stats module 6 homework question 16 urgent help please

Question

A survey of randomly selected university students found that 95 of the 109 first-year students and 100 of the 137 second-year students surveyed had purchased used textbooks in the past year.

(a) Construct a 98% confidence interval for the difference in the proportions of first-year and second-year university students who purchased used textbooks. (Round your answers to 4 decimal places, if needed.)

Answer: ( ,  )

(b) If a 99% confidence interval for the difference in proportion of first-year and second-year university students who purchased used textbooks is (0.0137, 0.2696). Give an interpretation of this confidence interval.

 We know that 99% of the first-year students bought between 1.37% and 26.96% higher than the proportion of second-year students who bought used textbooks.

 We are 99% confident that, at this university, the proportion of first-year students who bought used textbooks is between 1.37% and 26.96% higher than the proportion of second-year students who bought used textbooks.

 We are 99% confident that, at this university, the proportion of second-year students who bought used textbooks is between 1.37% and 26.96% higher than the proportion of first-year students who bought used textbooks.

 We are 99% confident that the first-year students at this university bought between 1.37% and 26.96% more used textbooks than second-year students.

 We know that 99% of all random samples done on the population will show that the proportion of first-year students who bought used textbooks is between 1.37% and 26.96% higher than the proportion of second-year students who bought used textbooks.

Answer 1

Hi Faith,

a. Two ways to do this: First way is by formula:

CI = (p^1 - p^2) +/- z* SE, SE= sqrt(p^1Q^1/n1 + p^2q^2/n2)

p^1 = Freshman Proportion who bought used = 95/109 = 0.8716

q^1= 1-p^1 = 0.1284

p^2 = Sophomore Proportion who bought used = 100/137 = 0.7299

q^2 = 1 - p^2 = 0.2701

SE= sqrt((0.8716*0.1284)/109) + (0.7299*0.2701)/137)) = 0.0497

z* = 2.326 (from z-table)

CI = (0.8716 - 0.7299) +/- [2.326 (0.0497)]

CI = (0.0261, 0.2573)

Second way is by TI-83 or 84 Plus calculator. Go to STAT-TESTS-2PropZInt--Enter 95 for x1, 100 for x2, 109 for n1, and 137 for n2. Interval will be slightly different due to rounding (0.0261, 0.2572).

b. I can see where this gave you trouble; all answers sound similar. Remember that most confidence interval interpretations begin with "We are x% confident that" and remember that we are dealing with difference between two population proportions here. Correct answer, given these, is:

We are 99% confident that, at this university, the proportion of first-year students who bought used textbooks is between 1.37% and 26.96% higher than the proportion of second-year students who bought used textbooks.

I hope this helps.