Question

Imagine that proportion p = 0.82 of adults in a large city are married, but that this is not known, and we would like to estimate p. Suppose that we survey a random sample of 250 adults in the city, and calculate ˆp, the sample proportion of respondents who are married.

(a) Are the requirements met for ˆp to be approximately normally distributed? Explain, showing any necessary calculations and/or computer commands.

(b) What is (approximately) the probability that ˆp will fall between 0.81 and 0.83? Show all calculations and/or computer commands.

(c) If we had sampled only 50 adults, would the requirements for ˆp to be approximately normal still hold? Explain.

Answer 1

a)

np = 250 * 0.82 = 205 > = 10

n ( 1 - p) = 250 ( 1 - 0.82) = 45 > 10

Since np > 10 and n ( 1 - p) > 10

requirements met for to be approximately normally distributed.

b)

Using normal approximation,

P( < p ) = P(Z < ( - p) / sqrt [ p ( 1- p) / n ]

So,

P( 0.81 < < 0.83) = P( < 0.83) - P( < 0.81)

= P(Z < ( 0.83 - 0.82) / sqrt [ 0.82 ( 1 - 0.82) / 250] - P(Z < ( 0.81 - 0.82) / sqrt [ 0.82 ( 1 - 0.82) / 250]

= P(Z < 0.41) - P(Z < -0.41)

P(Z < 0.41) - ( 1 - P(Z < 0.41) )

= 0.6591 - ( 1 - 0.6591) (From Z table)

= 0.3182

c)

If n = 50 ,

np = 50 * 0.82 = 41 > = 10

n ( 1 - p) = 50 ( 1 - 0.82) = 9 < 10

Since np > 10 and n ( 1 - p) < 10

requirements do not met for to be approximately normally distributed.