Question

In a sample of 848 children ages 3 to 5 it was found that \(3.7 \%\) of them had iron deficiency. \({ }^{17}\) Use these data to construct a \(90 \%\) confidence interval for the proportion of all 3 - to 5 -year-old children with iron deficiency.

Short answer

The 90% confidence interval for the proportion of all 3-to-5-year-old children with iron deficiency is calculated to be approximately (0.0281, 0.0459).

Step-by-step solution

Define the Known Values

Identify and note the sample size (n) and the sample proportion (\( \hat{p} \)) of children with iron deficiency. In this case, the sample size is 848 and the sample proportion is 3.7% or 0.037.

Calculate the Standard Error

Compute the standard error (SE) for the proportion using the formula SE = \( \sqrt{\frac{{\hat{p}(1 - \hat{p})}}{n}} \). Substitute the known values to find SE.

Find the Z-Score for the Confidence Level

Determine the z-score associated with a 90% confidence interval. This can often be found using a z-score table or calculator. For a 90% confidence interval, the z-score is typically around 1.645.

Calculate the Margin of Error

Calculate the margin of error (ME) using the formula ME = z * SE. Multiply the z-score by the standard error to find the margin of error.

Construct the Confidence Interval

Find the lower and upper bounds of the confidence interval using the sample proportion and the margin of error. The confidence interval is given by \( (\hat{p} - ME, \hat{p} + ME) \).

Key concepts

Sample Proportion

The sample proportion is a statistic that represents the fraction of individuals in a sample with a certain characteristic. In the context of our example, the characteristic of interest is iron deficiency among children aged 3 to 5. From the given exercise, a sample of 848 children was studied, and it was found that 3.7% of them had iron deficiency. This percentage represents the sample proportion, notated as \( \hat{p} \), which is crucial for making inferences about the larger population. To convert this percentage into a proportion, we divide by 100, resulting in \( \hat{p} = 0.037 \). Understanding the sample proportion is fundamental, as it forms the basis for calculating other statistics, such as the standard error and the margin of error, which are used to construct confidence intervals.

When interpreting the sample proportion, it's important to consider the size of the sample. Larger samples tend to give more accurate estimates of the population proportion, as they are less subject to random fluctuations. In the improvement of exercises, ensuring that the link between the sample proportion and the implications for the larger population is clear can be incredibly helpful for students grasping the concepts of inferential statistics.

Standard Error

Standard error (SE) is a measure of the amount of sampling variability one can expect in a statistic. It essentially gauges how much a sample statistic, like the sample proportion, would vary from one sample to another if multiple samples were taken from the same population.

In our exercise, the standard error for the sample proportion is computed using the formula \( SE = \sqrt{\frac{{\hat{p}(1 - \hat{p})}}{n}} \), where \( \hat{p} \) is the sample proportion and \( n \) is the sample size. This formula takes into account both the success rate (\( \hat{p} \)) and the failure rate (\( 1 - \hat{p} \)) in the sample, and adjusts for the size of the sample. The larger the sample size, the smaller the standard error, reflecting a more precise estimate of the population proportion.

Clarifying the standard error in the solutions and presenting it in a way that shows its relevance to the reliability of the estimate can help solidify students' understanding. Furthermore, examples or visuals that contrast the standard error for small versus large samples could aid in appreciation of why larger samples are more desirable in statistical analysis.

Margin of Error

The margin of error (ME) is a statistic that quantifies the uncertainty associated with the sample proportion. It provides a range above and below the sample proportion to express the precision of our estimate. The smaller the margin of error, the more confident we can be that the sample proportion closely approximates the true population proportion.

To calculate the margin of error, we use the formula \( ME = z \times SE \), where \( z \) is the z-score corresponding to our desired confidence level and \( SE \) is the standard error. The z-score reflects how many standard errors away from the sample proportion we extend our interval to achieve the specified level of confidence. In the exercise solution, for a 90% confidence interval, the z-score is 1.645, which is derived from the normal distribution table.

For students, understanding how the margin of error influences the width of a confidence interval is paramount. It's also beneficial to explain how the confidence level, typically 90%, 95%, or 99%, affects both the z-score and subsequently, the margin of error. Emphasizing that a larger margin of error gives us a wider confidence interval, implying less precision in our estimate, can help students grasp the trade-offs involved in statistical inference.

Z-Score

The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, expressed in units of the standard deviation. In the context of confidence intervals, a z-score represents the number of standard errors away from the sample proportion one would move to capture the central area of the normal distribution corresponding to the desired confidence level.

In the provided example, the desired confidence level is 90%, which corresponds to a z-score of approximately 1.645. This is because in a normal distribution, 90% of the area (or probability) lies within 1.645 standard deviations on either side of the mean. Using the z-score, we can then determine the margin of error and construct our confidence interval.

For educational purposes, it's essential to illustrate the connection between confidence levels and z-scores. Drawing a normal distribution and highlighting the area under the curve that corresponds to the desired confidence level can visually anchor the concept for students. Additionally, demonstrating how to use a z-score table to find the appropriate value for different confidence levels will aid in their comprehension of how theoretical concepts are applied in practice.