Question

Given a population in which p = 0.6 and a random sample from this population of size 100, find

a. the mean of the sampling distribution (for this proportion)

b. the standard deviation of the sampling distribution (for this proportion)

c. ?(?̂≤ 0.58) =

d. ?(?̂≥ 0.65) =

e. ?(0.58 ≤ ?̂ ≤ 0.65) =

Answer 1

Solution:

The population proportion is P = 0.6

Sample size (n) = 100

If nP and nQ both are greater than or equal to 10 then sampling distribution of sample proportion follows approximately normal distribution with mean P and variance PQ/n.

i.e.  p̂ ~ N(P, PQ/n)

(Where, p̂ is sample proportion, P is population proportion, Q = 1 - P and n is sample size.)

We have, P = 0.6, Q = 1 - 0.6 = 0.4 and n = 100

nP = 100×0.6 = 60 which is greater than 10.

nQ = 100×0.4 = 40 which is greater than 10.

Hence, sampling distribution of sample proportion follows normal distribution with mean 0.6 and variance (0.6×0.4)/100 = 0.0024.

a) The mean of the sampling distribution of sample proportion is 0.6.

b) The standard deviation of the sampling distribution of sample proportion is given by,

The standard deviation of sampling distribution of sample proportion is 0.04899.

c) We have to obtain Pr(p̂ ≤ 0.58).

We have, P = 0.6, Q = 1 - 0.6 = 0.4 and n = 100

We know that if p̂ ~ N(P, PQ/n) then

Using "pnorm" function of R we get, P(Z < -0.4082) = 0.3416

d) We have to obtain Pr(p̂ ≥ 0.65).

We have, P = 0.6, Q = 1 - 0.6 = 0.4 and n = 100

We know that if p̂ ~ N(P, PQ/n) then

Using "pnorm" function of R we get, P(Z ≥ 1.0206) = 0.1537