Question

In a survey of 1000 US adults, twenty percent say they never exercise. This is the highest level seen in five years. \(^{5}\) Find and interpret a \(99 \%\) confidence interval for the proportion of US adults who say they never exercise. What is the margin of error, with \(99 \%\) confidence?

Short answer

The 99% confidence interval for the proportion of US adults who never exercise is between 16.77% and 23.23%. The margin of error, at 99% confidence, is approximately 0.0326 or 3.26%.

Step-by-step solution

Calculate the Sample Proportion

Calculate the sample proportion ( \( p \) ) by dividing the number of adults who say they 'never' exercise by the total number of adults surveyed. Since there are 200 adults who 'never' exercise, out of 1000 respondents, \( p = 200 / 1000 = 0.20 \).

Determine the Z-Score

For a 99% confidence level, the corresponding Z-score value (denoted as \( Z_{0.01/2} \)), can be found in a standard normal distribution table as 2.58.

Calculate the Standard Error

Standard error of the sample proportion is calculated as \( SE = \sqrt{ (p*(1-p)) / n } \). Note that \( n \) is the sample size. After substituting \( p = 0.20 \) and \( n = 1000 \), the standard error becomes \( SE = \sqrt{ (0.20 * 0.80) / 1000 } = 0.01265 \).

Determine Confidence Interval

We use the formula \( CI = p \pm Z_{\alpha/2}*SE \). By substituting the values \( p = 0.20 \), \( Z_{\alpha/2} = 2.58 \), and \( SE = 0.01265 \) into the formula, the confidence interval becomes \( CI = 0.20 \pm 2.58 * 0.01265 \), which reduces to \( CI = [0.1677, 0.2323] \).

Calculate the Margin of Error

Margin of error is calculated as \( ME = Z_{\alpha/2}*SE \). Substitute the values of \( Z_{0.01/2} = 2.58 \) and \( SE = 0.01265 \) into the formula to obtain \( ME = 2.58 * 0.01265 = 0.0326 \)

Key concepts

Sample Proportion

Understanding the concept of sample proportion is paramount when conducting a survey or an experiment where you're gauging prevalence or frequency of a certain trait within a population.

For instance, imagine you've asked 1000 people about their exercise habits, and 200 people admit they never exercise. The sample proportion (\( p \)) is then the fraction representing this group, calculated by dividing the number who never exercise by the total number surveyed, which in this case is 200 out of 1000, or 0.20.

This proportion is a crucial building block for further statistical assessments, including confidence intervals and margin of error, because it offers a snapshot of the characteristic being measured within your sample that you will use to infer about the larger population.

Z-score

A Z-score is a statistical measure that tells us how many standard deviations an element is from the mean. In the context of confidence intervals, it's used to determine how confident we can be about the range within which the true population proportion lies, based on our sample data.

In our current example, a 99% confidence level requires a Z-score that corresponds to the area under the standard normal curve. We commonly refer to statistical tables, or nowadays, we use software that provides us with a Z-score of 2.58. This high Z-score reflects the high confidence level we are aiming for and widens the confidence interval accordingly. Consequently, a higher confidence level like 99% (compared to 95%, for instance) leads to a broader range of values, but with more certainty that this range includes the true population proportion.

Standard Error

The standard error (SE) is the estimate of the standard deviation of the sampling distribution of a statistic, most commonly of the mean or proportion. It's a measure of how much we expect the sample statistic to vary from one random sample to another.

In the context of a proportion, the formula for standard error is \( SE = \sqrt{ (p*(1-p)) / n } \), where \( p \) is the sample proportion and \( n \) is the sample size. As shown in our exercise, with a sample proportion of 0.20 and a sample size of 1000, the standard error is \( SE = \sqrt{ (0.20 * 0.80) / 1000 } = 0.01265 \). The smaller the standard error, the more precise our estimate of the population proportion is likely to be.

Margin of Error

A fundamental concept when communicating the precision of survey results is the margin of error (ME). It reflects the maximum expected difference between the true population parameter and a sample estimate of that parameter.

The margin of error is calculated using the standard error and the Z-score. In our given problem, by multiplying the standard error (0.01265) with the Z-score (2.58), we get a margin of error of about 0.0326. This means we can say with 99% confidence that the true proportion of US adults who do not exercise is within approximately 3.26 percentage points of the sample proportion (20%). Understanding and reporting this margin allows people to comprehend the possible variation in the data and assess the reliability of the survey results.

Statistical Significance

The concept of statistical significance is typically associated with hypothesis testing, but it's also relevant when discussing confidence intervals. It tells us whether the results from a study or experiment can be considered to reflect a true effect rather than just random chance.

When we calculate a 99% confidence interval, we are indicating that, if we were to take many samples and build a confidence interval from each sample, 99% of those intervals would include the true population proportion. In other words, there is only a 1% chance that our interval does not include the true proportion—leading us to believe that our results have statistical significance because the likelihood that they've occurred by chance is very low. Being aware of statistical significance is of immense value in assessing the validity of study outcomes.