Question

A poll of 793 adults aged 18 or older asked about purchases that they intended to make for the upcoming holiday season. One of the questions asked about what kind of gift they intended to buy for the person on whom they would spend the most. Clothing was the first choice of 482 people. Give a 90% confidence interval for the proportion of people in this population who intend to buy clothing as their first choice.

Answer 1

Answer:

(0.5793, 0.6363)

Step-by-step explanation:

Given that a  poll of 793 adults aged 18 or older asked about purchases that they intended to make for the upcoming holiday season. One of the questions asked about what kind of gift they intended to buy for the person on whom they would spend the most. Clothing was the first choice of 482 people.

i.e. sample proportion p= $\frac{482}{793} =0.608$

$q=1-p = 0.392$

Std error of p = $\sqrt{pq/n} =0.0173$

For 90% Z crtiical = 1.645

Hence margin of error = 1.645 * std error = 0.0285

Confidence interval 90%

= proportion ±margin of error

=$(0.5793, 0.6363)$