Business Statistics ? invloving confidence interval...PLEASE HELP?

In a survey of 894 respondents with salaries below 100,000 per year, 367 indicated that the primary reason for staying on their job was interesting job responsibilities.





A) Construct a 95% confidence interval for the proportion of all workers whose primary reason for staying on their job was interesting job responsibilities.





B) Interpret the interval constructed in part A.

Business Statistics ? invloving confidence interval...PLEASE HELP?
Let ^p = 367/894 and n = 100000 in this case.





A).


Since n^p = 100000*(367/894) = 41051.45414 and n(1-^p) = 100000*[1-(367/894)] = 58948.54586 are both greater than 5, then n= 100000 is considered to be large and hence we can use the following confidence interval formula in this case.





(1-∞)% = 95%


1-∞ = 0.95


-∞ = 0.95-1


-∞ = -0.05


∞ = 0.05


∞/2 = 0.05/2 = 0.025





A 95% confidence interval for the proportion of all workers whose primary reason for staying on their job was interesting job reponsibilities is


= {^p(+-) z∞/2√[^p(1-^p)]/(n-1)}


={(367/894) (+-) z[(0.05)/2] * √[(367/894)*(1-(367/894))]/(100000-1)}


= {(367/894) (+-) z0.025*√[(367/894)*(1-(367/894)]/99999}


= {(367/894) (+-)1.96*√[(367/894)*(1-(367/894))/99999...


= [0.410514541(+-) 0.00304901151]


= [0.407465529, 0.413563552] OR





= [0.407465529*100%, 0.413563552*100%]


= [40.7%, 41.4%]





B). The above interval indicates that we are 95% confident that truly between around 40.7% and 41.4% of all workers whose primary reason for staying on their job was interesting job responsibilities.