Examine the results of a study investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics 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 [tex]ng/mL[/tex]) in the urine: DEHP and DiNP.

Find a [tex]95\%[/tex] confidence interval for the difference, [tex]\mu_F - \mu_N[/tex], 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 3,095 participants who had eaten fast food was [tex]\bar{x}_F = 83.6[/tex] with [tex]s_F = 194.7[/tex], while the mean for the 5,782 participants who had not eaten fast food was [tex]\bar{x}_N = 59.1[/tex] with [tex]s_N = 152.1[/tex].

Round your answers to one decimal place.

The [tex]95\%[/tex] confidence interval is from [tex]\square[/tex] to [tex]\square[/tex].

Reference:
Zota, A.R., Phillips, C.A., Mitro, S.D., "Recent Fast Food Consumption and Bisphenol A and Phthalates Exposure among the U.S. Population in NHANES, 2003 - 2010," Environmental Health Perspectives, 13 April 2016.

To solve this problem of finding a 95% confidence interval for the difference in mean concentrations of DEHP between two groups (those who ate fast food and those who didn't), follow these steps:

1. Understand the Given Data:
- Group 1 (Fast Food Consumers):
- Sample size ([tex]$n_F$[/tex]) = 3095
- Mean concentration ([tex]$\bar{x}_F$[/tex]) = 83.6 ng/mL
- Standard deviation ([tex]$s_F$[/tex]) = 194.7 ng/mL

- Group 2 (Non-Fast Food Consumers):
- Sample size ([tex]$n_N$[/tex]) = 5782
- Mean concentration ([tex]$\bar{x}_N$[/tex]) = 59.1 ng/mL
- Standard deviation ([tex]$s_N$[/tex]) = 152.1 ng/mL

2. Calculate the Difference in Means:
[tex]\[
\text{Difference in means} = \bar{x}_F - \bar{x}_N = 83.6 - 59.1 = 24.5
\][/tex]

3. Calculate the Standard Error (SE) of the Difference in Means:
The SE is calculated using the formula:
[tex]\[
SE = \sqrt{\left(\frac{s_F^2}{n_F}\right) + \left(\frac{s_N^2}{n_N}\right)}
\][/tex]
Given the result, the calculated SE is approximately [tex]$4.03$[/tex].

4. Determine the Z-Score for a 95% Confidence Level:
For a 95% confidence interval, the Z-score is 1.96.

5. Calculate the Margin of Error:
[tex]\[
\text{Margin of Error} = Z \times SE = 1.96 \times 4.03 \approx 7.90
\][/tex]

6. Compute the Confidence Interval:
- Lower bound: [tex]$\text{Difference in means} - \text{Margin of Error} = 24.5 - 7.90 = 16.6$[/tex]
- Upper bound: [tex]$\text{Difference in means} + \text{Margin of Error} = 24.5 + 7.90 = 32.4$[/tex]

Therefore, the 95% confidence interval for the difference in mean concentrations of DEHP between people who have eaten fast food and those who haven't is approximately from 16.6 to 32.4 ng/mL.