Question

Problem 6: [7 points) A New York Times/CBS survey of 224 children of ages 9 to 17 was conducted shortly after the January 28, 1986, space shuttle disaster. Two-thirds of the children said that they would like to travel in space. (a) Give a point estimate of p, the true proportion of children who would like to space travel. [1] (b) Construct a 90% confidence interval for the proportion of children of ages 9 to 17 in 1986 who would like to experience space travel. [2] (c) Do you think most children in this age group in 1986 thought they would like to experience space travel? Why? [2] (d) For future studies, what sample size would be needed in order for the margin of error of the 90% confidence interval to be at most 0.01? [1] (e) Answer the same question in (d) if no prior study exists. [1]

Answer 1

a) The point estimate of the true population proportion of children who would like to space travel is equal to the sample proportion which is 2/3 here.

therefore p = 2/3 = 0.6667 is the required point estimate here.

b) From standard normal tables, we have here:
P(-1.645 < Z < 1.645) = 0.9

Therefore the confidence interval here is obtained as:

(2/3)0.0518

(0.6149.0.7185)

This is the required 90% confidence interval here.

c) As the whole confidence interval here is above 0.5, therefore we have sufficient evidence here at 90% confidence level, that most children in the age group in 1986 thought that they would like to experience space travel.

d) The margin of error here is computed as:

1 = 1.6452 * (2/31(1/3) 0.012 6014

Therefore 6014 is the required sample size here.

e) For no prior proportion value, we use p = 0.5 to get a conservative value of sample size which is computed here as:

n = 1.6452 * 0.5 * 0.5 0.012 6766

Therefore 6766 is the required sample size here.