/*! 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 14 Exercise 6.12 presents the resul... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Exercise 6.12 presents the results of a poll where \(48 \%\) of 331 Americans who decide to not go to college do so because they cannot afford it. (a) Calculate a \(90 \%\) confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context. (b) Suppose we wanted the margin of error for the \(90 \%\) confidence level to be about \(1.5 \%\). How large of a survey would you recommend?

Short Answer

Expert verified
(a) The 90% CI is [43.49%, 52.51%]. (b) A sample size of about 4514 is recommended.

Step by step solution

01

Identify the Key Variables

First, we need to identify the sample proportion (\(\hat{p}\)), sample size (\(n\)), and the desired confidence level. In this problem, \(\hat{p} = 0.48\), \(n = 331\), and the confidence level is \(90\%\).
02

Determine the Z-score for Confidence Level

For a \(90\%\) confidence interval, we use the Z-score that corresponds to this percent in a standard normal distribution. The Z-score for \(90\%\) is \(1.645\).
03

Calculate the Standard Error

The standard error (SE) for the sample proportion is calculated as: \[ SE = \sqrt{ \frac{\hat{p} (1 - \hat{p})}{n} } = \sqrt{ \frac{0.48 \times 0.52}{331} } \approx 0.0274. \]
04

Calculate the Margin of Error

The margin of error (ME) is calculated using the Z-score and the standard error: \[ ME = Z \times SE = 1.645 \times 0.0274 \approx 0.0451. \]
05

Find the Confidence Interval

The confidence interval is given by \(\hat{p} \pm ME\). Therefore, \[ 0.48 \pm 0.0451 = [0.4349, 0.5251]. \] This means the \(90\%\) confidence interval is approximately \([43.49\%, 52.51\%]\).
06

Interpret the Confidence Interval

We are \(90\%\) confident that the true proportion of Americans who decide not to go to college because they cannot afford it lies between \(43.49\%\) and \(52.51\%\).
07

Determine Required Sample Size for Given Margin of Error

To find the required sample size, use the formula \( n = \left( \frac{Z}{ME} \right)^2 \times \hat{p} (1 - \hat{p}) \). Here, \( ME = 0.015 \). Thus, \[ n = \left( \frac{1.645}{0.015} \right)^2 \times 0.48 \times 0.52 \approx 4514. \] Thus, a sample size of about 4514 would be needed.

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
Let's dive into what a sample proportion is and how it relates to statistics. When we talk about sample proportion, we are looking at the number of successes in a sample divided by the total number of observations in that sample. It's like taking a snapshot of a scenario and determining where your results fall. In our exercise, 48% of 331 Americans revealed they chose not to go to college for affordability reasons. Here, the sample proportion (\(\hat{p}\)) is 0.48.

Understanding this helps to realize that sample proportion is a statistic—an estimate of a population parameter. It's crucial because it lays the foundation for further calculations such as standard error and margin of error, which together paint a more complete picture of the data's reliability.
Calculating Standard Error
The standard error (SE) is a measure of the variability or spread of a sample proportion. Think of it as showing how much you should expect your sample proportion to fluctuate if you took multiple samples.

To calculate the standard error of a sample proportion, you use the formula:\[SE = \sqrt{ \frac{\hat{p} (1 - \hat{p})}{n} }\]In the given exercise, the resulting SE is approximately 0.0274.

The smaller the standard error, the closer your sample proportion is likely to be to the true population proportion. This measure is essential for constructing confidence intervals.
Defining Margin of Error
The margin of error (ME) tells you how much you can expect your sample results to vary from the true population proportion. It's a critical yardstick in surveys that show the range within which the true population parameter lies with a certain level of confidence.

Using the formula:\[ME = Z \times SE\]So, for the exercise, ME turns out to be approximately 0.0451 when calculated with a Z-score of 1.645 and standard error of 0.0274.

A larger margin of error implies less confidence in how accurately your sample proportion reflects the population parameter. Adjustments in sample size can influence this, as demonstrated when planning surveys.
Importance of Sample Size
Sample size (n) is the total number of observations or individuals considered in your sample. It plays a vital role in statistical analysis because it affects the precision of the estimation and the power of the tests that might be conducted.

In the given scenario, having a larger sample size reduced the margin of error, thereby producing a more accurate and reliable confidence interval. If you wish for a tighter margin of error, especially with a confidence level such as 90%, you're often required to increase the sample size.

For instance, to achieve a desired margin of error of 1.5%, calculations showed a sample size of about 4514 would be needed compared to the original 331. This adjustment ensures results are more aligned with the population reality.
Utilizing Z-score in Confidence Intervals
The Z-score is a statistical measure that describes a value's position relative to the mean of a group of values. In the context of confidence intervals, it determines how many standard deviations away you need to be to achieve your desired confidence level.

For a 90% confidence level, the Z-score is 1.645. This value factors into the calculations for both the margin of error and subsequently the confidence interval.
  • Higher confidence levels (e.g., 95% or 99%) correspond with larger Z-values.
  • Larger Z-scores result in wider confidence intervals, offering greater certainty that the true parameter lies within the interval but at the expense of precision.
The choice of Z-score is pivotal as it impacts the understanding of how reliably your sample statistic reflects the underlying population parameter.

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

The General Social Survey asked a random sample of 1,390 Americans the following question: "On the whole, do you think it should or should not be the government's responsibility to promote equality between men and women?" \(82 \%\) of the respondents said it "should be". At a \(95 \%\) confidence level, this sample has \(2 \%\) margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning. (a) We are \(95 \%\) confident that between \(80 \%\) and \(84 \%\) of Americans in this sample think it's the government's responsibility to promote equality between men and women. (b) We are \(95 \%\) confident that between \(80 \%\) and \(84 \%\) of all Americans think it's the government's responsibility to promote equality between men and women. (c) If we considered many random samples of 1,390 Americans, and we calculated \(95 \%\) confidence intervals for each, \(95 \%\) of these intervals would include the true population proportion of Americans who think it's the government's responsibility to promote equality between men and women. (d) In order to decrease the margin of error to \(1 \%\), we would need to quadruple (multiply by 4 ) the sample size. (e) Based on this confidence interval, there is sufficient evidence to conclude that a majority of Americans think it's the government's responsibility to promote equality between men and women.

A Kaiser Family Foundation poll for US adults in 2019 found that \(79 \%\) of Democrats, \(55 \%\) of Independents, and \(24 \%\) of Republicans supported a generic "National Health Plan". There were 347 Democrats, 298 Republicans, and 617 Independents surveyed. (a) A political pundit on TV claims that a majority of Independents support a National Health Plan. Do these data provide strong evidence to support this type of statement? (b) Would you expect a confidence interval for the proportion of Independents who oppose the public option plan to include \(0.5 ?\) Explain.

A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A \(95 \%\) confidence interval for the difference between the proportions of males and females whose favorite color is black \(\left(p_{\text {male }}-p_{\text {female }}\right)\) was calculated to be (0.02,0.06) . Based on this information, determine if the following statements are true or false, and explain your reasoning for each statement you identify as false. \(^{23}\) (a) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) lower to \(6 \%\) higher than the true proportion of females whose favorite color is black. (b) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) to \(6 \%\) higher than the true proportion of females whose favorite color is black. (c) \(95 \%\) of random samples will produce \(95 \%\) confidence intervals that include the true difference between the population proportions of males and females whose favorite color is black. (d) We can conclude that there is a significant difference between the proportions of males and females whose favorite color is black and that the difference between the two sample proportions is too large to plausibly be due to chance. (e) The \(95 \%\) confidence interval for \(\left(p_{\text {female }}-p_{\text {male }}\right)\) cannot be calculated with only the information given in this exercise.

The Marist Poll published a report stating that \(66 \%\) of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was \(3 \%\) using a \(95 \%\) confidence level. \(^{9}\) (a) Verify the margin of error reported by The Marist Poll. (b) Based on a \(95 \%\) confidence interval, does the poll provide convincing evidence that more than \(70 \%\) of the population think that licensed drivers should be required to retake their road test once they turn \(65 ?\)

Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) As the degrees of freedom increases, the mean of the chi-square distribution increases. (b) If you found \(\chi^{2}=10\) with \(d f=5\) you would fail to reject \(H_{0}\) at the \(5 \%\) significance level. (c) When finding the p-value of a chi-square test, we always shade the tail areas in both tails. (d) As the degrees of freedom increases, the variability of the chi-square distribution decreases.

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.