Question

Exercises 6.192 and 6.193 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 DEHP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=83.6\) with \(s_{F}=194.7\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=59.1\) with \(s_{N}=152.1\)

Short answer

The 95% confidence interval for the difference in mean concentration of DEHP between people who have eaten fast food in the last 24 hours and those who haven't is from 13.64 ng/mL to 35.36 ng/mL.

Step-by-step solution

Identify Given Data

First, you need to identify the given data. For the people who have eaten fast food, you have \( n_{F} = 3095, \bar{x}_{F} = 83.6, s_{F} = 194.7 \). For those who have not eaten fast food, you have \( n_{N} = 5782, \bar{x}_{N} = 59.1, s_{N} = 152.1 \). These values will be used to calculate the difference between the two means, along with the standard error for that difference.

Calculate Difference Between Means

Next, calculate the difference between the two sample means, \( \bar{x}_{F} - \bar{x}_{N} = 83.6 - 59.1 = 24.5 \) ng/mL. This is the observed difference in mean concentration of DEHP between the two groups.

Calculate Standard Error for Difference

The standard error for the difference between two means is given by the formula \[ SE = \sqrt { \frac {s_{F}^2}{n_{F}} + \frac {s_{N}^2}{n_{N}} } \], substituting the given values gives \[ SE = \sqrt { \frac {194.7^2}{3095} + \frac {152.1^2}{5782} } = 5.54 \] ng/mL.

Construct Confidence Interval

Lastly, construct the 95% confidence interval for the difference between the means using the formula \[ CI = (\bar{x}_{F} - \bar{x}_{N}) \pm Z*SE \] where Z is the Z-score for desired confidence level (1.96 for 95% confidence level). So \[ CI = 24.5 \pm 1.96*5.54 = 24.5 \pm 10.86 \]. Thus, the 95% confidence interval for the difference between the mean concentrations of DEHP for people who have eaten fast food and those who haven't ranges from 13.64 ng/mL to 35.36 ng/mL.