Question

A new vaccine was tested to see if it could prevent the ear infections that many infants suffer from. Babies about a year old were randomly divided into two groups. One group received​ vaccinations, and the other did not. The following​ year, only 329 of 2456 vaccinated children had ear​ infections, compared to 496 of 2448 unvaccinated children. Complete parts​ a) through​ c) below.

Let p1 be the sample proportion of success in the unvaccinated​ group, and let p2 be the sample proportion of success in the vaccinated group. Find the​ 95% confidence interval for the difference in rates of ear​ infection, p1−p2.

The confidence interval is (___%,___%).

there is a 95% confidence, based on these sample, that the proportion of American woman, age 65and older who suffer from arthritis is between ____% and ___% more or less than the proportion of American men of the same age who suffer

Answer 1

Sample proportion Sample Sample number Size of events Sample 1 n1 = 2448 X1 =496 P1 = X1/n1 =(496/2448) = 0.2026 Sample 2 n2 = 2456 X2 = 329 P2 = xz/n2 =(329/2456) = 0.134 P2(1-P2) Standard Error = P1(1-) ni n2 = sqrt((0.2026*(1-0.2020)/2448)+(0.134*(1-0.134)/2456)] = 0.0106 Confidence Level = 95 Significance Level = Q = (100-95)% = 0.05 Critical value = z* = 1.96 [Using Excel =ABS(NORMINV(0.05/2,0,1))] Margin of Error = z* (standard error )= 1.96*0.0106 = 0.02086 Lower Limit = (P1-P2) - Margin of Error = (0.2026 -0.134) -0.02086 = 0.04774 Upper Limit =(P1-P2) + Margin of Error = (0.2026 -0.134) + 0.02086 = 0.08946

Ans :

The confidence interval is (4.77 %, 8.95 %).

there is a 95% confidence, based on these sample, that the proportion of American woman, age 65and older who suffer from arthritis is between 4.77% and 8.95% more or less than the proportion of American men of the same age who suffer