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Chapter 22: Problem 19

How many American adults must be interviewed to estimate the proportion of all American adults who actively try to avoid drinking regular soda or pop within \(\pm 0.01\) with \(99 \%\) confidence using the large-sample confidence interval? Use \(0.5\) as the conservative guess for \(p\). (a) \(n=6765\) (b) \(n=9604\) (c) \(n=16590\)

Short Answer

Expert verified
Option (c): \(n = 16590\)

Step by step solution

01

Identify the formula

To find the minimum sample size required to estimate a population proportion with a given margin of error and confidence level, use the formula:\[n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right)n\]where:- \(n\) is the required sample size,- \(Z\) is the Z-value corresponding to the desired confidence level,- \(p\) is an estimate of the population proportion (often chosen as 0.5 for maximum variability), and- \(E\) is the desired margin of error.
02

Determine the Z-value

For a 99% confidence level, the Z-value is approximately 2.576. This value corresponds to the point on the standard normal distribution where 99% of the data falls within \(Z\) standard deviations of the mean.
03

Substitute the values into the formula

Substitute the given values into the formula:- \(p = 0.5\),- \(E = 0.01\),- \(Z = 2.576\)\[n = \left( \frac{2.576^2 \cdot 0.5 \cdot (1-0.5)}{0.01^2} \right) = \left( \frac{6.635776 \cdot 0.25}{0.0001} \right)\]
04

Calculate the sample size

Now, calculate the value:\[n = \frac{1.658944}{0.0001} = 16589.44\]Since you cannot interview a fraction of a person, round up to ensure the margin of error is within the specified range. Therefore, \(n = 16590\).
05

Verify the options

Compare the calculated sample size with the provided options to find the correct answer:- (a) \(n = 6765\)- (b) \(n = 9604\)- (c) \(n = 16590\)The correct answer is option (c): \(n = 16590\).

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Key Concepts

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

Confidence Interval
The confidence interval is a statistical range, calculated from the data, that is likely to contain the true population parameter (such as a population proportion). It provides an estimated range of values which is likely to include the unknown parameter.
A 99% confidence interval means that if we were to take 100 different samples and compute the confidence interval for each sample, 99 of those intervals would contain the true population proportion.
This concept is essential as it provides a buffer zone for estimation errors, assuring us about the accuracy of the sampling process.
When determining sample size, a higher confidence level (like 99%) means you'll need a larger sample size to ensure the estimate's precision.
Population Proportion
The population proportion is the fraction or percentage of the total population that has a particular characteristic of interest. It's an estimate derived from a sample of the broader population.
For instance, if we want to know how many people in a city like ice cream, the population proportion is the percentage of the entire city's population that likes ice cream.
  • This characteristic is often represented by the symbol \( p \) in formulas.
  • In cases where the exact proportion is unknown, a conservative estimate of 0.5 is used. This is because maximizes the required sample size, ensuring sufficient variability coverage.
This estimate ensures that our calculations for sample size cater to the broadest range of possible real-world proportions.
Margin of Error
The margin of error indicates the range that the population parameter is expected to vary. It's a buffer that allows researchers to have a degree of uncertainty in their estimations.
For example, a margin of error of \( \pm 0.01 \) means that the true population proportion is expected to be within 1% above or below the sample proportion.
  • In formulas, it's denoted by \( E \).
  • A smaller margin of error requires a larger sample size to ensure precise estimates.
This understanding helps researchers in deciding how precise they want their results to be and is a critical component in the sample size determination.
Z-value
The Z-value represents the number of standard deviations a data point is from the mean in a standard normal distribution. It is crucial for determining the confidence interval and sample size in statistics.
For a 99% confidence level, the Z-value is approximately 2.576. This number indicates that 99% of the population data lies within 2.576 standard deviations from the mean.
  • The Z-value helps cater to the desired accuracy of a confidence interval.
  • Higher confidence levels require larger Z-values, leading to larger sample sizes.
Understanding Z-values is key when ensuring that statistical estimations meet specific confidence requirements.

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Most popular questions from this chapter

Does the order in which wine is presented make a difference? Several choices of wine are presented one at a time and in sequence, and the subject is then asked to choose the preferred wine at the end of the sequence. In this study, subjects were asked to taste two wine samples in sequence. Both samples given to a subject were the same wine, although subjects were expecting to taste two different samples of a particular variety. Of the 32 subjects in the study, 22 selected the wine presented first, when presented with two identical wine samples. 29 (a) Do the data give good reason to conclude that the subjects are not equally likely to choose either of the two positions when presented with two identical wine samples in sequence? (b) The subjects were recruited in Ontario, Canada, via advertisements to participate in a study of "attitudes and values toward wine." Can we generalize our conclusions to all wine tasters? Explain

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. Only half of all mice complete one particular maze in less than 18 seconds. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures the proportion of 40 mice that completed the maze in less than 18 seconds with noise as a stimulus. The proportion of mice that completed the maze in less than 18 seconds is \(\mathrm{p}^{\wedge \hat{p}}=0.7\). The hypotheses for a test to answer the researcher's question are (a) \(H_{0}: p=0.5, H_{a}: p>0.5\). (b) \(H_{0}: p=0.5, H_{a}: p<0.5\). (c) \(H_{0}: p=0.5, H_{a}: p \neq 0.5\).

A random digit dialing telephone survey of 880 drivers asked, "Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?" Of the 880 respondents, 171 admitted that at least one light had been red. \({ }^{24}\) (a) Give a \(95 \%\) confidence interval for the proportion of all drivers who ran one or more of the last 10 red lights they met. (b) Nonresponse is a practical problem for this survey-only \(21.6 \%\) of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: do you think more or fewer than 171 of the 880 respondents really ran a red light? Why?

You are planning a survey of students at a large university to determine what proportion favor an increase in student fees to support an expansion of the student newspaper. Using records provided by the registrar, you can select a random sample of students. You will ask each student in the sample whether he or she is in favor of the proposed increase. Your budget will allow a sample of 100 students. (a) For a sample of size 100, construct a table of the margins of error for \(95 \%\) confidence intervals when \(\mathrm{p}^{\wedge \hat{p}}\) takes the values \(0.1,0.3,0.5,0.7\), and \(0.9\). (b) A former editor of the student newspaper offers to provide funds for a sample of size 500 . Repeat the margin of error calculations in part (a) for the larger sample size. Then write a short thank-you note to the former editor describing how the larger sample size will improve the results of the survey.

The proportion of drivers who use seat belts depends on characteristics such as age, sex, and ethnicity. As part of a broader study, investigators observed a random sample of 117 female Hispanic drivers in Boston. Suppose that, in fact, \(60 \%\) of all female Hispanic drivers in the Boston area wear seat belts. In repeated samples, the sample proportion \(p^{\wedge \hat{p}}\) would follow approximately a Normal distribution with mean (a) \(70.2\). (b) \(0.6\) (C) \(0.002\).
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