Can somebody please help me !!!!!!!!!47$4&
2 answers:
You might be interested in
Answer:
Step-by-step explanation:
Hello!
The objective is to test if people with less social contact get more colds than people who are involved in a variety of social activities.
For this, a sample of 276 healthy men and women was taken.
n₁= 96 had few social outlets
n₂= 105 had six or more activities
These two groups were exposed to the cold virus and the following proportions of individuals that got sick were obtained:
p'₁= 0.62 (62% of the people with few social outlets got sick)
p'₂= 0.35 (35% of the people with six or more activities got sick)
The study variables are X₁: number of people with few social outlets that got sick after being exposed to the cold virus, in a sample of 96, and X₂: number of people with six or more activities that got sick after being exposed to the cold virus, in a sample of 105.
You have to construct a 99% CI for the difference these two populations since there is no specific order given I'll estimate: p₁ - p₂
(p'₁ - p'₂) ±
(0.62-0.35) ± 2.586
[0.094; 0.045]
With a 99% confidence level, you'd expect that the difference between the proportion of people with few social outlets that got sick and the proportion of people with six or more social activities that got sick after being exposed to the cold virus.
I hope it helps!