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The article "Consumers Show Increased Liking for Diesel Autos" (USA Today, January 29,2003 ) reported that \(27 \%\) of U.S. consumers would opt for a diesel car if it ran as cleanly and performed as well as a car with a gas engine. Suppose that you suspect that the proportion might be different in your area. You decide to conduct a survey to estimate this proportion for the adult residents of your city. What is the required sample size if you want to estimate this proportion with a margin of error of 0.05 ? Calculate the required sample size first using 0.27 as a preliminary estimate of \(p\) and then using the conservative value of \(0.5 .\) How do the two sample sizes compare? What sample size would you recommend for this study?

Short Answer

Expert verified
The sample sizes that will be required for the margin error of 0.05 are 324 and 385, respectively for the proportions of 0.27 and 0.5. Due to the higher assurance of accuracy, the recommended sample size to be used in the survey is 385.

Step by step solution

01

Use the given proportion to compute for the first sample size

In this problem, the assumed proportion \(p\) is 0.27 and the desired margin of error \(E\) is 0.05. Since the problem does not specify a confidence level, you should assume a 95% confidence level, which corresponds to a Z-value (Z) of approximately 1.96. Plugging these values into the formula, we get: \(n = \frac{(0.27)(1-0.27)(1.96)^2}{(0.05)^2} \).
02

Simplify the calculation

Doing the calculation, the result is: \(n = 323.64\). Because we cannot have a fraction of a response, any fraction should be rounded up, giving us a sample size of 324 in order to achieve a 5% margin of error given the proportion of 0.27.
03

Use the conservative estimate to compute for the second sample size

Using the conservative estimate of \(p = 0.5\), we repeat the above calculation with this new value for \(p\). So we have \(n = \frac{(0.5)(1-0.5)(1.96)^2}{(0.05)^2}\). After solving this, the result should be 384.16, again rounded up to 385.
04

Compare the computed sample sizes and decide which one to use

Comparing the two computed sample sizes, we see that using the conservative estimate of \(p = 0.5\) yields a larger sample size. If you want to ensure a margin of error of 5% regardless of the actual proportion, it would be wise to use this larger sample size. If, however, you have reason to believe that the actual proportion is closer to \(0.27\), then the smaller sample size might suffice.

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Key Concepts

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

Margin of Error
When conducting surveys or research, the margin of error is a crucial metric that reflects the level of precision you can expect from your results. It is essentially an expression of the range within which the true value of the population parameter (e.g., proportion) is likely to fall, given the data from your sample.

The margin of error is linked to the 'confidence level', another fundamental aspect of statistical research. It decides how 'confident' you can be that the population parameter will fall within a certain range around your sample statistic. Commonly, the margin of error (MoE) is calculated using a formula that includes the sample proportion (\( p \)), the z-score corresponding to the chosen confidence level, and the sample size (\( n \)).

In the given exercise, the desired MoE was set at 0.05 (or 5%), which dictates how accurate the estimation of the population proportion is preferred to be. The initial calculation using a sample proportion of 0.27 leads to a sample size that will ensure that the true proportion is within 5% of the estimated 27%, 95 out of 100 times, under the assumption that the sample is a perfect random sample of the population.
Confidence Level
The confidence level, typically expressed as a percentage, represents how certain we are that the method we've used would capture the true parameter of the population if we were to repeat our study multiple times. For practical purposes, the 95% confidence level is often used, which implies that if we were to take 100 different samples and compute confidence intervals using the same method, approximately 95 of them would contain the true population parameter.

Understanding the confidence level is central to interpreting research findings correctly. It directly influences the z-score, which is a multiplier in the margin of error formula. Higher confidence levels correspond to higher z-scores, which, in turn, lead to larger sample sizes to maintain the margin of error. In the preamble to the exercise, it is presumed that a 95% confidence level is requested since it is not explicitly mentioned, corresponding to a z-score of 1.96.
Proportion Estimation
Proportion estimation is the task of estimating the value of a certain attribute within a population. For example, the original exercise intends to estimate the proportion of adults in a city who would prefer diesel cars, provided they performed as well as gasoline cars.

When estimating a proportion, choosing an initial value (known as a 'preliminary estimate') can greatly influence the required sample size. If no preliminary estimate is available, a conservative value that maximizes variability is used, typically set at 0.5. This choice ensures that the sample size will be large enough to achieve the desired margin of error for any proportion.

In our exercise, two different values of the proportion were used: an initial estimate of 0.27, and a conservative estimate of 0.5. The computations reveal that using the preliminary estimate leads to a smaller sample size compared to using the conservative value. However, the conservative value provides more 'security' as it will more likely result in a sufficiently large sample to estimate a wide range of potential true proportions with the desired precision.

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Most popular questions from this chapter

a. Use the given information to estimate the proportion of college students who use the Internet more than 3 hours per day. b. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in the context of this problem.Most American college students make use of the Internet for both academic and social purposes. What proportion of students use it for more than 3 hours a day? The authors of the paper "U.S. College Students" Internet Use: Race, Gender and Digital Divides" (Journal of Computer-Mediated Communication [2009]: 244-264) describe a survey of 7,421 students at 40 colleges and universities. The sample was selected to reflect general demographics of U.S. college students. Of the students surveyed, 2,998 reported Internet use of more than 3 hours per day.

The article "Hospitals Dispute Medtronic Data on Wires" (The Wall Street Journal, February 4, 2010) describes several studies of the failure rate of defibrillators used in the treatment of heart problems. In one study conducted by the Mayo Clinic, it was reported that failures within the first 2 years were experienced by 18 of 89 patients under 50 years old and 13 of 362 patients age 50 and older. Assume that these two samples are representative of patients who receive this type of defibrillator in the two age groups. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of patients under 50 years old who experience a failure within the first 2 years. b. Construct and interpret a \(99 \%\) confidence interval for the proportion of patients age 50 and older who experience a failure within the first 2 years. c. Suppose that the researchers wanted to estimate the proportion of patients under 50 years old who experience this type of failure with a margin of error of \(0.03 .\) How large a sample should be used? Use the given study results to obtain a preliminary estimate of the population proportion.

The formula used to calculate a large-sample confidence interval for \(p\) is $$ \hat{p} \pm(z \text { critial value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) b. \(98 \%\) c. \(85 \%\)

Thereport"2005 ElectronicMonitoring\& Surveillance \(\begin{array}{lll}\text { Survey: } & \text { Many Companies Monitoring, } & \text { Recording, }\end{array}\) Videotaping-and Firing-Employees" (American Management nesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet. Assume that this sample is representative of businesses in the United States. a. Estimate the proportion of all businesses in the U.S. that have fired workers for misuse of the Internet. What statistic did you use? b. Use the sample data to estimate the standard error of \(\hat{p}\). c. Calculate and interpret the margin of error associated with the estimate in Part (a). (Hint: See Example 9.3 )

Use the formula for the standard error of \(\hat{p}\) to explain why a. The standard error is greater when the value of the population proportion \(p\) is near 0.5 than when it is near \(1 .\) b. The standard error of \(\hat{p}\) is the same when the value of the population proportion is \(p=0.2\) as it is when \(p=0.8\)

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