Question

written on paper would be best

Question 3 (Essay Worth 10 points) (05.04 MC) Two opposing opinions were shown to a random sample of 1,744 US buyers of a particular political news app. The opinions, shown in a random order to each buyer, were as follows. Opinion A: Prescription drug regulation is more important than border security Opinion B: Border security is more important than prescription drug regulation Buyers were to choose the opinion that most closely reflected their own. If they felt neutral on the topics, they were to choose a third option of "Neutral." The outcomes were as follows: 30% chose Opinion A, 64% chose Opinion B, and 6% chose "Neutral." Part A Create and interpeet a 95% confidence interval for the proportion of all US buyers who would have chosen Opinion A (5 points) Part B. The number of buyers that chose Opinion A and the number of buyers that did not choose Opinion A are both greater than 10. Why must this inference condition be met? (5 point

Answer 1

(a) p = 0.30

n = 1744

z = 1.96

The 95% confidence interval will be:

= p ± z*(√p(1-p)/n)

= 0.30 ± 1.96*(√0.30(1-0.30)/1744)

= (0.2785, 0.3215)

We are 95% confident that the true proportion of all US buyers who would have chosen Opinion A is between 0.2785 and 0.3215.

(b) This is because we assume the number of failures and successes for a sample should be greater than 10 in order to meet the normal approximation condition.