Question

Examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DiNP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=10.1\) with \(s_{F}=38.9\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=7.0\) with \(s_{N}=22.8\)

Short answer

The 95% confidence interval for the difference in the mean concentration of DiNP between people who have eaten fast food and those who haven't is calculated using data from the study. The interval is computed using the provided sample means and standard deviations, following the steps of standard error calculation, margin of error calculation and finally constructing the confidence interval.

Step-by-step solution

Identify the data

First, let's identify the data given. We have \(n_{F}=3095\) (number of people who have eaten fast food), \(\bar{x}_{F}=10.1\) (mean concentration of DiNP for people who have eaten fast food) and \(s_{F}=38.9\) (standard deviation for people who have eaten fast food). Similarly, we have \(n_{N}=5782\) (number of people who have not eaten fast food), \(\bar{x}_{N}=7.0\) (mean concentration of DiNP for people who have not eaten fast food) and \(s_{N}=22.8\) (standard deviation for people who have not eaten fast food).

Calculate the Standard Error (SE)

The standard error (SE) for difference of means is calculated by the formula \(\sqrt{\(s_{F}^{2} / n_{F}\) + \(s_{N}^{2} / n_{N}\)}\). First calculate \(s_{F}^{2} / n_{F}\) and \(s_{N}^{2} / n_{N}\) separately and then sum those values. The square root of that value gives the standard error. Substitute the values into the formula to get the standard error.

Find the Margin of Error

Next, we need to find the margin of error for a 95% confidence interval. This is given by the z-score times the standard error (SE). For a 95% confidence interval, the z-score (which we get from the standard normal distribution) is approximately 1.96. Thus, the margin of error is 1.96 times the standard error.

Construct the Confidence Interval

Finally, we can construct the 95% confidence interval for the difference in means, which is \((\mu_{F} - \mu_{N})\), by taking the difference in sample means (\(\bar{x}_{F} - \bar{x}_{N}\)) and adding and subtracting the margin of error. This will give us the lower and upper bounds of the confidence interval.