In January, 2014, a research center found that 89% of 18-29 year olds in the U.S. used social media sites. That year I surveyed 500 Portlanders between the ages of 18-29, and found that 95% of them used social media sites. If Portland mirrors the US percentage, what is the probability that my sample of 500 would contain 95% successes?
for normal distribution z score =(p̂-p)/σ_{p} | |
here population proportion= p= | 0.8900 |
sample size =n= | 500 |
std error of proportion=σ_{p}=√(p*(1-p)/n)= | 0.0140 |
probability that my sample of 500 would contain 95% successes(greater than or equal to 95%):
probability =P(X>0.95)=P(Z>(0.95-0.89)/0.014)=P(Z>4.29)=1-P(Z<4.29)=1-1.0000=~ 0.0000 |
probability of above event is almost negligible.
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