interval will be (a) narrower and would involve a larger risk of being incorrect. (b) wider and would involve a smaller risk of being incorrect. (c) narrower and would involve a smaller risk of being incorrect. (d) wider and would involve a larger risk of being incorrect. (e) wider and would have the same risk of being incorrect.

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.