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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 )

Short Answer

Expert verified
Given the calculation above, a. the estimate of the proportion of all businesses in the U.S. that have fired workers for misuse of the internet is the sample proportion \(\hat{p}\). b. the standard error of \(\hat{p}\) can be estimated using the formula for standard error. c. The margin of error associated with the estimate can also be calculated using the formula for margin of error, which incorporates the standard error and the z-score corresponding to the desired level of confidence.

Step by step solution

01

Calculation of Sample Proportion

The sample proportion, often denoted by \(\hat{p}\), is calculated by dividing the number of successes in the sample by the total number of observations in the sample. In this case, the number of successes is the number of businesses that have fired workers for misuse of the Internet which is 137 and the total number of observations is the total number of businesses in the sample which is 526. Hence, \(\hat{p} = 137 / 526 \).
02

Estimation of Standard Error

The standard error of the sample proportion, often denoted by SE(\(\hat{p}\)), is given by the formula: SE(\(\hat{p}\)) = sqrt[\(\hat{p}(1 - \hat{p}) / n\)] where n is the total number of observations in the sample. Substituting the values of \(\hat{p}\) and n from the previous step, we can calculate SE(\(\hat{p}\)).
03

Calculation of Margin of Error

The margin of error, often denoted by E, is given by the formula: E = z * SE(\(\hat{p}\)) where z is the z-score, which corresponds to the desired level of confidence. In most cases, a 95% level of confidence is used, which corresponds to a z-score of approximately 1.96. Substituting the calculated value of SE(\(\hat{p}\)) and z = 1.96, we can get the value of E.

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Key Concepts

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

Standard Error Calculation
Understanding the standard error is crucial for interpreting how much the sample proportion, denoted as \(\hat{p}\), might differ from the actual population proportion. To calculate the standard error (SE) for \(\hat{p}\), you use the formula \(SE(\hat{p}) = \sqrt{\hat{p}(1 - \hat{p}) / n}\), where \(n\) is the sample size, \(\hat{p}\) is the sample proportion, and \(1 - \hat{p}\) represents the proportion of failures.

Let's simplify this with an example: if a survey shows that 137 out of 526 businesses fired employees for a certain reason, \(\hat{p}\) would be 137/526. Insert this value into our formula, and you get the standard error, which gives us an idea of how precisely our sample proportion estimates the true population proportion. The SE is smaller when the sample size is larger, which means we can be more confident in our estimate of \(\hat{p}\).
Margin of Error Interpretation
Once you determine the standard error, you can calculate the margin of error (E), which is essentially a buffer around our sample proportion estimate within which we are confident that the true population proportion lies. The margin of error formula is \(E = z \times SE(\hat{p})\), where \(z\) is the z-score that corresponds to our desired confidence level.

Interpreting Margin of Error

If our desired confidence level is 95%, we often use a \(z\)-score of approximately 1.96. The wider the margin, the less precise our estimate is, which may suggest we need a larger sample size for better precision. If a study finds that the margin of error is 5%, this means we are 95% confident that the true population proportion falls within 5% of our sample proportion — above or below. It's a measure of uncertainty in our estimate, due to the fact we're sampling only a fraction of the entire population.
Confidence Interval Estimation
Building off the standard error and margin of error, we come to the confidence interval, which is the range around our sample proportion where we believe the true population proportion lies, given a certain level of confidence. The confidence interval (CI) is calculated as \(CI = \hat{p} \pm E\).

Creating a Confidence Interval

To estimate a 95% CI, you add and subtract the margin of error from the sample proportion. This interval reveals that if we were to take numerous randomized samples and calculate their respective CIs, we'd expect about 95% of those intervals to contain the true population proportion. It's important to note, the level of confidence does not reflect the probability that the specific interval calculated from one sample contains the true population proportion. Rather, it's about the reliability of the method used to construct the interval over many samples.

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Most popular questions from this chapter

One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked "Do you think it is sometimes justified to lie, or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes OK. a. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who would say that lying is never justified. b. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who think that it is often or sometimes OK to lie to avoid hurting someone's feelings. c. Using the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses.

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?

Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1:(0.28,0.34) Interval 2:(0.31,0.33) (Hint: Consider the formula for the confidence interval given on page 401 ) a. Explain how it is possible that the two confidence intervals are not centered in the same place. b. Which of the two intervals conveys more precise information about the value of the population proportion? c. If both confidence intervals have a \(95 \%\) confidence level, which confidence interval was based on the smaller sample size? How can you tell? d. If both confidence intervals were based on the same sample size, which interval has the higher confidence level? How can you tell?

Based on data from a survey of 1,200 randomly selected Facebook users (USA Today, March 24, 2010), a \(90 \%\) confidence interval for the proportion of all Facebook users who say it is not OK to "friend" someone who reports to you at work is (0.60,0.64) . What is the meaning of the \(90 \%\) confidence level associated with this interval?

For estimating a population characteristic, why is an unbiased statistic with a small standard error preferred over an unbiased statistic with a larger standard error?

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