A new vaccine was tested to see if it could prevent the ear infections that many

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 foilowing year, only 325 of 2453 vaccinated children had ear infections, compared to 500 of 2449 unvaccinated children. Complete parts a) through e) below. a) Are the conditions for inference satisfied? O A Yes. The data were greated by a randomized experiment, less tan 10% of the population was sampled, te goups wer" dependert, and there were morehan 10 successes rd trh res n each goup. OB. "was not a fandom, sample C. No More than 10% of the population was sampled. O D. No The groups were not independent b) Lat p, bete sample proportion of success in the unwa nated group, andlat P2 be the sample proportion of success it evaconated goup. Find the 95% condenoe interval rte diference i rates o ear infection, P1 P The confidence interval is ( % Do not round until the final answer. Then round to one decimal place as needed.) e) Use your oonfdence interval to explain whether you think the vaccine is effective OA. No No condlusion can be made based on the ooridence interval OB. No we are 95% confident hat the rate of infecton of vaccinated babies could be as much as 5.1% higher compared to unvaconated babies O c. Yes. We are 95% oorftent that the rate of inde tion is 0.1 to 9.3% lower. This is a meaningful reduction, consid the 20% infecton rate among umvaccinated babes O D Yes we are 95% corfident that about 9.3% of unvaccinated babies wil get an ear infection, while only 5.1% of vaccinated babies i. This is a mea ngu reduction.

Explanation / Answer

(A) opion a is correct

(B) p^2 = 325/2453 = 0.1325

p^1 = 500/2449 = 0.20420

95% confidence interval = (p^1 - p^2) +- Z95%  sqrt [p^1 *(1-p^1)/n1 + p^2 *(1-p^2)/n2]

= (0.2042 - 0.1325) +- 1.96 * sqrt [0.1325 * 0.8675/2453 + 0.2042 * 0.7958/2449]

= 0.0717 +- 0.02085

= (0.0508, 0.0925) = 5.1% to 9.8%

(c) Option C is correct here.

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