Is Fast Food Messing With Your Hormones?

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]99\%[/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]99\%[/tex] confidence interval is [ ] to [ ].

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 determine a 99% confidence interval for the difference in mean concentration of DEHP between people who have eaten fast food and those who haven't, let's break it down step-by-step.

### Step 1: Identify the given data
- Mean concentration for people who ate fast food, [tex]$\bar{x}_F = 83.6 \, \text{ng/mL}$[/tex]
- Mean concentration for people who did not eat fast food, [tex]$\bar{x}_N = 59.1 \, \text{ng/mL}$[/tex]
- Standard deviation for fast food group, [tex]$s_F = 194.7 \, \text{ng/mL}$[/tex]
- Standard deviation for no fast food group, [tex]$s_N = 152.1 \, \text{ng/mL}$[/tex]
- Size of fast food group, [tex]$n_F = 3095$[/tex]
- Size of no fast food group, [tex]$n_N = 5782$[/tex]

### Step 2: Calculate the standard error of the difference in means
The formula to compute the standard error (SE) is:

[tex]\[
SE = \sqrt{\left(\frac{s_F^2}{n_F}\right) + \left(\frac{s_N^2}{n_N}\right)}
\][/tex]

### Step 3: Determine the critical value for a 99% confidence interval
For a 99% confidence level, we look at the Z-score for the two-tailed test. The Z-score is approximately 2.576.

### Step 4: Compute the margin of error
The margin of error (ME) is the product of the critical value and the standard error:

[tex]\[
ME = Z \times SE
\][/tex]

### Step 5: Calculate the confidence interval
The confidence interval for the difference in means [tex]$(\mu_F - \mu_N)$[/tex] is given by:

[tex]\[
CI = \left((\bar{x}_F - \bar{x}_N) - ME, (\bar{x}_F - \bar{x}_N) + ME\right)
\][/tex]

### Result
After completing the steps mentioned, the 99% confidence interval for the difference in mean concentration of DEHP between people who have and haven't eaten fast food is from 14.1 to 34.9 ng/mL.

This interval suggests that those who eat fast food have, on average, a higher concentration of DEHP in their bodies compared to those who do not.