If n=300 and X=120, construct a 90% confidence interval estimate for the population proportion. Round to four decimal places as needed |
sample success x = | 120 | |
sample size n= | 300 | |
sample proportion p̂ =x/n=120/300= | 0.400 | |
std error se= √(p*(1-p)/n) =√(0.4*0.6/300) = | 0.0283 | |
for 90 % CI value of z= | 1.645 | |
margin of error E=z*std error = | 0.0465 | |
lower bound=p̂ -E = | 0.3535 | |
Upper bound=p̂ +E = | 0.4465 |
from above 90% confidence interval for population proportion =(0.3535 , 0.4465) |
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