Question

Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children" (Pediatrics [2004]: 112-118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. a. Use the given information to estimate the proportion of American children who eat fast food on a typical day. b. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in the context of this problem.

Short answer

The estimated proportion of American children who eat fast food on a typical day is 27.7%. The conditions for the margin of error formula are met. The margin of error is 1.2%, which means we are 95% confident that the true proportion of American children who eat fast food on a typical day is between 26.5% and 28.9%.

Step-by-step solution

Estimating the proportion

To estimate the proportion, divide the number of children who ate fast food (1,720) by the total number of children in the sample (6,212). The resulting proportion is 1,720 / 6,212 = 0.277 or 27.7%.

Verifying conditions for margin of error formula

The conditions required for the margin of error formula to be appropriate are that the sample is a random sample, and that both np and n(1-p) are greater than 5. Here, the sample is given to be random. np = 6212 * 0.277 = 1720 and n(1-p) = 6212 * (1-0.277) = 4492, which are both greater than 5, so the conditions are met.

Calculating the margin of error

We calculate the margin of error using the formula \(E = z\sqrt{\frac{{p(1 - p)}}{{n}}}\), where E is the margin of error, z is the z-score for our confidence level (for a 95% confidence level, z is approximately 1.96), p is the sample proportion and n is the sample size. Plugging in our numbers, we get \(E = 1.96*\sqrt{(0.277*(1 - 0.277))/6212}\) = 0.012 or 1.2%.

Interpreting the margin of error

The margin of error gives us a range in which we expect the true proportion to lie, with a certain level of confidence. In this case, the margin of error is 1.2%, so we can say that we are 95% confident that the true proportion of American children who eat fast food on a typical day is between 26.5% and 28.9%.

Key concepts

Sample Proportion Estimation

When researchers aim to understand the behavior of a large group based on a sample, estimating the sample proportion is a crucial step. In the context of nutrition research, as with our example of children consuming fast food, this involves determining the fraction of participants exhibiting the behavior of interest out of the total sample.

To estimate the proportion, one simply divides the number of individuals with the characteristic (those children who ate fast food) by the total number of individuals in the sample. In the exercise, the sample proportion becomes \(\frac{1720}{6212} = 0.277\) or 27.7%. This represents an estimate of the entire population's behavior, providing a snapshot that researchers can use to infer broader trends.

Verifying Conditions for Margin of Error

To ensure the reliability of conclusions drawn from a sample, certain conditions must be met before calculating the margin of error. These are essentially checks to verify that the sample data will give meaningful results when generalized to the population.

Firstly, the sample must be random. This requirement ensures that every member of the population has an equal chance of being included, which helps prevent bias. Secondly, the conditions that both \(np\) and \(n(1-p)\) exceed 5 are necessary when using a normal approximation for the binomial distribution. In our example, with \(np = 1720\) and \(n(1-p) = 4492\), both conditions are satisfied, indicating the margin of error computation is suitable for our sample.

Calculating Margin of Error

The margin of error quantifies the uncertainty inherent in estimating a population parameter from a sample statistic. To calculate it, one uses the formula \(E = z\sqrt{\frac{p(1 - p)}{n}}\), where \(E\) indicates the margin of error, \(z\) corresponds to the z-score linked with the desired confidence level, \(p\) is the sample proportion (the estimate of the true population proportion), and \(n\) is the sample size.

For a 95% confidence level, the z-score is approximately 1.96. Plugging in the values from our exercise, the calculation provides a margin of error of 0.012, or 1.2%. This figure represents the range within which the true population proportion is likely to fall and is pivotal in assessing the precision of our sample proportion.

Confidence Level Interpretation

The confidence level reflects the degree of certainty one has regarding the estimation of the population proportion based on the sample data. Interpreting the confidence level in relation to the margin of error allows us to state how sure we are that the actual value lies within a certain range around our sample estimate.

In our fast-food study, with a margin of error of 1.2% and a 95% confidence level, we express that we can be 95% confident the true proportion of American children who eat fast food on a typical day is between 26.5% (27.7% - 1.2%) and 28.9% (27.7% + 1.2%). This interval is known as the confidence interval, offering a range, not just a singular estimate, to account for possible sample variations.