Question

Please show all your work so i can understand thank you!

Answer 1

The 99% confidence interval for the difference in mean DiNP concentration between people who ate fast food in the last 24 hours and those who didn't is 2.09 to 3.11 ng/mL. This means that there is a 99% chance that the true difference in mean DiNP concentration falls between these two values.

To find a 99% confidence interval for the difference in mean DiNP concentration between people who ate fast food in the last 24 hours and those who didn't, we can use the following formula:

`(\overline{x}_F - \overline{x}_N) \pm z_(\alpha/2) \sqrt{(s_F^2)/(n_F) + (s_N^2)/(n_N)}`

where:

`* $\overline{x}_F$ and $\overline{x}_N$` are the mean DiNP concentrations in the fast food and non-fast food groups, respectively

`* $s_F$ and $s_N$` are the standard deviations of DiNP concentrations in the fast food and non-fast food groups, respectively

`* $n_F$ and $n_N$` are the sample sizes of the fast food and non-fast food groups, respectively

`* $z_(\alpha/2)$` is the critical value of the standard normal distribution for a 99% confidence interval, which is 2.576

Plugging in the values that were given in the question, we get the following confidence interval:

`(10.1 - 7.0) \pm 2.576 \sqrt{(38.9^2)/(3095) + (22.8^2)/(5782)}= 3.11 \pm 0.56`

**Therefore, the 99% confidence interval for the difference in mean DiNP concentration between people who ate fast food in the last 24 hours and those who didn't is **2.09** to **3.11** ng/mL.

** This means that there is a 99% chance that the true difference in mean DiNP concentration falls between these two values.