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Chapter 6: Problem 16

As discussed in Exercise 6.10, the General Social Survey reported a sample where about \(61 \%\) of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a \(95 \%\) confidence interval to \(2 \%\), about how many Americans would we need to survey?

Short Answer

Expert verified
Approximately 2282 Americans need to be surveyed.

Step by step solution

01

Understanding the Problem

We need to find the sample size required to have a margin of error of 2% at a 95% confidence level for estimating the proportion of US residents who think marijuana should be made legal.
02

Formula for Sample Size

The formula to calculate the sample size for estimating a proportion is given by \( n = \left(\frac{z^2 \cdot p \cdot (1-p)}{E^2}\right) \) where \( z \) is the z-score for the desired confidence level, \( p \) is the estimated proportion, and \( E \) is the margin of error.
03

Determine the Z-Score

For a 95% confidence interval, the z-score (\( z \)) is approximately 1.96. This is derived from the standard normal distribution table.
04

Plug Values into Formula

Here, \( p = 0.61 \), \( E = 0.02 \), and \( z = 1.96 \). Plug these into the sample size formula: \[ n = \left(\frac{(1.96)^2 \cdot 0.61 \cdot (0.39)}{0.02^2}\right) \]
05

Calculate the Sample Size

Perform the calculations: \( n = \left(\frac{3.8416 \cdot 0.61 \cdot 0.39}{0.0004}\right) \)\( n = \left(\frac{0.9126792}{0.0004}\right) \)\( n \approx 2281.698 \).Since the number of people surveyed must be a whole number, round up to the nearest whole number: \( n \approx 2282 \).

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Key Concepts

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

Understanding Margin of Error
The **margin of error** is a crucial concept in statistics that reflects the expected range within which the true value of a population parameter is estimated to fall. It's often used in surveys and polls to express confidence in the derived data. When we say the margin of error is 2%, it means that the true proportion is expected to be within this range on either side of the estimate from the sample.
Let's break it down further:
  • A smaller margin of error means a more precise estimate, but it typically requires a larger sample size.
  • A larger margin of error can result from a smaller sample size or a high degree of variability in the data.
  • To achieve a certain margin of error, statisticians determine the necessary sample size by considering the estimated proportion and the desired confidence level.
The importance of margin of error is that it quantifies the uncertainty in sample estimates, giving users a way to understand possible deviations from the true population value.
Decoding Confidence Interval
A **confidence interval** provides a range of values within which we can be certain the true population parameter lies, to a specified degree of certainty. For a 95% confidence interval, like in our problem, this means that if we were to take 100 different samples and compute an interval estimate for each, approximately 95 of those intervals would contain the true proportion.
Here鈥檚 how confidence intervals are represented:
  • The confidence interval is centered around the sample proportion.
  • The width of the confidence interval is determined by the margin of error and the sample standard deviation.
  • A wider confidence interval implies more uncertainty about the estimate of the population parameter.
In essence, understanding confidence intervals allows individuals to see how precise the sample estimate is in relation to the entire population. It bridges the gap between sample data and population inferences.
Estimating Proportion
When we speak of **proportion estimation**, we refer to predicting the fraction of the population that has a particular attribute, based on sample data. In the context of our exercise, it's estimating the proportion of US residents who think marijuana should be made legal.
This involves several steps:
  • The estimated proportion, denoted as \( p \), is drawn from the sample data (in our case, 61%).
  • Using this initial estimate, along with the desired confidence level and margin of error, we calculate the necessary sample size to achieve our desired precision.
  • These calculations make use of statistical formulas and standard deviation scoring to ensure accurate predictions for the whole population.
Proportion estimation is pivotal in fields ranging from social sciences to health research, where determining public opinion or incidence rates is fundamental for decision-making and policy development.

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Most popular questions from this chapter

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. \(^{28}\) (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.

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?

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. \(^{16}\) (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 \(65 ?\)

A physical education teacher at a high school wanting to increase awareness on issues of nutrition and health asked her students at the beginning of the semester whether they believed the expression "an apple a day keeps the doctor away", and \(40 \%\) of the students responded yes. Throughout the semester she started each class with a brief discussion of a study highlighting positive effects of eating more fruits and vegetables. She conducted the same apple-a- day survey at the end of the semester, and this time \(60 \%\) of the students responded yes. Can she used a two-proportion method from this section for this analysis? Explain your reasoning.

Lymphatic filariasis is a disease caused by a parasitic worm. Complications of the disease can lead to extreme swelling and other complications. Here we consider results from a randomized experiment that compared three different drug treatment options to clear people of the this parasite, which people are working to eliminate entirely. The results for the second year of the study are given below: $$ \begin{array}{lcc} \hline & \text { Clear at Year 2 } & \text { Not Clear at Year 2 } \\ \hline \text { Three drugs } & 52 & 2 \\ \text { Two drugs } & 31 & 24 \\ \text { Two drugs annually } & 42 & 14 \\ \hline \end{array} $$ (a) Set up hypotheses for evaluating whether there is any difference in the performance of the treatments, and also check conditions. (b) Statistical software was used to run a chi-square test, which output: $$ X^{2}=23.7 \quad d f=2 \quad \text { p-value }=7.2 \mathrm{e}-6 $$ Use these results to evaluate the hypotheses from part (a), and provide a conclusion in the context of the problem.
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