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Hearing Loss in Teenagers A recent study" found that, of the 1771 participants aged 12 to 19 in the National Health and Nutrition Examination Survey, \(19.5 \%\) had some hearing loss (defined as a loss of 15 decibels in at least one ear). This is a dramatic increase from a decade ago. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error for the proportion is \(S E=0.009 .\) Find and interpret a \(90 \%\) confidence interval for the proportion of teenagers with some hearing loss.

Short Answer

Expert verified
The 90% confidence interval for the proportion of teenagers with some hearing loss is (0.18, 0.21). This means we can be 90% confident that the true proportion of teenagers with some level of hearing loss in the population is between 18.0% and 21.0%.

Step by step solution

01

Calculate the Margin of Error

The Margin of Error (E) for a confidence interval can be calculated using the formula \(E = Z * SE\) where Z is the Z-score. For a 90% confidence interval, the Z-score is 1.645. Given that the standard error (SE) is 0.009, the margin of error is \(E = 1.645 * 0.009 = 0.014805\). Rounded to three decimal places, E is 0.015.
02

Calculate The Confidence Interval

With the sample proportion (p̂) and Margin of Error (E) known, the confidence interval (CI) can be calculated using the formula \(CI = p̂ ± E\). Given that the sample proportion is 0.195, the confidence interval is \(CI = 0.195 ± 0.015\). This results in an interval of (0.180, 0.210).
03

Interpret The Confidence Interval

The confidence interval means that, given the sample data, we are 90% confident that the true population proportion of teenagers with some level of hearing loss lies between 18.0% and 21.0%.

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Key Concepts

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

Standard Error
The concept of standard error is crucial in understanding the precision of our sample statistics as estimates of the population parameters. The standard error (SE) quantifies the amount of variation or "spread" in the sampling distribution of a statistic, like a sample proportion. It tells us how much the sample statistics may vary from the true population statistics if different samples are taken. The smaller the standard error, the more accurate the sample proportion is as an estimate of the true population proportion.

In our original exercise with hearing loss, the standard error has been calculated based on the observed sample of participants. Specifically, the reported standard error is 0.009. This value conveys that the sample proportion is likely to vary by this much if different samples of the same size were drawn repeatedly from the same population. It gives us a sense of how much "wiggle room" there might be in the observed statistics, allowing for a more informed interpretation of confidence intervals.
Margin of Error
The margin of error is a measure that helps us understand the degree of uncertainty inherent in our sample estimate. When you see a confidence interval, the margin of error (E) represents the amount we add or subtract from the sample statistic to establish the interval. This range offers a buffer around the sample proportion to account for potential sampling errors.

To find the margin of error for a confidence interval, we multiply the standard error (SE) by the Z-score corresponding to our desired level of confidence. In the exercise, we are using a 90% confidence interval, so the Z-score is 1.645. Therefore, the calculation for the margin of error is as follows:
  • Formula:
    \(E = Z * SE\)
  • Using given values:
    \(E = 1.645 * 0.009 = 0.014805\)
  • Rounding gives us a margin of error of 0.015.
Understanding the margin of error provides insight into the precision of our confidence interval, indicating how much sampling variability is accounted for when estimating the true proportion of the population.
Normal Distribution
The normal distribution is a foundational concept in statistics, often referred to as the bell curve due to its shape. It's a probability distribution that is symmetric and depicts the distribution of many types of data. Most scores fall near the average (mean), and scores taper off symmetrically towards the extremes.

In the context of confidence intervals, the normal distribution is useful for its properties when estimating population parameters from sample statistics. It assumes that when we take a significant number of samples, the distribution of the sample means will resemble a normal distribution. This is why researchers can use the properties of the normal curve to make inferences about the population. Particularly, it allows us to determine the Z-score, which is essential for calculating margins of error and confidence intervals.

Our exercise made use of the normal distribution because the sample size was large enough, which enabled accurate approximation in the calculations of the confidence interval for the proportion of teenagers experiencing hearing loss. This illustrates the significance of the normal distribution in statistical analyses.

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Most popular questions from this chapter

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