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Chapter 9: Problem 67

Data from a representative sample were used to estimate that \(32 \%\) of all computer users in 2011 had tried to get on a Wi-Fi network that was not their own in order to save money (USA Today, May 16,2011 ). You decide to conduct a survey to estimate this proportion for the current year. 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 0.32 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 required sample sizes for the two cases are 866 and 961, respectively. Given the slight difference and for greater assurance in the results of the survey, a sample size of 961 is recommended.

Step by step solution

01

Find Formula for Sample Size

The formula for finding sample size when you want to estimate a proportion with a certain margin of error is given by \(n = p*(1-p)*(z/E)^2\). Here, 'n' is the sample size, 'p' is the preliminary estimate, 'E' is the margin of error, and 'z' is the z-score for the desired level of confidence.
02

Calculate Sample Size with p=0.32

In this scenario, 'p' is 0.32, 'E' is 0.05, and 'z' is 1.96 (z-score for 95% confidence level, assuming we want to be 95% confident in our estimate). Using these values in the sample size formula, we get \(n = 0.32*(1-0.32)*(1.96/0.05)^2\). Calculating this gives us \(n 鈮� 865.2\). Since we can't have a fractional sample size, we round up to the next whole number, resulting in \(n = 866\).
03

Calculate Sample Size with p=0.5

Now, we calculate the sample size using the conservative value of 'p'=0.5. Thus, our formula becomes \(n = 0.5*(1-0.5)*(1.96/0.05)^2\). Calculating this yields \(n 鈮� 960.4\). Again, rounding up to the next whole number, we get \(n = 961\).
04

Compare Sample Sizes and Give Recommendation

Note that the sample size required when using 'p'=0.5 is larger than when using 'p'=0.32. This is because a larger 'p' value implies greater variability, and hence, a larger sample size is needed for the same margin of error. The larger sample size can be viewed as a more conservative estimate. Considering the minimal difference in sample size and for greater assurance in the results, it is recommended to use the larger sample size of 961.

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Key Concepts

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

Estimate Proportion
Estimating a proportion means trying to understand what portion of a population behaves in a certain way or shares a specific characteristic. In this case, you want to know what percentage of computer users try to access Wi-Fi networks that aren't theirs.

To estimate a proportion effectively, sometimes you use existing data as a preliminary estimate. For example, past data suggest that 32% of users did this in 2011. This gives you a preliminary estimate of \(p = 0.32\). Even if things have changed over the years, using this kind of preliminary data can still be helpful.

If there's no preliminary estimate available, you can use a conservative proportion of 0.5. This is because 0.5 represents maximum variability, since it divides a group most evenly between two outcomes 鈥� those who do and those who don't exhibit a certain behavior.
Margin of Error
The margin of error represents the amount of error that you can tolerate in your estimate. In simpler terms, it's how much wiggle room you're allowing for your results to be off.

In the exercise above, the margin of error is set at 0.05. This margin tells us that if 32% is the true proportion, our final results could vary by 5%. It means our estimate could range from 27% to 37%.

The choice of margin of error directly impacts the required sample size. A smaller margin means you want more precision, which requires a larger sample. Conversely, allowing a larger margin means you're okay with less precision, so you can do with a smaller sample. This balance is crucial in planning your survey.
Confidence Interval
A confidence interval helps quantify the uncertainty of a proportion estimate. It's a range that likely contains the true proportion of the population you're studying.

The confidence interval depends on the margin of error and the confidence level. A confidence level, like 95% used in our exercise, expresses how sure we are that the interval includes the true proportion.

Here's how it works: if you conduct 100 surveys, a confidence interval at 95% means you expect 95 out of those 100 intervals to contain the true proportion.

By using the z-score associated with your chosen confidence level and your calculated margin of error, you can create a confidence interval. For example, with a z-score of 1.96 for 95% confidence, you can calculate how wide your interval needs to be using the margin of error calculation.

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Most popular questions from this chapter

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 )

A random sample will be selected from the population of all adult residents of a particular city. The sample proportion \(\hat{p}\) will be used to estimate \(p,\) the proportion of all adult residents who are registered to vote. For which of the following situations will the estimate tend to be closest to the actual value of \(p ?\) I. \(\quad n=1,000, p=0.5\) II. \(\quad n=200, p=0.6\) III. \(\quad n=100, p=0.7\)

In a survey of 1,000 randomly selected adults in the United States, participants were asked what their most favorite and least favorite subjects were when they were in school (Associated Press, August 17,2005\()\). In what might seem like a contradiction, math was chosen more often than any other subject in both categories. Math was chosen by 230 of the 1,000 as their most favorite subject and chosen by 370 of the 1,000 as their least favorite subject. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was their most favorite subject. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was their least favorite subject.

Consider taking a random sample from a population with \(p=0.40\) a. What is the standard error of \(\hat{p}\) for random samples of size \(100 ?\) b. Would the standard error of \(\hat{p}\) be larger for samples of size 100 or samples of size \(200 ?\) c. If the sample size were doubled from 100 to 200 , by what factor would the standard error of \(\hat{p}\) decrease?

USA Today (October 14,2002 ) reported that \(36 \%\) of adult drivers admit that they often or sometimes talk on a cell phone when driving. This estimate was based on data from a representative sample of 1,004 adult drivers. A margin of error of \(3.1 \%\) was also reported. Is this margin of error correct? Explain.
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