Answer :
To solve this problem, we need to find a 99% confidence interval for the difference in the mean concentration of DEHP between people who have eaten fast food in the last 24 hours and those who haven't. Here's how we would approach this step-by-step:
### Step 1: Gather the Information
We have two groups here:
1. Fast Food Group:
- Mean concentration ([tex]\(\bar{x}_F\)[/tex]) = 83.6 ng/mL
- Standard deviation ([tex]\(s_F\)[/tex]) = 194.7 ng/mL
- Sample size ([tex]\(n_F\)[/tex]) = 3095
2. Non-Fast Food Group:
- Mean concentration ([tex]\(\bar{x}_N\)[/tex]) = 59.1 ng/mL
- Standard deviation ([tex]\(s_N\)[/tex]) = 152.1 ng/mL
- Sample size ([tex]\(n_N\)[/tex]) = 5782
### Step 2: Calculate the Standard Error
The standard error (SE) is calculated using the formula for the difference between two means:
[tex]\[
SE = \sqrt{\left(\frac{s_F^2}{n_F}\right) + \left(\frac{s_N^2}{n_N}\right)}
\][/tex]
After plugging in the values, we get:
- SE ≈ 4.031
### Step 3: Determine the Critical Value
Because we're looking for a 99% confidence interval, we use a t-distribution to find the critical value. The degrees of freedom (df) are the smaller of [tex]\(n_F - 1\)[/tex] and [tex]\(n_N - 1\)[/tex], which in this case is 3094.
The critical value (t*) for a 99% confidence level with these degrees of freedom is approximately 2.577.
### Step 4: Calculate the Margin of Error
The margin of error (ME) is calculated as:
[tex]\[
ME = t^* \times SE
\][/tex]
Substituting the values, we get:
- ME ≈ 10.39
### Step 5: Compute the Confidence Interval
To find the confidence interval for the difference in means ([tex]\(\mu_F - \mu_N\)[/tex]), use:
[tex]\[
\text{Difference in means} = \bar{x}_F - \bar{x}_N
\][/tex]
The difference in means is:
[tex]\[
83.6 - 59.1 = 24.5
\][/tex]
Now, compute the confidence interval:
[tex]\[
\text{Lower bound} = 24.5 - 10.39 = 14.1
\][/tex]
[tex]\[
\text{Upper bound} = 24.5 + 10.39 = 34.9
\][/tex]
### Conclusion
The 99% confidence interval for the difference in the mean concentration of DEHP between those who have eaten fast food and those who haven't is [tex]\(14.1\)[/tex] to [tex]\(34.9\)[/tex] ng/mL. This interval suggests that fast food consumption is associated with a higher concentration of DEHP, within this range, with 99% confidence.
### Step 1: Gather the Information
We have two groups here:
1. Fast Food Group:
- Mean concentration ([tex]\(\bar{x}_F\)[/tex]) = 83.6 ng/mL
- Standard deviation ([tex]\(s_F\)[/tex]) = 194.7 ng/mL
- Sample size ([tex]\(n_F\)[/tex]) = 3095
2. Non-Fast Food Group:
- Mean concentration ([tex]\(\bar{x}_N\)[/tex]) = 59.1 ng/mL
- Standard deviation ([tex]\(s_N\)[/tex]) = 152.1 ng/mL
- Sample size ([tex]\(n_N\)[/tex]) = 5782
### Step 2: Calculate the Standard Error
The standard error (SE) is calculated using the formula for the difference between two means:
[tex]\[
SE = \sqrt{\left(\frac{s_F^2}{n_F}\right) + \left(\frac{s_N^2}{n_N}\right)}
\][/tex]
After plugging in the values, we get:
- SE ≈ 4.031
### Step 3: Determine the Critical Value
Because we're looking for a 99% confidence interval, we use a t-distribution to find the critical value. The degrees of freedom (df) are the smaller of [tex]\(n_F - 1\)[/tex] and [tex]\(n_N - 1\)[/tex], which in this case is 3094.
The critical value (t*) for a 99% confidence level with these degrees of freedom is approximately 2.577.
### Step 4: Calculate the Margin of Error
The margin of error (ME) is calculated as:
[tex]\[
ME = t^* \times SE
\][/tex]
Substituting the values, we get:
- ME ≈ 10.39
### Step 5: Compute the Confidence Interval
To find the confidence interval for the difference in means ([tex]\(\mu_F - \mu_N\)[/tex]), use:
[tex]\[
\text{Difference in means} = \bar{x}_F - \bar{x}_N
\][/tex]
The difference in means is:
[tex]\[
83.6 - 59.1 = 24.5
\][/tex]
Now, compute the confidence interval:
[tex]\[
\text{Lower bound} = 24.5 - 10.39 = 14.1
\][/tex]
[tex]\[
\text{Upper bound} = 24.5 + 10.39 = 34.9
\][/tex]
### Conclusion
The 99% confidence interval for the difference in the mean concentration of DEHP between those who have eaten fast food and those who haven't is [tex]\(14.1\)[/tex] to [tex]\(34.9\)[/tex] ng/mL. This interval suggests that fast food consumption is associated with a higher concentration of DEHP, within this range, with 99% confidence.
