Question

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 interpret 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 points)

Answer 1

Answer:

95% confidence interval for the proportion of all US buyers who would have chosen Opinion A is [0.28 , 0.32].

Step-by-step explanation:

We are given that Two opposing opinions were shown to a random sample of 1,744 US buyers of a particular political news app.

The outcomes were as follows: 30% chose Opinion A, 64% chose Opinion B, and 6% chose "Neutral."

Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;

                         P.Q. =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of US buyers who chose Opinion A = 0.30

           n = sample of US buyers = 1,744

           p = population proportion

Here for constructing 95% confidence interval we have used One-sample z test for proportions.

(a) SO, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                               of significance are -1.96 & 1.96}  

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [$\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

  = [ $0.30-1.96 \times {\sqrt{\frac{0.30(1-0.30)}{1,744} } }$ , $0.30+1.96 \times {\sqrt{\frac{0.30(1-0.30)}{1,744} } }$ ]

  = [0.28 , 0.32]

Therefore, 95% confidence interval for the proportion of all US buyers who would have chosen Opinion A is [0.28 , 0.32].

(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, that is;

Number of buyers that did chose Opinion A =  $n \times p$

                                                                          =  $1,744 \times 0.30$ = 523.2

Number of buyers that did not chose Opinion A =  $n \times (1-p)$

                                                                                =  $1,744 \times 0.70$ = 1220.8    

Now, this inference condition should have been met because if these were not met then we were not able to use the normal distribution statistics above because binomial distribution can be converted to normal distribution only when  $np$  and  $n(1-p)$ both are greater than 10.