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Lying about age Pew Research, in November 2011 polled a random sample of 799 U.S. teens about Internet use. \(49 \%\) of those teens reported that they had misrepresented their age online to gain access to websites and online services. a) Find the margin of error for this poll if we want \(95 \%\) confidence in our estimate of the percent of American teens who have misrepresented their age online. b) Explain what that margin of error means. c) If we only need to be \(90 \%\) confident, will the margin of error be larger or smaller? Explain. d) Find that margin of error. e) In general, if all other aspects of the situation remain the same, would smaller samples produce smaller or larger margins of error?

Short Answer

Expert verified
The 95% margin of error is 3.47%. The 90% margin of error is 2.92%. Smaller samples produce larger margins of error.

Step by step solution

01

Understanding the Confidence Level

The confidence level is the probability that the confidence interval contains the true population parameter. For part (a), this is 95%, and for part (c), it's 90%.
02

Calculate the Margin of Error for 95% Confidence

The formula for the margin of error (ME) at a given confidence level is:\[ ME = z \sqrt{\frac{p(1-p)}{n}} \]where \( p = 0.49 \), \( n = 799 \), and \( z \) is the z-score corresponding to the desired confidence level. For 95% confidence, \( z \approx 1.96 \). Plug in the values:\[ ME = 1.96 \sqrt{\frac{0.49 \times 0.51}{799}} \approx 0.0347 \]Thus, the margin of error is approximately 3.47%.
03

Interpretation of Margin of Error

The margin of error of 3.47% means that we are 95% confident that the true proportion of American teens who have misrepresented their age online is within 3.47% of 49%.
04

Predicting Margin of Error for 90% Confidence

With lower confidence (90%), the margin of error will be smaller because the z-score for 90% confidence (approximately 1.645) is smaller than that for 95% confidence.
05

Calculate the Margin of Error for 90% Confidence

Using the same formula with a z-score of approximately 1.645:\[ ME = 1.645 \sqrt{\frac{0.49 \times 0.51}{799}} \approx 0.0292 \]Therefore, the margin of error is approximately 2.92%.
06

Understanding Effect of Sample Size on Margin of Error

Larger samples generally produce smaller margins of error because the sample provides a more accurate estimate of the population parameter. Thus, smaller samples lead to larger margins of error.

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Key Concepts

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

Confidence Level
When conducting surveys or polls, the confidence level tells us how sure we are that the interval calculated contains the true population parameter. Imagine a large number of samples from the population. A 95% confidence level means that about 95% of those samples will contain the true average or proportion. The higher the confidence level, the more certain we are about our estimates. Common confidence levels are 90%, 95%, and 99%. In the exercise, a 95% confidence level was used to determine the percentage of teens misrepresenting their age online. This means there's a 95% chance that the real percentage lies within the calculated margin of error around the sample percentage of 49%. Lowering the confidence level to 90% means accepting a higher risk of the confidence interval not containing the true population value. But it comes with the benefit of a smaller margin of error.
Sample Size
Sample size plays a critical role in determining the accuracy of survey results. Larger samples generally provide more reliable estimates of the population parameter since they tend to capture a wider variety of responses. For instance, in the exercise, a sample of 799 teens was used. If the sample size were smaller, it would mean each individual's data weighs more heavily on the result, increasing the variability and hence the margin of error. On the other hand, increasing the sample size reduces the margin of error because it tends to balance out any anomalies or biases in individual responses. This concept shows why larger surveys or polls, though more expensive and time-consuming, yield more accurate estimates.
Z-Score
The z-score is a critical statistical measure used to determine the margin of error. It represents how many standard deviations an element is from the mean. The z-score changes with the confidence level. In the context of the exercise, for a 95% confidence level, the z-score is 1.96. For a 90% confidence level, it decreases to approximately 1.645. This z-score is plugged into the margin of error formula. A higher z-score means a larger margin of error because the confidence interval has to account for more variation. Conversely, a smaller z-score means a smaller interval, reflecting less variation. Understanding and correctly using z-scores is essential in statistics to ensure accurate confidence intervals.
Population Parameter
The population parameter is a value that describes something about the entire population you're studying. It could be a mean, proportion, or variance. In the given exercise, the parameter of interest is the true proportion of teenagers who lie about their age online. This parameter is unknown, which is where sample statistics come into play. We use data from a sample to estimate it with a certain level of confidence. In this case, the percentage calculated from the sample was 49%, and the confidence intervals give a range in which the true parameter likely falls given the derived margin of error. The whole purpose of deriving margins of error, confidence intervals, and using sample data is to provide insights into the population parameter without surveying the entire population, which is often impractical or impossible.

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