Answer:
a) p = 28/100 p = 0,28 or p = 28%
b) SE = 0,0448
c) CI 95 % = ( 0.1921 ; 0.368 )
We can support with 95 % of confidence that the proportion of people with a College degree would be found within these limits
Step-by-step explanation:
a) Sample proportion: Is the number of positive success divide by sample size
That is from 100 ( individuals) ( sample size ) it was found that 28 had College degree ( positive success)
Then p = 28/100 p = 0,28 or p = 28%
b) The standard error is
SE = (√p*q)/n
p = 0,28 then q = 1 - 0,28 q = 0,72
SE = √ (0.28)*(0.72)/100
SE = √0.002016
SE = 0,0448
c) CI = 95 % then significance level is α = 5 % α = 0.05
α/2 = 0.025
From z- table we find z(c) = 1.96
Then CI 95 % = p₀ ± z(c) * SE
CI 95 % = 0.28 ± 1.96*0.0448
CI 95 % = ( 0.28 ± 0.088 )
CI 95 % = ( 0.1921 ; 0.368 )
d) We can support with 95 % of confidence that the proportion of people with college degree would be found within these limits
