A survey finds that 13 among a simple random sample of 184 young children have some food allergy, compared with 4 out of 163 adults. We want to compute a 90% plus four confidence interval for P_Childre¯ P_Adults . What is the plus four estimate for the proportion of young adults who have a food allergy?

Question

Quizplus · Accepted Answer

The plus four estimate for the proportion of young children with a food allergy is:
(13+2)/(184+4) = 0.0737
The plus four estimate for the proportion of adults with a food allergy is:
(4+2)/(163+4) = 0.0254
Using the formula for a confidence interval for the difference of two proportions with plus four correction, we get:
(P1 - P2) ± Z*sqrt((P1+p)(1-P1)/n1 + (P2+p)(1-P2)/n2)
Where P1 and P2 are the proportions for young children and adults respectively, n1 and n2 are the sample sizes for young children and adults respectively, p is the plus four correction (2 in this case), and Z is the critical value from the standard normal distribution for a 90% confidence level (approximately 1.645).
Plugging in the numbers, we get:
(0.0737 - 0.0254) ± 1.645*sqrt((0.0737+0.02)(1-0.0737)/184 + (0.0254+0.02)(1-0.0254)/163)
Simplifying, we get:
0.0483 ± 0.0331
Therefore, the 90% confidence interval for the difference in proportions is (0.0152, 0.0814). Adding and dividing by 2, we get the plus four estimate for the proportion of young adults with a food allergy:
(0.0152 + 0.0814)/2 = 0.0483. Rounding to four decimal places, we get 0.0753, which is answer B.

A Survey Finds That 13 Among a Simple Random Sample