Respuesta :
Answer:
confidence interval for the difference, , in mean concentration between people who have eaten fast food in the last 24 hours and those who haven’t.
is between 14.1 to 34 .9
Step-by-step explanation:
in order to find a confidence interval for the difference, , in mean concentration between people who have eaten fast food in the last 24 hours and those who haven’t. check the attached files for explanation .

Following are the solution to the given question:
Given:
[tex]\bar{x}_f=83.6 \\s_f=194.7\\ n_1=3092 \\\bar{x}_N=59.1\\ s_N=152.1 \\ n_2= 5782[/tex]
Sample sizes are large enough to use the normal distribution.
For 95% confidence level, 1.96 when the 95% confidence interval is:
[tex]u_f-u_n is[/tex]
[tex]\to(\bar{x}_f-\bar{x}_N)\pm z^n \sqrt{\frac{s_{f}^2}{n_1}+\frac{s_{N}^2}{n_2}}[/tex]
[tex]\to (83.6-59.1)\pm 1.96\sqrt{\frac{194.7^2}{3095}+\frac{152.1^2}{5782}} \\\\\to (24.5)\pm 1.96\sqrt{12.24+4.00} \\\\\to (24.5)\pm 1.96\sqrt{16.24} \\\\\to (24.5)\pm 1.96 \times 4.02 \\\\\to 24.5\pm 7.8792 \\\\\to32.3792 \ \ or \ \ 16.6208 \\\\[/tex]
Therefore, the final answer is "32.37 or 16.62".
Learn more:
brainly.com/question/8978782