Question

A survey of n=90 families showed that x=40 owned at least one television set. Find the 95% confidence interval of the true proportion of families own at least one television set.

Answer 1

u have to find a confidence interval for p(probability of success)

Answer 2

?

Answer 3

ur observed p = (4/9), but that is not the true p.u have to find a CI for the true p

Answer 4

define p1 at the value of p that satisfies the following: \[\sum_{j=y}^{n}\left(\begin{matrix}n \\ j\end{matrix}\right)\] p^j (1-p)^n-j = 0.25

Answer 5

check out question 14

Answer 6

they defined y as the number of success, so where they have 6 you will have 40 and where they have 75 you will have 90

Answer 7

sorry, i meant question 11.4

Answer 8

That helps but im still not sure if im doing it right

Answer 9

ok.since n>30, i suggest you use method 11.4 a)...the confidence interval will be p'\[\pm\] 1.96sqrt(p'(1-p')/90))...where p'=(4/9)

Answer 10

method 11.4b) sorry.not a)

Answer 11

the idea is to use the normal approxiamtion to the binomial distribution..if you still confused, here is the answer: [0.33 ; 0.56]