Question

Examine the results of a study1 investigating whether fast food consumption increases one’s concentration of phthalates, an ingredient in plastics which has been linked to multiple health problems including hormone disruption. The study included 8877 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 ng/mL) in the urine: DEHP and DiNP. Find a 99 % confidence interval for the difference, μ F - μ N , in mean concentration between people who have eaten fast food in the last 24 hours and those who haven’t. The mean concentration of DiNP in the 3095 participants who had eaten fast food was x ¯ F = 10.1 with s F = 38.9 while the mean for the 5782 participants who had not eaten fast food was x ¯ N = 7.0 with s N = 22.8 .

Round your answers to two decimal places.

Answer 1

Sample Mean 1 (X1) = 10.1 Sample Standard Deviation 1 (91) = 38.9 Sample Size 1 (N1) = 3095 Sample Mean 2 (X2) = Sample Standard Deviation 2 (82) = 22.8 Sample Size 2 (N2) = 5782 Based on the informationa provided, we assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows: df = (sỉ/nı + să/n2)2 (si/nı) (sz/n2)2 + 2 = 4260.166 ni +1 n2+1 The critical value for a = 0.01 and df = 4260.166 degrees of freedom is to t1-a/2;n–1 = 2.577. The corresponding confidence interval is computed as shown below: Since we assume that the population variances are unequal, the standard error is computed as follows: si s3 se = ni n2 38.92 3095 22.82 5782 = 0.761 Now, we finally compute the confidence interval: CI (X - X) - te X se, X - X +t X se) (10.1 – 7 – 2.577 x 0.761, 10.1 – 7 + 2.577 x 0.761) = (1.139,5.061)

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