Question

An opinion poll asks an SRS of 1500 adults, "Do you happen to jog?" Suppose that the population proportion who jog (a parameter) is p = 0.15. To estimate p, we use the proportion p-hat in the sample who answer "Yes." Justify the use of a normal approximation and find the following probabilities. a) P(p > 0.16) b) P(0.14 < p < 0.16)

Answer 1

a) P(p > 0.16) means find the probability that the sample proportion is greater than 0.16
b) P(0.14 < p < 0.16) means find the probability that the sample proportion lies between 0.14 and 0.16

Use z distribution for answering the question
z = (O - E) / SEp
O denotes observed or sample proportion or p-hat
E denotes expected or population proportion or p
p = 0.15 and n = 1500
SEp denotes standard error of the sample proportion = sqrt (p*(1-p)/n)
= sqrt (0.15*0.85/1500)
= 0.0092
a) z = (0.16 - 0.15) / 0.0092 = + 1.09
Required probability = P(p > 0.16) = P(z > 1.09)
The area under the standard normal curve right to the z value indicates the required probability. This area lies in the extreme right tail of the normal curve.
= 0.5000 (total area on the right side of p) - 0.3621(area between p and z)
= 0.1379

b) z1 = (0.14 - 0.15) / 0.0092 = - 1.09
z2 = (0.16 - 0.15) / 0.0092 = + 1.09
Required probability = P(0.14 < p < 0.16) = P(- 1.09 < z < 1.09)
The area under the standard normal curve between the two z values indicate the required probability.
= 0.3621 (area corresponding to z1 which lies on the left side of p) + 0.3621 (area corresponding to z2 which lies on the right side of p)
= 0.7242