(20.47 S-AQ) A sample survey interviews an SRS of 252 college women. Suppose (as is roughly true) that 71% of all college women have been on a diet within the past 12 months.

Question

Using Normal approximation, what is the probability (±0.001) that 77% or more of the women in the sample have been on a diet? .

Answer 1

n = 252
phat = 0.77
p = 0.71
q = 1-p = 0.29
standard deviation = sqrt((pq/n))

= sqrt(0.7)(0.29)/252)

= 0.02858

z score = (phat -p)sd/sqrt(n)

z = (0.77-0.71)/0.02858

z = 2.10

P(phat > 0.77) = p(z < 2.10)

1- 0.9821 = 0.0179