We want to know what percentage of the population have college degrees. We take a sample of 100 individuals and find that 28 have college degrees. Answer the following, rounding your answers to 2 decimal places. a. What is the sample proportion? b. Find the SE. c. Construct a 95% confidence interval. d. What does this 95% confidence interval tell us about our population proportion p? (One sentence is the perfect length for an 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

exactly you go it

step-by-step explanation: