Question

A survey given to consumers leaving a supermarket three days before Thanksgiving asked whether turkey would be part of the Thanksgiving meal. Of the 182 people surveyed, 26 said "no." Give a 95% confidence interval for the population proportion who claimed not to be eating turkey.

Answer 1

(0.1392,0.1466)

Explanation:

Given that a survey given to consumers leaving a supermarket three days before Thanksgiving asked whether turkey would be part of the Thanksgiving meal

People surveyed = 182

claimed not eating turkey = 26

Sample proportion =
`(26)/(182) \\=0.1429`

Std error =
`\sqrt{(pq)/(n) } \\=0.0260`

Margin of error = 1.96*SE=
`0.00378`

Confidence interval =
`(0.1429-0.0037,0.1429+0.0037)\\=(0.1392,0.1466)`