Question

The Social Media and Personal Responsibility Survey found that 69% of parents are "friends" with their children on Facebook. A random sample of 140 parents was selected.

a. Calculate the standard error of the proportion.

b. What is the probability that 100 or more parents from this sample are "friends" with their children on Facebook?

c. What is the probability that between 96 and 105 parents from this sample are "friends" with their children on Facebook?

d. If 81 parents responded that they are "friends" with their children on Facebook, does this result support the findings of the Social Media and Personal Responsibility Survey?

Answer 1

(a) The standard error of the proportion (SE) = Sqrt[p * (1-p)/n] = Sqrt( 0.69 * 0.31/140) = 0.0391

b) Let the sample size be n.

If X denotes the number of parents out of n who are "friends" with their children on facebook then,

We need to find the probability that 100 or more parents from the sample of n are "friends" with their children on Facebook.

Then, , this value can be calculated using MsExcel function BINOMDIST()

(c) Z = ( - p)/SE

Therefore P(96 < X < 105) = P(p < 105/140) - P(p < 96/140)

For P(p < 105/140) where 105/140 = 0.75

Z = (0.75- 0.69)/0.0391 = 1.53. The p value = 0.9370

For P(p < 96/140) where 96/140 = 0.6857

Z = (0.6857 - 0.69)/0.0391 = -0.11. The p value = 0.4562

Therefore the required probability = 0.9370 - 0.4562 = 0.4808